Mastering Biology Chp. 13 HW

PART A – Analyzing the data

Tables like the one shown here are useful for organizing sets of data representing a common set of values (in this case, percentages of A, G, C, and T) for a number of different samples (in this case, species).

Does the distribution of bases in sea urchin DNA and salmon DNA follow Chargaff’s rules?

Yes, because the %A approximately equals the %T and the %G approximately equals the %C in both species.

PART B – Calculating missing data

You can use Chargaff’s rules to predict the percentage of one or more bases in the DNA of a species if at least one value is known.

What is the %T in wheat DNA?

Approximately 28%

PART C

Use Chargaff’s rules to predict the missing values for E. coli, human, and ox DNA. Round your answers to the nearest whole number.

E.coli – Cytosine: 25.3 – Thymine: 24.0 Human – Guanine: 19.8 – Cytosine: 19.8 Ox – Guanine: 21.0 – Cytosine: 21.0 – Thymine: 29.0

PART D – Evaluating a hypothesis

If Chargaff’s equivalence rule is valid, then hypothetically we could extrapolate this to the combined genomes of all species on Earth (as if there were one huge Earth genome). In other words, the total amount of A in every genome on Earth should equal the total amount of T in every genome on Earth. Likewise, the total amount of G in every genome on Earth should equal the total amount of C in every genome on Earth.

Calculate the average percentage for each base in your completed table. Do Chargaff’s equivalence rules still hold true when you consider those six species together?

Yes, the average for A approximately equals the average for T, and the average for G approximately equals the average for C. [The averages from the table support Chargaff’s A = T and G = C equivalence relationships.]

Complete the following vocabulary exercise related to DNA replication.

Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column.

1. OKAZAKI FRAGMENTS are the short sections of DNA that are synthesized on the lagging strand of the replicating DNA. 2. After replication is complete, the new DNAs, called DAUGHTER DNA, are identical to each other. 3. During DNA replication, an open section of DNA, in which a DNA polymerase can replicate DNA, is called a REPLICATION FORK. 4. The enzyme that can replicate DNA is called DNA POLYMERASE. 5. The new DNA strand that grows continuously in the 5′ to 3′ direction is called the LEADING STRAND. [DNA replication is a central process in biochemistry and must be done with the highest precision. If a mistake is made in the replication of DNA, the error will propagate into future generations. An error that occurs during replication is called a mutation.]

PART A – The experimental technique: Density-gradient centrifugation

When a solution of cesium chloride (CsCl) is subjected to high-speed centrifugation, a stable density gradient is formed. Meselson and Stahl found that when cell contents were subjected to centrifugation with a CsCl solution, a band of DNA formed at the CsCl density that matched the density of the DNA. This technique is called density-gradient centrifugation.
The test tubes below show the results of density-gradient centrifugation of five different DNA samples.

Drag the description of each DNA sample to the appropriate location to identify the expected appearance of the DNA band(s) after density-gradient centrifugation.

-Test tube 1: b. DNA from E. coli cells grown in 14N -Test tube 2: e. DNA containing one strand of 15N-DNA and one strand of 14N-DNA -Test tube 3: a. DNA from E. coli cells grown in 15N -Test tube 4: c. A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N -Test tube 5: d. A 1:1 mixture of DNA from cells grown in 14N and 15N, heated (to disrupt hydrogen bonds) and cooled (to allow reannealing). [The densities of 14N/14N, 14N/15N, and 15N/15N double helices differ from each other and thus form bands in different positions. 14N/14N forms a band toward the top, 14N/15N in the middle, and 15N/15N toward the bottom. DNA from cells grown in 15N contains only 15N/15N double helices. DNA from cells grown in 14N contains only 14N/14N double helices. A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N contains both 14N/14N and 15N/15N double helices. A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N, heated and then cooled, contains 14N/14N, 14N/15N, and 15N/15N double helices. DNA containing one strand of 15N-DNA and one strand of 14N-DNA contains only 14N/15N double helices.]

PART B – Experimental results: Banding pattern predictions for each model of replication

Meselson and Stahl designed an experiment that would allow them to discern whether DNA replication occurs in a dispersive, semiconservative, or conservative manner.
-They started with E. coli that had been growing for many generations in medium containing 15N.
-They then transferred the bacteria into medium containing only 14N, and allowed the bacteria to undergo two rounds of DNA replication.
-After each round of replication, the scientists performed density-gradient centrifugation of the DNA.
The scientists reasoned that each of the three models would predict different DNA banding patterns after the two rounds of replication.

