6.2 DNA Replication

1. AIt doubles the amount of DNA, and each daughter cell then receives half of the parent cell's genes.
   A student who thinks that dividing a cell's DNA halves its genetic information picks this. Replication doubles the DNA so that each daughter cell can receive a complete copy of every gene, not half of them.
2. BIt makes a second complete copy of the DNA, so each daughter cell receives a full set of the parent's genes. Correct
   Replication produces two copies of every DNA molecule, each with the same base sequence. Mitosis then gives one copy of each molecule to each daughter cell, so both receive the complete genetic information of the parent cell.
3. CIt keeps each original DNA molecule intact and makes an all-new copy for the other daughter cell.
   A student who thinks replication works like photocopying picks this. Replication is semiconservative: each DNA molecule produced has one parental strand and one new strand, so neither daughter cell receives all-new DNA.
4. DIt builds each new strand as an identical copy of the strand it pairs with, so no information changes.
   A student who thinks a new strand is an identical copy of its template picks this. Each new strand is complementary to its template (A with T, G with C); the double helix as a whole is copied faithfully.

CED 6.2.A.1 · Read this in Fix

1. AAt end Q, its 3′ end, so the strand grows to the right Correct
   The new strand is antiparallel to its template. The template's 3′ end is on the left, so the new strand's 5′ end is at P and its 3′ end is at Q. DNA polymerase adds nucleotides only to the 3′ end, so the strand grows from Q toward the right, in the 5′ to 3′ direction.
2. BAt end P, its 3′ end, so the strand grows to the left
   A student who draws the new strand running in the same direction as its template picks this, placing the new strand's 3′ end at P, on the same side as the template's 3′ end. The strands are antiparallel, so the new strand's 3′ end is at Q.
3. CAt end P, its 5′ end, so the strand is extended to the left
   A student who reads '5′ to 3′' as 'nucleotides are added at the 5′ end' picks this. P is the 5′ end, but nucleotides are added at the 3′ end, Q, so the strand grows to the right.
4. DAt both ends, P and Q, so it grows in two directions
   A student who thinks DNA polymerase can extend either end of a strand picks this. DNA polymerase adds nucleotides only to the 3′ end, so the strand grows in one direction, from Q.

CED 6.2.A.1.i · Read this in Fix

1. A1 of 8
   A student who thinks the original double helix stays intact and every copy is all new picks this: 1 all-¹⁵N molecule and 7 all-¹⁴N molecules. Replication is semiconservative, so the two ¹⁵N strands end up in two different molecules.
2. B8 of 8
   A student who thinks every molecule keeps one of the original strands picks this. The original molecule has only two strands; strands made in rounds 1 and 2 act as the parental strands in later rounds, so only 2 of the 8 molecules contain ¹⁵N.
3. C2 of 8 Correct
   In round 1 the two ¹⁵N strands separate and each pairs with a new ¹⁴N strand. In every later round each strand is a template, but the two original ¹⁵N strands stay intact and each ends up in one molecule. After round 3 there are 8 molecules: 2 contain an original ¹⁵N strand and 6 contain only ¹⁴N.
4. D4 of 8
   A student who thinks 'semiconservative' means that half of the molecules are original picks this. The term describes each molecule (one parental and one new strand), and after three rounds only 2 of the 8 molecules contain a ¹⁵N strand.

Working Round 1: the two ¹⁵N strands separate and each pairs with a new ¹⁴N strand, giving 2 molecules, each with one ¹⁵N strand and one ¹⁴N strand. Round 2: each of the 4 strands is a template; 4 molecules form, 2 containing an original ¹⁵N strand and 2 made only of ¹⁴N strands. Round 3: 8 molecules form; the two original ¹⁵N strands are still intact and each is in one molecule, so 2 molecules contain ¹⁵N and 6 contain only ¹⁴N. Answer: 2 of 8. Distractors: conservative replication (original double helix kept intact, all copies new) gives 1 of 8; keeping an original strand in every molecule gives 8 of 8 (2³ = 8); reading 'semiconservative' as 'half the molecules are original' gives 8/2 = 4 of 8.

CED 6.2.A.1.ii · Read this in Fix

1. AATGCCA, from 5′ to 3′
   A student who thinks a new strand is an identical copy of its template picks this. The new strand is complementary to the template: opposite A is T and opposite G is C.
2. BTACGGT, from 5′ to 3′
   A student who thinks a new strand runs in the same direction as its template picks this. The bases are right, but the strands are antiparallel, so opposite the template's 5′ end is the new strand's 3′ end.
3. CUACGGU, from 3′ to 5′
   A student who uses the RNA pairing rule picks this. DNA contains thymine, so A on the template pairs with T, not U, in the new DNA strand.
4. DTACGGT, from 3′ to 5′ Correct
   Each template base pairs with its complement (A with T, G with C), giving TACGGT. The new strand is antiparallel to the template, so the end lined up under the template's 5′ end is the new strand's 3′ end: reading from left to right, the new strand is TACGGT from 3′ to 5′.

