Overall

Question 1

Describe what is meant by the term “end replication problem”. How do germ line cells (eggs, sperm) avoid this “problem”?

Answer 1

DNA is synthesized with an RNA primer. After synthesis is completed the RNA primer is degraded leaving an exposed 3’ end. This 3’end is degraded which results in an overall shortening of the double stranded molecule. Germ line cells encode the enzyme, telomerase, which includes an RNA template that is complementary to the telomere sequence. The telomerase RNA serves as a template for the addition of nucleotides, extending the 3’ end. The gap in the complementary strand is filled by the polymerase a- primase.

Question 2

Methylation levels on genes vary widely from zygote through development and into adulthood. Describe the relative methylation levels for each of these stages of development and elaborate on why it is critical for these levels to change throughout the different stages of development.

Answer 2

The DNA of the zygote is substantially methylated. The level of methylation declines as the zygote develops. Blastocyst has low level of methylation.

At the time of embryo implantation, the DNA becomes highly methylated. These methylation “tags” are maintained in somatic cells, but primordial germ line cells are demethylated.

methylation is mark of inactivation
fully differentiated cells are highly methylated

Question 3

Chromatin remodeling complexes facilitate the “opening up” of the promoter region to permit access by RNA polymerase II and general transcription factors. Describe four models that explain how the promoter is made accessible.

Answer 3

1) Nucleosome sliding: The nucleosome slides along the DNA which exposes the TATA site.
2) Reorganization of the histone octamer in which the DNA forms a loop or bulge that makes the site more accessible.
3) Replacement of histone proteins with variant histone that is correlated with active transcription e.g. H2A.Z
4) Displace the histone octamer from the DNA entirely.

Question 4

What is a “masked” mRNA? How does the fertilization of the egg result in the unmasking of mRNAs?

Answer 4

mRNAs are kept in an inactive state by binding of the protein “maskin”. Maskin binds to the protein “CPEB” which is bound to the DNA in the 3’UTR and to the eIF4E cap binding protein which is bound to the 5’ methyl guanosine cap. Following fertilization, CPEB is phosphorylated which displaces maskin. The phosphorylated maskin recruits CPSF which recruits poly(A) polymerase that elongates the poly(A) tail. The elongated poly(A) tail serves as a binding site for PABP which recruit eIF4G, an initiator factor required for translation.

Question 5

miRNAs play a role in translational regulation. Describe three mechanisms by which miRNAs regulate translation.

Answer 5

1) Deadenylation:miRNA as part of miRNP complex promotes deadenylation of poly(A) tail followed by decapping and degradation
2) Initiation block: miRNP blocks initiation of translation by repressing cap recognition or 60S subunit joining.
3) Degradation: Degradation of nascent polypeptide.
4) Elongation Block: slowed elongation or ribosome drop off

Question 6

The glucocorticoid receptor transcription factor activates transcription by binding to the glucocorticoid response element (GRE) on select genes. Describes the steps following the binding of the transcription factor to the GRE and how the gene is converted from a repressed state to an activated transcription pre-initiation complex.

Answer 6

Glucocorticoid receptor binds to GRE and recruits CPB coactivator. Histone acetyltransferase activity of CPB acetylates histone protein lysine residues. Acetylated histone proteins recruit SWI/SNF chromatin remodeling complex alters and opens up the chromatin. TFIID binds to the open region of DNA which also acetylates nucleosomes to allow transcription initiation.

1. DNA in repressed state (deacetylated)
2. GR binds to GRE and recruits CPB coactivator (histone acetyltransferase)
3. Lysine residues acetylated
4. Acetylated histones recruit SW1/SNF remodeling complex
5. TFIID binds to open region of DNA- TAF 250 and TBP acetylate additional nucleosomes
6. RNA Pol II binds to promoter

Question 7

If a DNA polymerase were discovered that was able to synthesize DNA without the requirement of a free 3’-hydroxyl group, which feature of DNA replication would be least likely to be needed?

