Final Material Cell Bio

Women born with Turner's syndrome are X0 and only have one X chromosome. They are lacking a second sex chromosome. These women need hormone replacement therapy for the development of normal female sexual characteristics and to prevent diseases like heart disease and osteoporosis. Most of these women are infertile with only 2% able to have a spontaneous pregnancy. If a woman with Turner's syndrome were to get pregnant (assume her partner is a normal 46-chromosome male), what proportion of the offspring will be expected to have Turner's syndrome?

1/4

Women born with Turner's syndrome are X0 and only have one X chromosome. They are lacking a second sex chromosome. These women need hormone replacement therapy for the development of normal female sexual characteristics and to prevent diseases like heart disease and osteoporosis. Most of these women are infertile with only 2% able to have a spontaneous pregnancy. If a woman with Turner's syndrome were to get pregnant (assume her partner is a normal 46-chromosome male), what proportion of the offspring will be normal XY male? Group of answer choices

1/4

An organism that has four homologous pairs of chromosomes can produce, in the absence of recombination, __________________ genetically different types of gametes. Recombination will __________________ the number of genetically different gametes possible.

16, increase

The new strand of DNA is synthesized in the

5' - 3' direction

Which of the following mutations is likely to prevent normal cell‐cycle progression?

A mutation that removed the phosphorylation sites on the Rb protein.

Which of the following would be effective ways of blocking progression of the cell cycle and arresting the cell in G1? Choose all that apply. Highly phosphorylate Rb Add a phosphatase that removes phosphates from MAP kinases Block binding of cyclins to Cdk proteins Activate Ras

Add a phosphatase that removes phosphates from MAP kinases Block binding of cyclins to Cdk proteins

What happens to the level of apoptosis in tBid treated cells when both Bak and Bax are deleted in the same cells?

Apoptosis levels are highly decreased compared to wildtype cells - When both Bax and Bak are deleted, the levels of apoptosis remain very low even when treated with tBid to induce apoptosis. This indicates that there are no homo or heterodimer channels being formed when Bax and Bak are deleted.

What happens to the level of apoptosis in tBid treated cells when only Bax or Bak are deleted?

Apoptosis levels stay about the same as wildtype cells - When only Bax or Bak are deleted, the levels of apoptosis still stays relatively high. This is because Bax or Bak can form homodimer channels that still let cytochrome C out from the mitochondria.

Intracellular protein that inhibits apoptosis by blocking Bax, Bak and Bad

Bcl-2

You are completing an UROP project over the summer and look at cells in the microscope. What are the two reasons you can tell this cell is undergoing meiosis and not mitosis. Choose two answers.

Crossing over has occurred & The homologs are paired

Place the following events into the correct order. A. Rb releases E2F B. S Cdk triggers S phase C. E2F activates transcription/translation of G1/S and S cyclin D. G1 Cdk phosphorylates Rb

D, A, C, B

Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base?

DNA in both daughter cells would be radioactive

During the G2 checkpoint in eukaryotic cells, the cell is assessed with regard to

DNA replication.

Jack and Diane are having their first baby together and they find out the baby has Down syndrome at their screening completed at 15 weeks of pregnancy. Which of the following gametes (A-E) could be a normal gamete produced by Jack or Diane?

E

What do these results tell you about the requirements for Bax and Bak in cytochrome c induced apoptosis in these cells?

Either Bak or Bax can function to induce apoptosis - Cells need to have at least one of the two proteins to make the channel for cytochrome c in the mitochondrial membrane. Cytochrome c release must be required for inducing apoptosis in these cells.

Before chromosome segregation in M phase, the chromosomes and the segregation machinery must be appropriately prepared. Indicate whether the following statement is TRUE or FALSE. If FALSE, change a single noun to make the statement TRUE: Cohesins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes.

False: Condensins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes.

Nerve cells lose their ability to undergo mitosis. Instead, they are permanently stuck in:

G0 phase

Cells that are serum starved (lack mitogens)

G1 phase

Most of the diversity in the rates of cell division in the adult body lies in variations in the time that each cell spends in which phase?

G1 phase

Which of the following definitions is incorrect? Mitogens - stimulate cell division (proliferation) Growth factors - stimulate cell division (proliferation) Survival factor - inhibit apoptosis

Growth factors - stimulate cell division (proliferation)

Type of cells that are adult cells that have been reprogrammed into pluripotent stem cells.

Induced pluripotent stem cells

What must happen for a Cdk to be active?

It must bind to its cyclin partner.

