pierceessentials3e

WORKED PROBLEMS

253

WORKED PROBLEMS

Problem 1

The following diagram represents the template strands of a replication bubble in a DNA molecule. Draw in the newly synthesized strands and identify the leading and lagging strands.

Solution Strategy

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What information is required in your answer to the problem?

The diagram above with the newly synthesized strands drawn in and the leading and lagging strands labeled.

Recall: The two strands of DNA are antiparallel, so the newly synthesized strand should have the opposite polarity (direction) from the template strand.

What information is provided to solve the problem?

A diagram of the template DNA with 5′ and 3′ ends labeled.

For help with this problem, review:

Direction of Replication in Section 9.2 and Figure 9.8.

Solution Steps

To determine the leading and lagging strands, first note which end of each template strand is 5′ and which end is 3′. With a pencil, draw in the strands being synthesized on these templates, and identify their 5′ and 3′ ends.

Recall: DNA synthesis is always 5′ to 3′.

Next, determine the direction of replication for each new strand, which must be 5′→3′. You might draw arrows on the new strands to indicate the direction of replication. After you have established the direction of replication for each strand, look at each fork and determine whether the direction of replication for a strand is the same as the direction of unwinding. The strand on which replication is in the same direction as that of unwinding is the leading strand. The strand on which replication is in the direction opposite that of unwinding is the lagging strand.

Hint: Each replication fork should have one leading and one lagging strand.

Problem 2

Consider the experiment conducted by Meselson and Stahl in which they used ¹⁴N and ¹⁵N in cultures of E. coli and equilibrium density gradient centrifugation. Draw pictures to represent the bands produced by bacterial DNA in the density-gradient tube before the switch to medium containing ¹⁴N and after one, two, and three rounds of replication after the switch to the medium containing ¹⁴N. Use a separate set of drawings to show the bands that would appear if replication were (a) semiconservative; (b) conservative; (c) dispersive.

Solution Strategy

What information is required in your answer to the problem?

Drawings that represent the bands produced by bacterial DNA in density-gradient tubes before the switch to medium containing ¹⁴N and after one, two, and three rounds of replication following the switch to the medium containing ¹⁴N; thus, you should have drawings of four tubes for each model of replication. You will need a separate set of drawings for semiconservative, conservative, and dispersive replication.

What information is provided to solve the problem?

The bacterial DNA was originally labeled with ¹⁵N, and then the bacteria were switched to a medium with ¹⁴N. Original DNA will have ¹⁵N. Newly synthesized DNA will have ¹⁴N.
Equilibrium density gradient centrifugation was performed before switching to ¹⁴N and after one, two, and three rounds of replication following the switch.

For help with this problem, review:

Meselson and Stahl’s experiment in Section 9.2.

Hint: Review the distribution of new and old DNA in semiconservative, conservative, and dispersive replication in Figure 9.1.

Solution Steps

DNA labeled with ¹⁵N will be denser than DNA labeled with ¹⁴N; therefore, ¹⁵N-labeled DNA will sink lower in the density-gradient tube. Before the switch to medium containing ¹⁴N, all DNA in the bacteria will contain ¹⁵N and will produce a single band in the lower end of the tube.

With semiconservative replication, the two strands separate, and each serves as a template on which a new strand is synthesized. After one round of replication, the original template strand of each molecule will contain ¹⁵N and the new strand of each molecule will contain ¹⁴N; so a single band will appear in the density-gradient tube halfway between the positions expected of DNA with ¹⁵N and of DNA with ¹⁴N. In the next round of replication, the two strands again separate and serve as templates for new strands. Each of the new strands contains only ¹⁴N; thus, some DNA molecules will contain one strand with the original ¹⁵N and one strand with new ¹⁴N, whereas the other molecules will contain two strands with ¹⁴N. This labeling will produce two bands, one at the intermediate position and one at a higher position in the tube. Additional rounds of replication should produce increasing amounts of DNA that contains only ¹⁴N; so the higher band will get darker.
With conservative replication, the entire molecule serves as a template. After one round of replication, some molecules will consist entirely of ¹⁵N, and others will consist entirely of ¹⁴N; therefore, two bands should be present. Subsequent rounds of replication will increase the fraction of DNA consisting entirely of new ¹⁴N; thus the upper band should get darker. However, the original DNA with ¹⁵N will remain, and so two bands should be present.
In dispersive replication, both nucleotide strands break down into fragments that serve as templates for the synthesis of new DNA. The fragments then reassemble into DNA molecules. After one round of replication, all DNA should contain approximately half ¹⁵N and half ¹⁴N, producing a single band that is halfway between the positions expected of DNA labeled with ¹⁵N and of DNA labeled with ¹⁴N. With further rounds of replication, the proportion of ¹⁴N in each molecule increases; so a single hybrid band should remain, but its position in the density-gradient tube should move upward. The band should also get darker as the total amount of DNA increases.