CED 6.2.A.1.ii · Read this in Fix

1. AReplication would stop, because helicase is what adds nucleotides to the new strands.
   A student who thinks helicase builds the new strands picks this. DNA polymerase, which the drug does not affect, adds the nucleotides; replication stops because helicase no longer separates the parental strands, so no new template is exposed.
2. BReplication would continue, because topoisomerase would separate the parental strands.
   A student who swaps the roles of helicase and topoisomerase picks this. Topoisomerase relaxes overwinding ahead of the fork; it does not separate the two strands, so it cannot take over helicase's job.
3. CReplication would stop, because the parental strands would no longer be separated. Correct
   Helicase breaks the hydrogen bonds between paired bases, separating the two parental strands so that each can act as a template. Without helicase no new template is exposed, so DNA polymerase has nothing new to copy and replication stops.
4. DReplication would continue more slowly, as the parental strands would come apart unaided.
   A student who thinks blocking an enzyme only slows a process picks this. In a living cell the paired strands of DNA do not separate on their own at a useful rate, so without helicase replication stops.

CED 6.2.A.1.iii · Read this in Fix

1. AIt kept making new DNA, at a steady rate, from 8 minutes to 20 minutes.
   A student who reads the height of the line as the rate picks this. From 8 to 20 minutes the line is flat, so its slope, the rate of DNA synthesis, is zero.
2. BIt made new DNA for the first 8 minutes, and then it made no more. Correct
   The drug-treated line rises from 0 to 10 μg over the first 8 minutes and is then flat at 10 μg until 20 minutes. A flat line on a graph of the total new DNA means the total is no longer increasing, so no more new DNA was made after 8 minutes.
3. CIts amount of new DNA went down after reaching a peak at 8 minutes.
   A student who thinks a graph that levels off shows a decrease picks this. The line stays at 10 μg after 8 minutes; it does not slope downward, so the amount of new DNA did not fall.
4. DIt made new DNA more slowly than the control did, from the start.
   A student who expects a blocked enzyme only to slow the process picks this. In the first 2 minutes the drug-treated culture made 4 μg, the same 2 μg per minute as the control; its synthesis slowed and stopped only later.

CED 6.2.A.1.iv · Read this in Fix

1. AReplication would not begin, as the parental strands could not be separated.
   A student who thinks topoisomerase separates the strands picks this. Helicase, which the compound does not block, separates the strands, so replication can begin; it is the movement of the forks that is halted as overwinding builds up.
2. BReplication would be normal, but the new DNA could not be wound around histones.
   A student who equates the supercoiling that topoisomerase relaxes with the coiling of DNA around histones picks this. In replication, topoisomerase relaxes overwinding ahead of the fork, so blocking it stops the forks rather than leaving replication normal.
3. CReplication would continue at a steady but lower rate until all the DNA was copied.
   A student who thinks blocking an enzyme only slows a process picks this. Without topoisomerase the overwinding is not relaxed at a useful rate; it keeps building until the forks stop, so replication is not completed.
4. DReplication forks would stall as the DNA ahead of them became overwound. Correct
   As helicase separates the strands, the DNA ahead of each fork becomes overwound (supercoiled). Topoisomerase normally relaxes this overwinding. Without it, the overwinding builds up until the strands ahead can no longer be separated and the forks stop moving.

CED 6.2.A.1.iv · Read this in Fix

1. ANo new DNA strand could be started, because DNA polymerase needs an RNA primer to begin. Correct
   DNA polymerase can add nucleotides only to an existing 3′ end, so every new strand, and every lagging-strand fragment, begins with a short RNA primer. If RNA nucleotides cannot be joined, no primers are made and DNA polymerase cannot begin any new strand.
2. BOnly the leading strand would be made, because only lagging-strand fragments need primers.
   A student who thinks primers are needed only on the lagging strand picks this. The leading strand also begins with an RNA primer, so it cannot be started either.
3. CBoth new strands would be made, because DNA polymerase can start a strand on a bare template.
   A student who thinks DNA polymerase starts strands as RNA polymerase does picks this. DNA polymerase cannot begin a strand on a bare template; it needs a primer with a free 3′ end.
4. DBoth new strands would be made, because DNA replication does not involve RNA in any step.
   A student who thinks RNA is made only in transcription picks this. Each new DNA strand begins with an RNA primer, so blocking RNA synthesis blocks the start of DNA synthesis.

