The "speed" of DNA polymerase
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The "speed" of DNA polymerase
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[post_date] => 2024-12-23 07:42:46
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[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
During DNA replication, DNA polymerase adds bases that pair to the bases already present in the template DNA strand. The average distance between consecutive base pairs in a DNA strand is 3.4 Angstroms (1 Angstrom = 10⁻¹⁰ m). DNA polymerase can replicate a DNA strand even as it follows the twists and bends of the DNA strand.
Figure 1 DNA polymerase replicating template strand
The DNA polymerase molecule can form new base pairs quickly. However, quantifying its maximum replication rate is difficult, because individual polymerase molecules can accelerate, decelerate, pause, or fall off the template strand. Polymerase replication rates are often reported in base pairs per second (bp/s).
Researchers determined the maximum replication rate of individual DNA polymerase molecules by measuring the elasticity of single strands of DNA as they were being replicated. The total number of bases in the DNA molecule, N(total), was known. The elasticity of the molecule at each point in time was compared with the known elasticities of single-stranded and double-stranded DNA of length N(total), in order to infer the percentage of the DNA molecule that had been replicated, p(t). Researchers used p(t) to determine the number of base pairs replicated at any given time, as shown in Equation 1.
Equation 1 N(t) = p(t) x N(total)
Results of this analysis are shown below.
Figure 2 Activity of a single DNA polymerase molecule
[post_title] => The "speed" of DNA polymerase
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[question] => If a single stranded DNA template is 6.8 micrometers in length and a single DNA polymerase requires ~4,000 seconds to replicate it what is the average replication rate of the polymerase?
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[answer] => 2
[description] => Reason for Correct Answer:
From the passage, we know that replication rate is normally expressed base pairs per second. To find this, we need to determine two things: (1) the number of base pairs replicated, and (2) the time it took to do this in seconds, which is given in the question stem (4,000 seconds).
The question stem gives the length of the DNA strand (6.8 𝜇m). Remember that a 𝜇m is 10⁻⁶ m. From this we can determine the number of base pairs, because the passage tells us “the average distance between consecutive base pairs in a DNA strand is 3.4 Angstroms (10⁻¹⁰ m).”
This is the math for converting length to number of base pairs:
So, base pairs per second equals:
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[answers] => Array
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[0] => Array
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[each_answer] => A. 1.4 bp / s
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[1] => Array
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[each_answer] => B. 5.0 bp / s
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[each_answer] => C. 1.4 x 103 bp / s
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[each_answer] => D. 5.0 x 103 bp / s
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[1] => Array
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[quiz_unique_key] => 3243476205
[question] => A scientist allows a single-stranded DNA molecule of known length to undergo replication with a DNA polymerase for a known time interval. At the end of this interval, she records the elasticity of the DNA molecule. Based on the method described in the passage, which quantity can be calculated for the DNA polymerase molecule?
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[answer] => 1
[description] => Reason for Correct Answer:
The passage says that researchers can use the elasticity of the DNA molecule at a given time, compared to DNA molecules of the same total length, to determine how much of the DNA molecule has been replicated.
The researcher will therefore be able to determine the length of DNA replicated (number of base pairs) following the time interval.
She can divide this length (in bp) by the time interval (s) to come up with an AVERAGE speed of replication (bp/s). However, she won’t be able to determine instantaneous speed or the time point of any pauses during this interval.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Average speed of base pair generation
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[1] => Array
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[each_answer] => B. Instantaneous speed of base pair generation
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[each_answer] => C. Average time interval between pauses in base pair generation
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[each_answer] => D. Total number of pauses in base pair generation
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[2] => Array
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[quiz_unique_key] => 2187790141
[question] => Which feature of Figure 2 represents the maximum replication rate of a single DNA polymerase molecule?
[value] => Array
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[answer] => 3
[description] => Reason for Correct Answer:
The replication rate is the number of base pairs generated per second (bp/s), or N(t)/s.
The y-axis in Figure 2 is N(t), the number of base pairs synthesized. The x-axis is t, the number of seconds that have passed.
To find the number of base pairs per second, you would need to divide the number of base pairs that had been replicated (the “rise” of the y-axis) by the number of seconds that had passed (the “run” of the x-axis). In other words, the replication rate of the polymerase can be found by dividing the rise by the run of the graph.
