Question 1

Several members of the same family were diagnosed with the same kind of cancer when they were unusually young.Which one of the following is the most likely explanation for this phenomenon? It is possible that the individuals with the cancer have inherited

Accepted Answer

The most likely explanation for several members of the same family being diagnosed with the same kind of cancer at an unusually young age is that they have inherited a mutation in a gene required for mismatch repair (C). This type of mutation can lead to an increased rate of mutations throughout the genome, including in genes that control cell growth, which can result in cancer. Mismatch repair is crucial for correcting errors that occur during DNA replication, and defects in this process can lead to a predisposition to certain types of cancer, which can be inherited across generations.

Question 2

Recombination has occurred between the chromosome segments shown in Figure 6-40.The genes A and B, and the recessive alleles a and b, are used as markers on the maternal and paternal chromosomes, respectively.After alignment and homologous recombination, the specific arrangements of A, B, a, and b have changed.  Figure 6-40
Which of the choices below correctly indicates the gene combination from the replication products of the maternal chromosome?

Accepted Answer

The answer of Recombination has occurred between the chromosome segments...

Question 3

Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated.This gives rise to mutation.If the cytosine in the sequence TCAT is deaminated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication?

Accepted Answer

Deamination of cytosine converts it to uracil, which pairs with adenine during DNA replication. After one round of replication, the original strand with uracil (U) will produce a complementary strand with adenine (A) opposite the uracil. In the subsequent round of replication, the strand with adenine (A) will produce a complementary strand with thymine (T) opposite the adenine, resulting in the sequence TTAT.

Question 4

Even though DNA polymerase has a proofreading function, it still introduces errors in the newly synthesized strand at a rate of 1 per 10⁷ nucleotides.To what degree does the mismatch repair system decrease the error rate arising from DNA replication?

Accepted Answer

The mismatch repair system significantly decreases the error rate arising from DNA replication by approximately 100-fold, further enhancing the fidelity of DNA replication beyond the proofreading capability of DNA polymerase.

Question 5

Sickle-cell anemia is an example of an inherited disease.Individuals with this disorder have misshapen (sickle-shaped) red blood cells caused by a change in the sequence of the β-globin gene.What is the nature of the change?

Accepted Answer

The change in the sequence of the β-globin gene in sickle-cell anemia is a base-pair change, specifically a substitution of a single nucleotide (adenine replaced by thymine) in the DNA sequence.

Question 6

Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?

(a)

(b)

(c)

(d)

Figure 6-25

Accepted Answer

The answer of Which of the choices below represents the...

Question 7

In somatic cells, if a base is mismatched in one new daughter strand during DNA replication, and is not repaired, what fraction of the DNA duplexes will have a permanent change in the DNA sequence after the second round of DNA replication?

Accepted Answer

After the first round of DNA replication, one of the two daughter strands will have a mismatched base. In the second round of replication, the strand with the mismatch will produce one mutated strand and one correct strand (since the original template strand is correct), while the original correct strand will produce two correct strands. Thus, out of the four strands produced in the second round, only one will have a permanent change, making the fraction 1/4.

Question 8

Telomeres serve as caps protecting the ends of linear chromosomes.Which of the following is FALSE regarding the replication of telomeric sequences?

Accepted Answer

The leading strand does not double back on itself to form a primer for the lagging strand during telomere replication. Instead, a specialized enzyme called telomerase adds the repeated sequences to the template strand, allowing for the lagging strand telomeres to be fully replicated.

Question 9

You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons.You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit.What is the most likely explanation for this?

Accepted Answer

Mutations in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences. Therefore, natural selection is likely to be more stringent in preserving the coding sequence, but less strict in preserving the surrounding noncoding sequences. This could explain why the two organisms have very similar coding sequences but variable noncoding sequences.

Question 10

Select the option that best completes the following statement: Nonhomologous end joining is a process by which a double-stranded DNA end is joined

Accepted Answer

Nonhomologous end joining is a process where the ends of a broken double-stranded DNA molecule are joined together, without regard for sequence homology. In this process, the broken ends are simply joined to the nearest available complementary DNA end. There is no sequence filling, repairing of mismatches or sequence interaction with the complementary chromosome involved in this process.

Question 11

Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated.This gives rise to mutation.If the adenosine in the sequence TCAT is depurinated and not repaired, which of the following is the point mutation that you would observe after this segment has undergone two rounds of DNA replication?

Accepted Answer

Depurination of adenosine in TCAT leads to a missing base. During replication, a random base is inserted opposite the missing base, often resulting in a mutation. After two rounds of replication, this can lead to the sequence TCT.