CED 6.2.A.1.v · Read this in Fix

1. ATube 2
   A student who thinks a control is a tube with nothing added picks this. Tube 2 contains no fragments and no ligase, so no joining can happen in it whatever X does; it cannot show the effect of X.
2. BTube 3
   A student who thinks the control for an inhibitor is a reaction without the enzyme picks this. Tube 3 shows how much joining happens with no ligase, but it contains X, so it cannot show whether X reduces joining by ligase.
3. CTube 4
   A student who thinks a repeat of the treatment is a control picks this. Tube 4 has the same contents as the experimental tube, so it is a replicate: it shows how much the results vary, not what X does.
4. DTube 1 Correct
   Tube 1 is identical to the experimental tube except that it lacks X: it contains the fragments, the ligase and the same buffer. If less joined DNA forms in the experimental tube than in tube 1, the difference can be attributed to X.

CED 6.2.A.1.vii · Read this in Fix

1. AIn fragments, each made 3′ to 5′ and growing toward the fork
   A student who thinks the lagging strand is made 3′ to 5′ picks this. Every fragment of Y grows 5′ to 3′, away from the fork; Y as a whole extends toward the fork only because new fragments keep being started.
2. BIn fragments, each made 5′ to 3′ and growing away from the fork Correct
   Y's template, the bottom strand, runs 5′ to 3′ from left to right, so Y, being antiparallel, has its 3′ end on the left. Y can be extended only at its 3′ end, so it grows to the left, away from the fork. As the fork opens, the newly exposed template near the fork must be copied by a new fragment, so Y is made discontinuously: it is the lagging strand.
3. CIn one piece, made 5′ to 3′ and growing toward the fork
   A student who draws Y running in the same direction as its template picks this, which puts Y's 3′ end toward the fork. Y is antiparallel to its template, so its 3′ end points away from the fork and it cannot be made in one piece.
4. DIn fragments, but made only after strand X has been completed
   A student who reads 'lagging' as 'made later' picks this. Y is made in fragments, but at the same fork and at the same time as X, not after X is finished.

CED 6.2.A.1.vi · Read this in Fix

1. ABoth carry the same base sequence, but only one of them contains DNA from the original molecule.
   A student who thinks the original double helix stays intact during replication picks this. Each chromatid's DNA contains one strand of the original molecule and one new strand.
2. BEach carries half of the chromosome's genes, so the two together hold one full set of genes.
   A student who thinks dividing replicated DNA halves its genetic information picks this. Each chromatid contains a complete double helix with all of the chromosome's genes.
3. CEach original strand acted as a template, so each chromatid has one original strand and one new strand. Correct
   Replication is semiconservative: the chromosome's DNA molecule is copied into two double helices, one in each chromatid, and each consists of one parental strand and one new strand. The two chromatids therefore carry the same hereditary information.
4. DEach is one strand of the original DNA molecule, so the two have complementary sequences.
   A student who equates the two sister chromatids with the two strands of a double helix picks this. Each chromatid contains a whole double-stranded DNA molecule, and the two chromatids carry the same, not complementary, sequences.

CED 6.2.A.1 · Read this in Fix

1. AEach leading strand would be continuous, but each lagging strand would stay as separate fragments. Correct
   DNA polymerase, which is unaffected, still makes the leading strand continuously and the lagging strand as fragments. Ligase normally joins the lagging-strand fragments by forming a covalent bond between adjacent fragments; with ligase inactive, the fragments stay unjoined.
2. BBoth new strands would be complete, but they would not stay paired with their template strands.
   A student who thinks ligase forms the hydrogen bonds between a new strand and its template picks this. Base pairing holds each new strand to its template whether or not ligase works; ligase joins fragments end to end.
3. CNeither new strand would be made, because nucleotides could not be linked to the growing strands.
   A student who thinks ligase adds the nucleotides to growing strands picks this. DNA polymerase adds nucleotides; ligase joins fragments that DNA polymerase has already made.
4. DThe lagging-strand fragments would still be joined into one strand, though more slowly than at 30 °C.
   A student who thinks blocking an enzyme only slows a process picks this. Without active ligase the fragments are not joined at a useful rate, so they stay separate.

CED 6.2.A.1.vii · Read this in Fix

1. AIt separates the two parental strands by breaking the covalent bonds in their backbones.
   A student who confuses the bonds between the strands with the bonds within a strand picks this. Each strand's sugar–phosphate backbone stays intact; only the hydrogen bonds between paired bases are broken.
2. BIt separates the two parental strands by breaking the hydrogen bonds between their bases. Correct
   Helicase unwinds the double helix at the replication fork by breaking the hydrogen bonds that hold paired bases together. This separates the two strands so that each can serve as a template.
3. CIt relaxes the overwinding that builds up in the double helix ahead of the replication fork.
   A student who swaps the roles of helicase and topoisomerase picks this. Relaxing the overwinding ahead of the fork is the job of topoisomerase.
4. DIt builds each new strand by adding nucleotides that pair with the bases on the template strand.
   A student who thinks helicase makes the new strands picks this. DNA polymerase adds nucleotides to the new strands; helicase separates the parental strands.

CED 6.2.A.1.iii · Read this in Fix