Can you identify the banding patterns predicted by each model after the first round of replication? (Then, in Part C, you will identify the banding patterns predicted after the second round of replication.)

Drag the test tubes to the appropriate locations in the table to show the banding patterns that each model predicts. Test tubes may be used once, more than once, or not at all.

[Notice that after one round of replication, the dispersive and semiconservative models predict identical results, whereas the conservative model predicts different results. By continuing the experiment through a second round of replication, would all three models lead to different predictions?]

PART C – Experimental results: Banding pattern predictions after the second round of replication

Can you identify the banding patterns predicted by each model after the second round of replication? (Note: The dispersive model’s prediction is already shown; see Hint 1 if you need help understanding it.)

Drag the test tubes to the appropriate locations in the table to show the banding patterns that each model predicts. Test tubes may be used once or not at all.

[When Meselson and Stahl performed this experiment, their results were consistent with the pattern predicted by semiconservative replication, confirming that as the correct model. When DNA replicates, the two parent strands separate, and each strand serves as a template for the synthesis of a new DNA strand. The figure shows semiconservative replication of DNA double helix. There is DNA double helix labeled as superscript 15 N slash superscript 15 N at the beginning of the experiment. There are two DNA double helices labeled each as superscript 14 N slash superscript 15 N after the first replication. Four DNA double helices are formed after the second replication. Two of them are labeled as superscript 14 N slash superscript 15 N. Other two DNA are labeled as superscript 14 N slash superscript 14 N. Note that Meselson and Stahl were able to rule out the conservative model based on the results after one round of replication. However, they could not rule out the dispersive model at that point because that model predicts the same pattern as the semiconservative model. But, the results after two rounds did enable them to rule out the dispersive model.]

PART D – Experimental prediction: Predicting the results after additional rounds of replication

To confirm the semiconservative model of replication, it was important for Meselson and Stahl to quantify the amount of DNA in each band produced by density-gradient centrifugation. To accomplish this, they took advantage of the fact that DNA absorbs ultraviolet light, and used UV light to photograph each tube.

By scanning the UV photographs with a microdensitometer, graphs like the ones below were produced. The height of each peak in the graph is directly proportional to the concentration of DNA in the corresponding band. Also, the position of each peak reflects the 14N and 15N content of the band.

The figure shows two graphs and two tubes with a sample of DNA. The first graph and the first tube correspond to the beginning of the experiment. The second graph and the second tube correspond to the first replication. The first microdensitometer graph shows one peak, which corresponds to the position of the superscript 15 capital letter N slash superscript 15 capital letter N DNA band. There are three marks at the first tube. A sample of DNA is located at the third (from top to bottom) mark labeled as superscript 15 capital letter N slash superscript 15 capital letter N. After one round of replication in superscript capital letter N medium, the microdensitometer graph shows one peak, but at a position of lower density (representing superscript 14 capital letter N slash superscript 15 capital letter N-DNA). There are three marks at the second tube. A sample of DNA is located at the second (from top to bottom) mark labeled as superscript 14 capital letter N slash superscript 15 capital letter N.

Suppose that the scientists analyzed the same amount of DNA (10 units) by density-gradient centrifugation after two, three, and four rounds of replication in 14N medium. What would you predict the microdensitometer graph would look like after each round?
Drag each microdensitometer graph to the appropriate bin.

[After transfer to 14N medium, new DNA strands are made from 14N only. So, with each round of replication, the amount of 14N/14N-DNA increases relative to the amount of 14N/15N-DNA. Thus, the height of the 14N/14N peak (at position 3, closer to the top of the tube) increases with each round of replication, while the 14N/15N peak (at position 5) decreases.]

PART A – The chemical structure of DNA and its nucleotides

The DNA double helix is composed of two strands of DNA; each strand is a polymer of DNA nucleotides. Each nucleotide consists of a sugar, a phosphate group, and one of four nitrogenous bases. The structure and orientation of the two strands are important to understanding DNA replication.

Drag the labels to their appropriate locations on the diagram below. Pink labels can be used more than once.