Cells with an activated Ras protein

M

Cells with a damaged mitotic spindle

M phase

The copies of chromosome 21 for both Jack and Diane are shown. Describe whether nondisjunction occurred in Jack or Diane and in meiosis I or meiosis II. Choose all possibilities.

Meiosis 2 of the father

In mammals, there are two sex chromosomes, X and Y, which behave like homologous chromosomes during meiosis. Normal males have one X chromosome and one Y chromosome, and normal females have two X chromosomes. Males with an extra Y chromosome (XYY) occasionally are found. Which of the following could give rise to such an XYY male? (only 1 nondisjunction event occurred.)

Nondisjunction in the second meiotic division of spermatogenesis; normal meiosis in the mother.

Use this information to answer questions 2-4 Dyskeratosis congenita is a genetic disorder caused by an inactivating mutation in either the protein or RNA component of the telomerase enzyme. Symptoms of dyskeratosis congenita include premature graying of the hair, anemia, osteoporosis, and pulmonary fibrosis (scarring of the lungs). Cancer on the other hand often has activating mutations in telomerase. You compare telomerase in normal stem cells and non-stem cells to the telomerase from a cancer sample and a person with dyskeratosis congenita. You want to determine the nature of the defect by assaying if the protein (hTERT enzyme) or RNA (hTR RNA) components are altered. Which of the following are correctly matched? Choose all that apply. Northern blot: Measure RNA levels of hTERT Northern blot: Measure RNA levels of hTR Western blot: Measure protein levels of hTERT Western blot: Measure protein levels of hTR

Northern blot: Measure RNA levels of hTR Western blot: Measure protein levels of hTERT

Type of stem cells that can become any tissue in the body.

Pluripotent

Crossing over occurs during

Prophase I

As a stem cell divides, one daughter cell differentiates, the other daughter cell

Remains a stem cell

Meiosis II is similar to mitosis in that:

Sister chromatids separate during anaphase

What happens if MPF (mitosis-promoting factor) is introduced into immature frog oocytes that are arrested in G2?

The cells enter mitosis

What is the defect in the liver cancer cells?

The cells make more hTERT than normal Lane one shows that normal stem cells express both the hTR and hTERT. Lane three shows that normal adult cells lack the hTERT and so would not have telomerase activity. Lane four shows the protein levels of hTERT and RNA levels of hTR in the liver cancer cells. The western blot shows that hTERT is being expressed in the cancer cells when it normally should not be.

Once a cell completes mitosis, molecular division triggers must be turned off. What happens to MPF during mitosis

The cyclin component of MPF is degraded

What problem with replication of linear chromosomes does telomerase address?

The lagging strand stops short of the 3' end during replication, so chromosomes would shorten in each replication cycle without telomerase.

What is the defect in the dyskeratosis congenita family?

Their cells do not make hTERT Lane two shows the protein levels of hTERT and RNA levels of hTR in stem cells from the dyskeratosis congenita patient. The northern blot shows a normal level of hTR RNA in the dyskeratosis congenital patient compared to the healthy stem cells in lane one. The western blot shows decreased levels of hTERT enzyme in the dyskeratosis congenital patient's stem cells compared to the normal stem cells in lane one. The defect in the dyskeratosis congenital patient may arise from a decrease in the amount of hTERT protein.

When cells in G0 are exposed to mitogenic growth factors like EGF, they enter S phase about 20 hours after stimulation. Entry into S phase can be measure by BrdU incorporation into new DNA during DNA synthesis. The time on the graph shows when the control antibody or anti-G1 cyclin antibodies were added after the stimulation with mitogenic growth factors. The anit-G1 cyclin antibodies block the G1 cyclin from binding to the G1-Cdk. If antibodies to the G1 cyclin are injected into cells up to 12 hours after adding growth factors, very few of the injected cells incorporate BrdU. By contrast, anti-G1 cyclin antibodies have little effect on BrdU incorporation when injected more than 14 hours after addition of growth factors. What critical event in G1 do antibodies against the G1 cyclin block? Why do antibodies against the G1 cyclin have no effect after 14 hours?