Rise over run is the definition of slope. Therefore the slope represents the replication rate of the polymerase. The highest slope will be the maximum replication rate of the polymerase.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Highest N(t) value
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[1] => Array
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[each_answer] => B. Highest t value
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[2] => Array
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[each_answer] => C. Highest slope value
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[each_answer] => D. Highest plateau length
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[3] => Array
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[quiz_unique_key] => 1579828684
[question] => Which of the following would best explain the plateaus seen in Figure 2?
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[answer] => 4
[description] => Reason for Correct Answer:
The question asks you to focus on the plateaus–or flat areas–of the graph.
During the plateaus, time is passing but DNA polymerase is not adding any new bases. Therefore the polymerase has paused during the plateaus.
Though the rate of DNA replication may depend on the concentration of nucleotides, dwindling concentrations of nucleotides would not cause an abrupt stop in DNA polymerase, which was just moving at a constant rate. It is likely that the reaction vessel includes excess nucleotides so that they are not a limiting factor – which also would explain the constant rates during periods of replication.
The most likely answer is that DNA contains specific sequences that cause the polymerase to pause. These sequences could result in temporary hairpin structures in the DNA template that can disrupt the smooth progression of DNA polymerase during replication, leading to pauses and stalls in the replication process.
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[answers] => Array
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[0] => Array
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[each_answer] => A. DNA structure is highly regular, allowing the polymerase to move at a constant speed.
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[1] => Array
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[each_answer] => B. DNA polymerase has reached its maximum velocity and can no longer accelerate.
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[2] => Array
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[each_answer] => C. The replication rate of DNA polymerase depends on the surrounding concentrations of nucleotides.
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[each_answer] => D. Certain sequences in the DNA form secondary structures, such as hairpins, which impede the progress of DNA polymerase.
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[4] => Array
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[quiz_unique_key] => 1498560436
[question] => At approximately what time does the DNA polymerase molecule in Figure 2 experience the greatest positive acceleration in its rate of base pair generation?
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
Acceleration is normally defined as change in velocity (speed or direction). Since we are ignoring change in direction (DNA has twists and bends in it), we are looking for the greatest positive acceleration in terms of speed increase.
This graph is quite similar to a distance vs. time graph, except that the y-axis [N(t)] indicates “distance” in terms of how many base pairs the polymerase has traveled along the DNA strand. Therefore, the slope of the graph will tell us the speed of the polymerase in base pairs per second.
The CHANGE IN SLOPE will tell you about the acceleration (change in speed).
At 75 seconds, the slope decreases. The polymerase decelerates.
Between 350 and 400 seconds – so, around 375 seconds, the slope changes from 0 (the line is flat) to a large positive number (the line is very steep). Therefore, positive acceleration is occurring. The greatest positive acceleration occurs at 375 seconds.
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[answers] => Array
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[each_answer] => A. 75 s
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[each_answer] => B. 375 s
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[each_answer] => C. 425 s
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[each_answer] => D. 550 s
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[5] => Array
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[quiz_unique_key] => 3413571627
[question] => In prokaryotes, which DNA polymerase is primarily responsible for DNA replication (i.e. elongating a growing strand of DNA)?
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[answer] => 3
[description] => Reason for Correct Answer:
In prokaryotes, DNA polymerase I is involved in removing RNA primers during DNA replication and replacing them with DNA nucleotides. After the RNA primers are removed, Pol I fills the resulting gaps with the appropriate DNA nucleotides.
Pol I is also involved in DNA repair processes, such as base excision repair. It can remove damaged or incorrect nucleotides and replace them with the correct ones.
DNA polymerase II is primarily involved in DNA repair processes, especially in the repair of DNA damage caused by UV radiation. It participates in the DNA damage tolerance pathway and is important for maintaining genome stability under conditions of DNA damage.
DNA polymerase V is an error-prone polymerase involved in a process called translesion synthesis, which allows replication to proceed over damaged DNA templates. It is not important to know about it for the MCAT.
DNA polymerase III is the main polymerase responsible for the elongation of the leading and lagging strands during DNA replication in prokaryotes. It is highly processive, meaning it can add multiple nucleotides to the growing DNA strand without dissociating from the template.
Pol III is highly accurate and has proofreading capabilities. It can correct errors made during replication, enhancing the fidelity of DNA synthesis.
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[answers] => Array
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[each_answer] => A. DNA polymerase I
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[each_answer] => B. DNA polymerase II
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[2] => Array
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[each_answer] => C. DNA polymerase III
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[each_answer] => D. DNA polymerase V
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Figure 1
Figure 2