[The DNA double helix is constructed from two strands of DNA, each with a sugar-phosphate backbone and nitrogenous bases that form hydrogen bonds, holding the two strands together. Each DNA strand has two unique ends. The 3′ end has a hydroxyl (-OH) group on the deoxyribose sugar, whereas the 5′ end has a phosphate group. In the double helix, the two strands are antiparallel, that is, they run in opposite directions such that the 3′ end of one strand is adjacent to the 5′ end of the other strand.]

PART B – The role of DNA polymerase III

In DNA replication in bacteria, the enzyme DNA polymerase III (abbreviated DNA pol III) adds nucleotides to a template strand of DNA. But DNA pol III cannot start a new strand from scratch. Instead, a primer must pair with the template strand, and DNA pol III then adds nucleotides to the primer, complementary to the template strand. Each of the four images below shows a strand of template DNA (dark blue) with an RNA primer (red) to which DNA pol III will add nucleotides.

In which image will adenine (A) be the next nucleotide to be added to the primer?

[In the example above, DNA pol III would add an adenine nucleotide to the 3′ end of the primer, where the template strand has thymine as the next available base. You can tell which end is the 3′ end by the presence of a hydroxyl (-OH) group. The structure of DNA polymerase III is such that it can only add new nucleotides to the 3′ end of a primer or growing DNA strand (as shown here). This is because the phosphate group at the 5′ end of the new strand and the 3′ -OH group on the nucleoside triphosphate will not both fit in the active site of the polymerase.]

PART C – The replication bubble and antiparallel elongation

DNA replication always begins at an origin of replication. In bacteria, there is a single origin of replication on the circular chromosome, as shown in the image here. Beginning at the origin of replication, the two parental strands (dark blue) separate, forming a replication bubble. At each end of the replication bubble is a replication fork where the parental strands are unwound and new daughter strands (light blue) are synthesized. Movement of the replication forks away from the origin expands the replication bubble until two identical chromosomes are ultimately produced.

Diagram showing DNA replication in a circular chromosome. For simplicity, the double-stranded DNA is shown as two concentric circles. There is one origin of replication, where the two parental strands, shown in dark blue, separate, forming a replication bubble. At each end of the replication bubble is a replication fork, indicated by a pink arrow, where the parental strands are unwound and new daughter strands, shown in light blue, are synthesized. The replication forks move away from the origin and expand the replication bubble. As the light blue strands elongate, the two double-stranded circles that are forming start to peel off from each other. The end result is two separate, identical daughter DNA molecules, each composed of one parental strand, dark blue, and one new strand, light blue.

In this activity, you will demonstrate your understanding of antiparallel elongation at the replication forks. Keep in mind that the two strands in a double helix are oriented in opposite directions, that is, they are antiparallel.

Drag the arrows onto the diagram below to indicate the direction that DNA polymerase III moves along the parental (template) DNA strands at each of the two replication forks. Arrows can be used once, more than once, or not at all.

[DNA polymerase III can only add nucleotides to the 3′ end of a new DNA strand. Because the two parental DNA strands of a double helix are antiparallel (go from 3′ to 5′ in opposite directions), the direction that DNA pol III moves on each strand emerging from a single replication fork must also be opposite. For example, in the replication fork on the left, the new strand on top is being synthesized from 5′ to 3′, and therefore DNA pol III moves away from the replication fork. Similarly, the new strand on the bottom of that same replication fork is being synthesized from 5′ to 3′. But because the bottom parental strand is running in the opposite direction of the top parental strand, DNA pol III moves toward the replication fork. In summary, at a single replication fork, one strand is synthesized away from the replication fork, and one strand is synthesized toward the replication fork. When you look at both replication forks, note that a single new strand is built in the same direction on both sides of the replication bubble.]

PART D – Unwinding the DNA

As DNA replication continues and the replication bubble expands, the parental double helix is unwound and separated into its two component strands. This unwinding and separating of the DNA requires three different types of proteins: helicase, topoisomerase, and single-strand binding proteins.

Sort the phrases into the appropriate bins depending on which protein they describe.

Helicase -binds AT the replication fork -breaks H-bonds between bases Topoisomerase -binds AHEAD the replication fork -breaks covalent bonds in DNA Single-strand binding protein -binds AFTER the replication fork -prevents H-bonds between bases [At each replication fork, helicase moves along the parental DNA, separating the two strands by breaking the hydrogen bonds between the base pairs. (This makes the two parental DNA strands available to the DNA polymerases for replication.) As soon as the base pairs separate at the replication fork, single-strand binding proteins attach to the separated strands and prevent the parental strands from rejoining. As helicase separates the two parental strands, the parental DNA ahead of the replication fork becomes more tightly coiled. To relieve strain ahead of the replication fork, topoisomerase breaks a covalent bond in the sugar-phosphate backbone of one of the two parental strands. Breaking this bond allows the DNA to swivel around the corresponding bond in the other strand and relieves the strain caused by the unwinding of the DNA at the helicase.]