They block the ability of G1-Cdk to phosphorylate Rb

Which of the following is NOT expected to impair anaphase A? a. An antibody against myosin b. Taxol (microtubule depolymerization inhibitor) c. ATPγS, a nonhydrolyzable ATP analog that binds to and inhibits ATPases d. An antibody against motor proteins that move from the plus end of microtubules to the minus end

a. An antibody against myosin

Sister chromatids separate in

anaphase

A current focus of molecular medicine is to trigger or prevent apoptosis in specific cells. How would treatment of cells with antisense oligonucleotides targeted against the Bcl‐2 gene effect apoptosis (antisense oligonucleotides bind to mRNAs, preventing their translation into protein)?

apoptosis would be stimulated

Consider an animal cell that has eight chromosomes (four pairs of homologous chromosomes) in G1 phase. Consider how many of each of the following features the cell will have at mitotic prophase. Which is NOT correct for mitotic prophase? a. 16 Sister chromatids b. 16 centrosomes c. 16 kinetochores d. 16 centromeres

b. 16 centrosomes

Which answer is NOT correct. The cleavage furrow a. is a puckering of the plasma membrane caused by constriction of a ring of filaments attached to the membrane b. forms by motor proteins that pull on microtubules attached to the cell cortex c. forms perpendicular to the interpolar microtubules d. will not begin to form in the absence of a mitotic spindle.

b. forms by motor proteins that pull on microtubules attached to the cell cortex

Which of the following is NOT an effect of MCdk/cyclin involved in moving a cell into M phase? a. phosphorylation of condensins, initiating chromosome condensation b. phosphorylation of APC, triggering cyclin degradation c. phosphorylation of lamins, initiating breakdown of the nuclear membrane d. phosphorylation of microtubule associated proteins, triggering formation of the mitotic spindle

b. phosphorylation of APC, triggering cyclin degradation

Which of the following cell cycle stages and descriptions is NOT correctly matched. a. S phase ‐ This phase is part of interphase b. M phase ‐ Sister chromatids separate from each other c. G2 phase ‐ Cells that will never divide again are likely to be arrested in this phase d. G1 phase ‐ Chromosomes are present as diffuse, extended chromatin

c. G2 phase ‐ Cells that will never divide again are likely to be arrested in this phase

Apoptosis differs from necrosis in that necrosis

causes cells to swell and burst while apoptotic cells shrink and condense.

During metaphase I of meiosis, what appears to hold homologs together as a bivalent?

chiasmata

Protein released from the mitochondria that promotes formation of the apoptosome.

cytochrome c

Chromosomes decondense in

cytokinesis

Apoptosis program that is triggered by the binding of an extracellular signal protein

extrinsic

Which of the following is NOT a source of genetic diversity? Fertilization Separation of sister chromatids Independent assortment of homologs Crossing over

fertilization

Is the M-Cdk activity in mutant cells less than, greater than or equal to M-Cdk activity in wild-type cells. Inactivating mutations in wee1 and cdc25

greater than Increase M-Cdk activity because wee1 does not add inhibitory phosphate. Do not need cdc25 to remove the phosphate if it is never added.

Is the M-Cdk activity in mutant cells less than, greater than or equal to M-Cdk activity in wild-type cells. Inactivating mutation in wee1

greater than Without wee1, the inhibitory phosphate is not added and M-Cdk activity would be increased.

At the end of meiosis I, cells are

haploid

A cyclin _____

is present in similar concentrations throughout the cell cycle.

Cyclin-dependent kinase (Cdk) is _____.

kinase the triggers cell cycle events.

Is the M-Cdk activity in mutant cells less than, greater than or equal to M-Cdk activity in wild-type cells. Inactivating mutation in CAK

less than Decreased activity because the activating phosphate would not be added to M-Cdk.

Is the M-Cdk activity in mutant cells less than, greater than or equal to M-Cdk activity in wild-type cells. Inactivating mutation in M-cyclin

less than If the M-cyclin is mutant, the M-Cdk cannot be activated.

Is the M-Cdk activity in mutant cells less than, greater than or equal to M-Cdk activity in wild-type cells. Inactivating mutation in cdc25

less than Without cdc25, the inactivating phosphate would not be removed and the M-Cdk would stay inactive

Chromosomes align in the center of the cell in

metaphase

To pass the G2 to M phase checkpoint, the DNA must be properly replicated. If DNA is unreplicated, proteins inhibit the:

phosphatase that activates MCdk

Mitogenic or cell division stimulating signaling pathways

phosphorylate and inactivate Rb

Cells with mutated and inactive Cdc25

proliferating

Microtubules attach to kinetochores in

prometaphase

DNA replication is

semiconservative

A serine to alanine mutation in the phosphorylation site of a lamin protein will

stabilize the nuclear lamina and thereby prevent its disassembly during mitosis.

Which enzyme is used to extend the 3' end of the DNA so chromosomes do not shorten every round of DNA replication?

telomerase

Levels of cyclin‐dependent kinase (Cdk) activity change during the cell cycle partially as a result of

the Cdks binding cyclins to become active.