Exam II Chapter Questions

Question 0

Explain where chromosomal DNA is located in prokaryotic vs eukaryotic cells

Answer 0

It’s located within the nucleoid in prok cells and and the nucleus for euk cells. The nucleus, unlike the nucleoid, is membrane bound and typically occupies a small fraction of cell volume.

Question 1

List three properties that differ bt chromosome makeup in E. coli compared to human cells

Answer 1

Q

Question 2

Explain genome size and organism complexity

Answer 2

A

Question 3

Inter genetic sequences make up >60% of the human genome. Where do hear inter genetic sequences come from and what are some of their functions?

Answer 3

Transposition elements.they may encode miRNAs, serve as regulatory seqs for transcription, or may be nonfunctional seqs such as pseudo genes.

Question 4

Why does each grim isomer in a eukaryotic cell contain multiple origins of replication but includes only one centromere?

Answer 4

Q

Question 5

How does the sister chromatid cohesion ensure that each daughter cell receives one copy of each chromosome?

Answer 5

Cohesion so holds sister chromatid together during S phase and early stages of mitosis. During anaphase, cohesion is eliminated so that the microtubules attached to the kinetochore which assemble at the centromere separate the sister chromatid pairs into the daughter cells

Question 6

From a diploid human cell tell how many copies of each chromosome are present in each cell (or soon to be daughter cell): 
1. Start of mitosis
2 end of mitosis
3 start of meiosis
4 end of meiosis
5 end of meiosis II

Answer 6

A

Question 7

For humans, what cells undergo mitosis? What cells undergo meiosis?

Answer 7

All cells that grow and divide undergo mitosis (somatic and germ cells), only cells that produce eggs and sperms undergo meiosis (germ cells)

Question 8

Describe the components of a nucleosome

Answer 8

D

Question 9

What types of bonds occurs between histone proteins and DNA and the region of DNA where these bonds form? Are these actions seq specific? Explain why/why not

Answer 9

H bonds form bt histone proteins and phosphodiester backbone near the minor groove and additionally bt the bases of the minor groove. These interactions are not seq specific. DNA throughout the genome wraps around the histone proteins. Proteins that interact w the minor groove of DNA are much less likely to interact in a seq specific manner. In contrast, interactions w major groove of the DNA generally make seq specific interactions.

Question 10

Explain why stored negative superhelicity from packaging dan into nucleosomes is advantageous for cellular functions

Answer 10

D

Question 11

Which protein domains recognize acetylation of histone amino-terminal tails? Which protein domains recognize methylated ones?

Answer 11

Bromodomains recognize acetylation. Chromodomains, TUDOR-domains, and PHD fingers recognize methylation. SANT domains recognize unmodified histone tails.

Question 12

Predict where each of the following would migrate on a gel:
1 relaxed cccDNA
2 initiate nucleosome assembly without topoisomerase and treat w detergent before running on a gel
3 add topoisomerase to previous rxn but prevent additional nucleosome assembly. Then add detergent before running on gel
4. Add topoisomerase to rxn 2 but allow additional nucleosome assembly. Then add detergent before running on a gel.

Answer 12

1
2
3
4

Question 13

Name two substrates for DNA synthesis and explain why each is needed

Answer 13

D

Question 14

List mechanistic steps of DNA synthesis starting w the primed template and deoxynucleoside triphosphates

Answer 14

Synthesis begins with the h bond dependent interaction of the incoming nuc to the DNA template. After an appropriate base pair is formed the 3’OH of the primer initiates a nucleophilic attack of the alpha phosphate of the incoming nuc. Pyrophosphate is released and hydrolyzed to two phosphatases by pyrophosphate se. The incoming nuc is now base pair to the template and covalently linked to the primer DNA strand.

Question 15

Explain why DNA synthesis is coupled to hydrolysis of pyrophosphate

Answer 15

S

Question 16

Why is magnesium chloride added to PCR buffer

Answer 16

A

Question 17

Hypothesize why some DNA polymerases lack exonuclease activity without significantly contributing to the number of mismatches introduced during DNA replication

Answer 17

Some DNA polymerases are only used during special processes like DNA repair. They aren’t very processing and don’t carry out the bulk of DNA synthesis in the cell. Proof reading is less imp for these polymerases that will insert a small number of nucs relative to the leading or lagging strand DNA polymerases.