PART A – Comparing the leading and lagging strands

As the two parental (template) DNA strands separate at a replication fork, each of the strands is separately copied by a DNA polymerase III (orange), producing two new daughter strands (light blue), each complementary to its respective parental strand. Because the two parental strands are antiparallel, the two new strands (the leading and lagging strands) cannot be synthesized in the same way.

Diagram showing a replication bubble with replication forks on the left and the right of the origin of replication. The top parental DNA strand goes 5 prime to 3 prime and the bottom parental DNA strand goes 3 prime to 5 prime. At each replication fork, two DNA polymerases 3 are synthesizing two new daughter strands. At the left replication fork, the top strand is the leading strand and the bottom strand is the lagging strand. At the right replication fork, the top strand is the lagging strand and the bottom strand is the leading strand. Replication proceeds to the left and to the right.

Drag each phrase to the appropriate bin depending on whether it describes the synthesis of the leading strand, the synthesis of the lagging strand, or the synthesis of both strands.

Leading strand -made continuously -daughter strand elongates toward replication fork -only one primer needed Lagging strand – multiple primers needed -daughter strand elongates away from replication fork -made in segments both strands -synthesized 5′ to 3′ [Because DNA polymerase III can only add nucleotides to the 3′ end of a new DNA strand and because the two parental DNA strands are antiparallel, synthesis of the leading strand differs from synthesis of the lagging strand. -The leading strand is made continuously from a single RNA primer located at the origin of replication. DNA pol III adds nucleotides to the 3′ end of the leading strand so that it elongates toward the replication fork. -In contrast, the lagging strand is made in segments, each with its own RNA primer. DNA pol III adds nucleotides to the 3′ end of the lagging strand so that it elongates away from the replication fork. In the image below, you can see that on one side of the origin of replication, a new strand is synthesized as the leading strand, and on the other side of the origin of replication, that same new strand is synthesized as the lagging strand. The leading and lagging strands built on the same template strand will eventually be joined, forming a continuous daughter strand.]

PART B – RNA primers on the leading and lagging strands

The diagram below shows a replication bubble with synthesis of the leading and lagging strands on both sides of the bubble. The parental DNA is shown in dark blue, the newly synthesized DNA is light blue, and the RNA primers associated with each strand are red. The origin of replication is indicated by the black dots on the parental strands.

Diagram showing a replication bubble. The top parental DNA strand goes 5 prime to 3 prime and the bottom parental DNA strand goes 3 prime to 5 prime. Four segments of new DNA are being synthesized inside the bubble – two as leading strands and two as lagging strands. One leading strand is in the top left, and its single primer is labeled (a) near the origin of replication. The other leading strand is in the bottom right, and its single primer is labeled (h) near the origin of replication. One lagging strand is in the top right, and its three primers are labeled left to right (b), (c), and (d). The other lagging strand is in the bottom left, and its three primers are labeled left to right (e), (f), and (g).

Rank the primers in the order they were produced. If two primers were produced at the same time, overlap them.

a, b, c, d h, g, f, e (on top of each other just like this) [As soon as the replication bubble opens and the replication machinery is assembled at the two replication forks, the two primers for the leading strands (primers a and h) are produced. The production of the first primers on the lagging strands (those closest to the origin of replication, b and g) is delayed slightly because the replication forks must open up further to expose the template DNA for the lagging strands. After completion of the first segments of the lagging strands, additional template DNA must be exposed before the second primers (c and f) can be produced. And after completion of the second segments, additional template DNA must be exposed before the third primers (d and e) can be produced. In summary, because of the way the replication bubble expands, the lagging strand primers near the origin of replication were produced before the primers near the replication forks.]

PART C – Synthesis of the lagging strand

In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B.

Diagram showing a lagging strand being synthesized on a dark blue parental strand running 5 prime to 3 prime. The replication fork is off to the right. Fragment A (light blue) is on the far left and is the most recently synthesized Okazaki fragment. Primer A (red) sets just to the right of fragment A. Then there is a space for fragment B. Also, there is primer B (red) at the right end of the diagram, before the replication fork.