Question 12

Beside the distortion in the DNA backbone caused by a mismatched base pair, what additional mark is there on eukaryotic DNA to indicate which strand needs to be repaired?

Accepted Answer

A nick in the new strand, also known as an "Okazaki fragment," indicates where repair is needed on the newly synthesized strand. This nick is created by the DNA polymerase during replication, and is later repaired by specialized enzymes. There is no specific chemical modification or sequence gap that indicates repair is needed.

Question 13

Which of the following statements is NOT an accurate statement about thymine dimers?

Accepted Answer

Thymine dimers are actually covalent bonds between adjacent thymine bases on the same DNA strand, not between thymidines on opposite strands.

Question 14

Nonhomologous end joining can result in all but which of the following?

Accepted Answer

Nonhomologous end joining is a process that repairs double-stranded DNA breaks by directly ligating the ends of the broken strands. This process is error-prone and can result in the loss of nucleotides at the site of repair or the translocation of DNA fragments to different parts of the genome. However, it does not typically result in the recovery of lost nucleotides on a damaged DNA strand. The interruption of gene expression can also occur if the break occurs within a gene or its regulatory region, and the resulting repair process disrupts the gene or its regulation.

Question 15

Human beings with the inherited disease xeroderma pigmentosum have serious problems with lesions on their skin and often develop skin cancer with repeated exposure to sunlight.What type of DNA damage is not being recognized in the cells of these individuals?

Accepted Answer

Xeroderma pigmentosum involves a deficiency in DNA nucleotide excision repair, which specifically recognizes and repairs damage caused by ultraviolet radiation. Chemical damage and X-ray irradiation damage can also be repaired by this pathway, but the deficiency in xeroderma pigmentosum only affects the recognition and repair of UV irradiation-induced damage. Mismatched bases are recognized and repaired by a different pathway called mismatch repair.

Question 16

The DNA duplex consists of two long covalent polymers wrapped around each other many times over their entire length.The separation of the DNA strands for replication causes the strands to be "overwound" in front of the replication fork.How does the cell relieve the torsional stress created along the DNA duplex during replication?

Accepted Answer

Topoisomerases break the covalent bonds of the backbone, allowing the local unwinding of DNA ahead of the replication fork. This process relieves the torsional stress created along the DNA duplex during replication.

Question 17

In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes.During which process does swapping occur?

Accepted Answer

Swapping of segments of parental chromosomes occurs during meiosis, specifically during the process of homologous recombination. DNA replication and DNA repair involve the copying or repair of existing genetic material, while transposition involves the movement of genetic elements within a genome, but not the swapping of segments between chromosomes.

Question 18

Homologous recombination is an important mechanism in which organisms use a "backup" copy of the DNA as a template to fix double-strand breaks without loss of genetic information.Which of the following is NOT necessary for homologous recombination to occur?

Accepted Answer

Homologous recombination requires 3′ DNA strand overhangs, not 5′ overhangs, to initiate the process. These overhangs invade the homologous DNA sequence to begin repair. Sequence similarity is necessary for the strands to align correctly, and nucleases are needed to process the DNA ends to generate the 3′ overhangs.

Question 19

The events listed below are all necessary for homologous recombination to occur properly:
A.Holliday junction cut and ligated
B.strand invasion
C.DNA synthesis
D.DNA ligation
E.double-strand break
F.nucleases create uneven strands
Which of the following is the correct order of events during homologous recombination?

Accepted Answer

The correct order for homologous recombination starts with a double-strand break (E), followed by the action of nucleases to create uneven strands (F), then strand invasion (B), DNA synthesis (C), DNA ligation (D), and finally, the Holliday junction is cut and ligated (A).

Question 20

The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process.Which choice below describes the known sequence of events in this process?

Accepted Answer

The known sequence of events in the repair of mismatched base pairs or damaged nucleotides in a DNA strand involves several steps. Firstly, the damage is recognized. Secondly, the newly synthesized strand is identified by an existing nick in the backbone. Thirdly, a segment of the new strand is removed by repair proteins. Fourthly, the gap is filled by DNA polymerase. Lastly, the strand is sealed by DNA ligase. This process is known as nucleotide excision repair. Choice A accurately describes this process. Choice B is incorrect because there is no such thing as a "DNA repair polymerase." Choice C is incorrect because an exonuclease is not involved in this process. Choice D is incorrect because it oversimplifies the process and does not accurately describe the steps involved.

Quiz 6: DNA Replication and Repair