Answer 7

The DNA polymerase would not require the synthesis of an RNA primer by RNA primase

Question 8

What enzyme activity are the Mcm proteins (Mcm2 – Mcm7) thought to exhibit once they have been loaded onto the replication origin and associated into a ring-shaped complex? What role does this enzyme activity play in replication and to what protein in E. coli is it analogous to?

Answer 8

Mcm proteins are thought to possess helicase activity. The helicase helps unwind the DNA of the oppositely directed replication forks. The Mcm proteins are analogous to the DNAb proteins of E. coli.

Question 9

Meselson and Stahl devised an experiment to determine the mode of DNA replication. Describe the experimental design and how their results demonstrated the mode of DNA replication.

Answer 9

Meselson and Stahl grew bacteria in a medium containg heavy 15N nitrogen and then transferred the bacteria to light 14N nitrogen containing medium. The bacteria were permitted to replicate and samples were removed every 20 minutes and subjected to density gradient centrifugation to determine the density of the replicated DNA. The density gradient revealed a DNA band of intermediate density which confirmed the semi conservative mode of replication. As rounds of replication proceeded the proportion of DNA shifted towards light DNA

Question 10

Describe how Okazaki discovered the short segments of DNA that now bare his name (i.e. Okazaki fragments) and how these fragments can be explained based on the replication mechanism.

Answer 10

Okazaki performed a DNA labeling experiment. Bacteria were incubated in 3H-thymidine for a brief period of time and immediately killed. The DNA in the bacteria was mostly small DNA fragments of 1000 to 2000 nucleotides. If bacteria were labeled for a longer period of time the shorter fragments would be transformed into longer fragments. Okazaki reasoned that the shorter fragments were being ligated together.

Question 11

DNA polymerase contains three distinct enzymatic properties. What are these three properties and what roles do they play in DNA replication?

Answer 11

5’ to 3’ polymerase for DNA replication.
5’ to 3’ exonuclease for digestion and replacement of RNA primer
3’ to 5’ exonuclease for removal of mispaired nucleotides (proofreading)

Question 12

What type of mutation is repaired by the nucleotide excision pathway? Describe the steps involved in nucleotide excision repair.

Answer 12

The nucleotide excision pathway removes bulky lesions such as thymidine dimers. There is a transcriptionally coupled pathway and a global pathway. Damage recognition is mediated by the XPCcontaining protein complex. In the transcriptionally-coupled pathway, the RNA polymerase stalls at the site of the lesion with the help of the CSB protein. The XPB and XPD proteins act as helicases and unwind the DNA surrounding the lesion. Excisions are made on the 5’ and 3’ sides of the lesion by XPC-ERCC1 and XPG respectively. DNA repair synthesis is done by DNA polymerase d/e. DNA is ligated by DNA ligase I.

Question 13

A syncytium is a “cell” that contains more than one nucleus; examples are a skeletal muscle fiber and a blastula of a fly embryo. These two types of syncytia arise by very different pathways. What two mechanisms can you envision that could result in the formation of such syncytia? What does this tell you about the relationship between mitosis and cytokinesis?

Answer 13

By cell fusion and by mitosis without cytokinesis. This tells one that mitosis and cytokinesis are independent processes that do not have to occur together.

Question 14

If you were able to block the function of Cdk kinase in early prophase, what might happen to the nuclear membrane and why?

Answer 14

The Cdk kinase would be unable to phosphorylate nuclear lamins, the nuclear lamina filaments would not disassemble and thus the nuclear membrane would be unlikely to break down.

Question 15

The formation of the spindle pole in some organisms (e.g. plants) occurs in the absence of centrosomes. Explain how this process works.

Answer 15

Spindle fibers nucleate at the chromosomes instead of the centrosome. The minus ends of the microtubules are brought together by minus-end directed motor proteins.

Question 16

Chromosomes oscillate back and forth along the microtubules after their attachment to the spindle. What is thought to cause this oscillatory movement?