Question 18

You want to set up an assay starting with a sliding clamp bound to DNA what special property must the DNA have to establish binding bt the sliding clamps and DNA? What other protein components are needed in the rxn to ensure binding?

Answer 18

H

Question 19

Explain how the time req to complete rep of the E. coli genome is 40 mins yet the cells can divide every 20 min

Answer 19

E. coli cells only initiate DNA rep once per cell division but when E. coli divides rapidly, initiation of the next round of rep starts before the previous round of rep is complete. Under these circumstances, the time for cell division Can be as low as 20 mins

Question 20

Why is telomerase not needed in E. coli cells?

Answer 20

Their genome is circular thus E. coli doesn’t have the prob of chromosome length shortening after each round of rep in the absence of telomerase bc the genome can be completely replicated.

Question 21

Describe the role of a DNA helicase at the rep fork

Answer 21

D

Question 22

As a result of DNA helicase activity, topoisomerases are also req during rep. Explain how topoisomerases help DNA helicase to function more properly.

Answer 22

D

Question 23

Why does helicase not have to be added to a PCR rxn

Answer 23

D

Question 24

In E. coli DNA pol I possesses 5’ exonuclease and 3’ exonuclease activity, whereas DNA pol III only possesses 3’ exonuclease activity. Why is this functional?

Answer 24

The 3’ exonuclease activity of each DNA pol allows it to remove incorrect nucs during synthesis. DNA pol I has the additional 5’ exonuclease activity to remove nucs ahead of the DNA pol. Specifically, this function helps the DNA pol remove RNA primers on the lagging strand of DNA.

Question 25

Why are defects in the RNA component of telomerase associated with diseases?

Answer 25

S

Question 26

While performing incorporation assays to measure DNA synthesis using 32P-labeled dNTPs (where 32P replaces the alpha phosphate of the dNTP)..

1. If you use dNTPs labeled at the beta or gamma P, you don’t detect any radioactivity, why?
2. How does gel electrophoresis serve as a means to separate unincorporated 32P-labeled dNTPs from the newly synthesized DNA strand?
3. For the filter binding, describe a negative control that would ensure that you’re filter is separating the unincorporated 32P-labeled dNTPs from the DNA.

Answer 26

F

Question 27

DNA polymerase mistakenly inserts a C across from a T during replication. Assuming that proofreading and the mismatch repair do not correct the mismatch, is the resulting mutation a transition or transversion after the next round of replication? Explain

Answer 27

A

Question 28

Explain why the deamination of 5-methylcytosine leads to hot spots for spontaneous mutations more than the deamination of cytosine in DNA does.

Answer 28

This demethylation produces uracil. A specific glycosylase in base excision repair recognizes uracil as not belonging to DNA. If U remains, a mutation could occur after the next round of rep. Deamination of 5’-methylcytosine will produce thymine which is not recognized as a mistake by DNA repair molecules. Thus, following the next round of DNA rep, the thymidine will base pair with an A, yielding a transition mutation. Therefore the cell removed Us to prevent mutations, but does not remove thyminez to prevent deamination.

Question 29

The following terms describe the general steps of a DNA repair pathway. Place the steps in the correct order. Name the protein(s) that complete each of the steps in E. coli for the mismatch repair, base excision, repair, and nucleotide excision repair pathways; ligation, DNA synthesis, recognition, excision

Answer 29

A

Question 30

Calculate the number of mismatches that could occur in one human cell during one round of rep. Assume the size of the human genome is 3.2 billion base pairs.

Answer 30

W

Question 31

Calculate the number of mismatches that occur in one human cell during one round of rep in the absence of mismatch repair

Answer 31

W

Question 32

Give a loss of function mutant for dam (the gene encoding the Dam synthesis) in E. coli, predict the phenotype one would observe with respect to spontaneous mutagenesis. Explain

Answer 32

Without dam methylated the parent strand wouldn’t be methylated during rep. Without this, MutH would have no way of distinguishing which strand is parental vs which is newly synthesized. MutH will nick the incorrect strand at some frequency. Mismatch repair of the parent strand would lead to an increase in spontaneous mutagenesis (not induced by an exogenous agent).