Drag the labels to their appropriate locations in the flowchart below, indicating the sequence of events in the production of fragment B. (Note that pol I stands for DNA polymerase I, and pol III stands for DNA polymerase III.)

1. pol III binds to 3′ end of primer B 2. pol III moves 5′ to 3′, adding DNA nucleotides to primer B 3. pol I binds to 5′ end of primer A 4. pol I replaces primer A with DNA 5. DNA ligase links fragments A and B [Synthesis of the lagging strand is accomplished through the repetition of the following steps. Step 1: A new fragment begins with DNA polymerase III binding to the 3′ end of the most recently produced RNA primer, primer B in this case, which is closest to the replication fork. DNA pol III then adds DNA nucleotides in the 5′ to 3′ direction until it encounters the previous RNA primer, primer A. Step 2: DNA pol III falls off and is replaced by DNA pol I. Starting at the 5′ end of primer A, DNA pol I removes each RNA nucleotide and replaces it with the corresponding DNA nucleotide. (DNA pol I adds the nucleotides to the 3′ end of fragment B.) When it encounters the 5′ end of fragment A, DNA pol I falls off, leaving a gap in the sugar-phosphate backbone between fragments A and B. Step 3: DNA ligase closes the gap between fragments A and B. These steps will be repeated as the replication fork opens up. Try to visualize primer C being produced to the right (closest to the replication fork). Fragment C would be synthesized and joined to fragment B following the steps described here.]

PART A – The mechanism of DNA replication

The diagram below shows a double-stranded DNA molecule (parental DNA).

Drag the correct labels to the appropriate locations in the diagram to show the composition of the daughter DNA molecules after one and two cycles of DNA replication. In the labels, the original parental DNA is blue and the DNA synthesized during replication is red.

[During DNA replication, each strand in the parental DNA serves as the template for the production of a daughter strand by complementary base pairing. Therefore, one cycle of replication will produce two daughter DNA molecules, each with one parental strand and one newly synthesized strand. During a second cycle of replication, all four strands in the two DNA molecules will serve as templates, resulting in four molecules (eight strands of DNA).]

PART B – Processes occurring at a bacterial replication fork

The diagram below shows a bacterial replication fork and its principal proteins.

Drag the labels to their appropriate locations in the diagram to describe the name or function of each structure. Use pink labels for the pink targets and blue labels for the blue targets

[During replication, DNA synthesis occurs in the 5′ to 3′ direction along both template strands. -On one template strand, synthesis proceeds continuously toward the replication fork, generating the leading strand. -On the other template strand, DNA is synthesized away from the replication fork in segments called Okazaki fragments, generating the lagging strand. Several proteins are involved in DNA replication, including the following: -Topoisomerase cuts, swivels, and rejoins DNA strands ahead of the replication fork. -Helicase breaks the hydrogen bonds between the parental DNA strands and unwinds the double helix. -Single-strand binding proteins bind to the single strands of DNA, preventing them from re-forming hydrogen bonds with each other and allowing synthesis to occur on both strands. -DNA polymerase III synthesizes the new strands, but it requires an existing 3′ hydroxyl (—OH) group to add nucleotides. -Primase creates short RNA primers, initiating DNA synthesis on both template strands. -DNA polymerase I removes the RNA primers and replaces them with DNA nucleotides. -On the lagging strand, DNA ligase joins Okazaki fragments by forming phosphodiester bonds between them, thus completing DNA replication.]

DNA is a self-replicating molecule. What accounts for this important property of DNA?

The nitrogenous bases of the double helix are paired in specific combinations: A with T and G with C. [Because of the specificity of the pairing of the nitrogenous bases, each strand of the double helix specifies the matching sequence of bases on the other strand. During replication, when the strands separate, each strand serves as a template for the replication of other strand. Read about base pairing in DNA.]

Nucleotides are added to a growing DNA strand as nucleoside triphosphates. What is the significance of this fact?

Hydrolysis of the two phosphate groups (P-Pi) and DNA polymerization are a coupled exergonic reaction. [Read about DNA polymerization.]

Select the most accurate statement describing DNA replication complexes.

DNA replication complexes are grouped into factories, which are anchored to the nuclear matrix. [Read about the DNA replication complex.]