Answer 16

The back-and-forth movements are thought to result from both minus end-directed and plus end-directed motors residing in the kinetochores and especially along the chromosome arms.

Question 17

Describe the two types of meiotic nondisjunction. Why is it believed that one of these mechanisms is more prevalent i.e. is responsible for a greater proportion of meiotic
nondisjunctions.

Answer 17

Primary meiotic nondisjunction is more common than secondary nondisjunction. This may reflect the fact that oocytes of older women have remained arrested in meiosis I for a very long period within the ovary. The meiotic spindles of older oocytes are less able to hold together weakly constructed bivalents. Another possibility is that sister chromatid cohesion is not fully maintained over an extended period allowing homologues to separate prematurely.

Question 18

Cell cycle check points require the participation of three types of proteins. What are these three protein types?

Answer 18

• Sensors that detect abnormalities and emit signals
• Transmitters that send signal along pathway
• Effectors that respond to signal and inhibits cell cycle machinery

Question 19

The movement of chromosomes toward the spindle pole during anaphase is associated with the depolymerization of the microtubules. Describe an experiment that I discussed in class that supports this mechanism.

Answer 19

The experiment uses a protozoan basal body as a site of microtubule initiation which grow outward from the basal body. Once the microtubule formed, condensed mitotic chromosomes are introduced which bind to the microtubule. The concentration of tubulin is reduced in the media which causes the depolymerization of the microtubule. The shrinkage of the microtubule is associated with the movement of the chromosome.

Question 20

How can people who inherit one copy of the mutant retinoblastoma gene have a strong probability of exhibiting retinoblastoma if it is a tumor-suppressor gene that characteristically acts recessively?

Answer 20

When they develop retinoblastoma, it means that they developed a somatic mutation of the normal allele in one cell. This resulted in a double recessive situation in this cell, which can now develop into a tumor.

Question 21

Mutations of the p53 gene are linked to a high proportion of cancers (>50%). What is the normal role of p53 and why does it have such an influential role in cells becoming malignant?

Answer 21

P53 monitors for DNA damage. The p53 transcription factor is phosphorylated by a checkpoint kinase that is activated by the ATM protein. P53 in turn effects the transcription of p21 which inhibits cyclin dependent kinase. Mutations in p53 result in cells not having the ability to repair DNA before DNA replication (i.e. no checkpoint control).

Question 22

Mutation of the rb gene is known to be associated with development of tumors and the cancer retinoblastoma. What is role of the wildtype rb gene product and why does its mutation lead to a proliferative growth?

Answer 22

The protein product of the rb gene, pRB, binds to the E2F protein. The pRB-E2F protein complex binds to the promoters of genes involved in cell proliferation, acting as a transcriptional repressor and blocking gene transcription. Activation of cyclin dependent kinase leads to the phosphorylation of pRB which can no longer bind E2F. Loss of bound pRB converts E2F to a transcriptional activator. If the rb gene is mutated and the pRB protein loses its ability to bind to E2F, the cell can no longer control its entry into S phase and will enter prematurely therefore promoting proliferative growth.

Question 23

How might an altered growth factor receptor cause cancer if it becomes constitutively active? As part of your answer explain how the growth factor receptor acts in normal cells.

Answer 23

Normal growth factor receptors are inactive until they bind to the growth factor. When that happens, the receptor is turned on (activated) and initiates the series of events that results in cell proliferation. If the receptor is operating constitutively, it is turned on whether the growth factor is present or not. Thus, the receptor will initiate these events
under all circumstances and it cannot be turned off.

Question 24

What domains within the p53 gene product (protein) are the most frequently found to be mutated in cancers? How do these mutations affect the function of the p53 protein and which specific amino acids are most commonly mutated?

Answer 24

The most frequently characterized mutations in the p53 gene are associated with DNA binding domains. These are frequently in DNA binding regions and are normally arginine residues. The positively charged arginine residues bind to the negatively charged DNA backbone. This is required as the p53 gene is a transcription factor.