Question 33

Describe a possible advantage and disadvantage of repairing 3-methyladenine through base excision repair relative to repairing O6-methylguanine through direct reversal by a methyltransferase

Answer 33

G

Question 34

Why is the intrastrand cross link bt two adjacent guanines in DNA is a better candidate for nucleotide excision repair rather than for base excision repair

Answer 34

Because a cross link bt two guanines distorts the DNA helix similar to a thymine dimer. This allows NER proteins to recognize the distortion in order to excise the stretch of DNA including the cisplatin-induced cross link. Base excision repair only excises one nuc. Also, a specific DNA glycosylase must recognize the DNA lesion. No glycosylase recognizes cisplatin-induced cross links.

Question 35

Predict the immediate consequence to a cell in which the system of transcription-coupled nucleotide excision repair stopped functioning properly

Answer 35

W

Question 36

Aside from DNA damage tolerance, name the repair pathway that potentially introduces mutations. Describe how this pathway introduces mutations

Answer 36

Non homologous end joining (NHEJ) repairs double strand breaks DSBs at the cost of introducing mutations. The NHEJ enzymes process the free ends of a DSB. Through this processing, DNA seq is lost or added before the two strands are lighted together.

Question 37

Explain the difference bt DNA repair and DNA damage tolerance

Answer 37

S

Question 38

What medium must be used in the selective plate as part of the Ames test. Explain how mutation gives rise to a revertant in this experiment?

Answer 38

D

Question 39

Give an ex of how a revertant can arise in the absence of an added mutagen

Answer 39

D

Question 40

Explain two ways in which DNA replication can introduce double strand breaks (DSBs) in a DNA template

Answer 40

D

Question 41

You’re considering two alleles for one specific gene. Describe what feature, with respect to the DNA, distinguishes one allele from the other. Are the two alleles homologous?

Answer 41

They differ from each other by their minor seq variation. The majority of the seq for the gene remains the same, so the alleles are homologous.

Question 42

What is heteroduplex DNA?

Answer 42

D

Question 43

List the different enzymatic activities that RecBCD catalyzes and describe the significance of each activity in the steps of homologous recombination (via the DSB-repair pathway)

Answer 43

RecBCD includes DNA helicase and nuclease activity. Specifically, RecBCD acts as a 3’-5’ DNA helicase and a nuclease.
RecD acts as a 5’-3’ DNA helicase.
RecC helps to improve the efficiently of the other two. RecC recognizes and binds to the x site to stop the nuclease activity on the 3’ tail. RecBCD plays a critical role in processing the dsDNA at a break to produce ssDNA for invasion.

Question 44

explain why RecA-dependent strand exchange cannot occur bt two homologous, double stranded covalently closed circular DNAs, but can occur bt the two dsDNA molecules

Answer 44

D

Question 45

Propose an assay that researchers may have used to determine RuvA protein binding to the DNA substrate. Propose a mod to the DNA substrate to use as a neg control that demonstrates an aspect of the specificity of RuvA binding

Answer 45

D

Question 46

Explain the most significant role of homologous recombination in euks that isn’t found in proks

Answer 46

D

Question 47

Describe how cells generate the DSB required to initiate meiotic homologous recombination in euks

Answer 47

Spo11 mediates double stranded cleavage of the DNA. A tyrosine in Spo11 attacks the phosphodiester backbone to cut the DNA. Spo11 store the energy from breaking the phosphodiester bond by forming a covalent high energy intermediate within the broken DNA.

Question 48

Define gene conversion and give an example of a mechanism explaining how gene conversion occurs

Answer 48

A

Question 49

Compare and contrast synthetic dependent strand annealing (SDSA) used in mating type switching with DSB-repair homologous recombination

Answer 49

Like DSB-repair homologous recombination, SDSA starts w a DSB at the recombination site, 5’-3’ resection, and invasion of the 3’ end to serve as primer for DNA synthesis.
SDSA differs from DSB-repair homologous recombination in that there is n resolution via cleavage of a Holliday junction. Following strand invasion, a complete rep fork forms in SDSA. The ‘ end that doesn’t invade is removed in SDSA. The newly synthesized DNA in displaced, and a second DNA synthesis event completes the process resulting in a gene conversion.

Question 50

Explain why DSB-repair homologous recombination can occur bt any two DNA molecules that share homologous rather than only bt two DNA molecules that carry a specific sequence

Answer 50

D