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13 Feb 2019

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A cell’s progression through the cell cycle requires the activity of several cyclin-dependent kinases (G1-Cdk, G1/S-Cdk, S-Cdk, M-Cdk) How are they activated, how are they inactivated, when do you see their levels rise and what are their targets? How does this activity regulate the cell cycle?

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Sixta Kovacek
Sixta KovacekLv2
15 Feb 2019

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Related questions

Different cyclin-dependent protein kinases (CDKs) regulate the transition of eukaryotic cells from one phase of the cell cycle to another: from G1 to S, S to M, or M to G1. Describe the different mechanisms by which the enzyme activity of the CDKs is regulated over time during the cell cycle and how such mechanisms are altered in different cancers. In some cells such as trophoblasts (cells of the placenta) and megakaryocytes (precursors to blood platelets), cells will undergo repeated rounds of G1 and S phase without G2 or M phases – what is the advantage of this cell cycle for such cells?

Question 1


Which of the following provides the best description of adifference between prokaryotic and eukaryotic cells?.l.14.3>

The cell membrane of prokaryotic cells contains proteins, whilethe cell membrane of eukaryotic cells contains only lipids andcarbohydrates.

The DNA in eukaryotic cells is enclosed in the nucleus, whilethe DNA in prokaryotic cells is not separated from the rest of thecell’s contents.

Eukaryotic cells have cell walls that consist mainly of sugars,while prokaryotic cells do not.

Prokaryotic cells contain chloroplasts and mitochondria, whileeukaryotic cells do not contain specialized organelles.

Question 2

Cellular respiration is a process involving the breakdown ofglucose molecules in order to generate energy for cellularactivities. Given the general reaction below, which of thefollowing best describes the products labeled x andy?

C6H12O6 + O2 → _x___+ __y___ + energy.l.18.8>

The products are CO (carbon monoxide) andH2O2 (hydrogen peroxide), which thendecompose into carbon dioxide (CO2) and water(H2O).

The products are H2O (water) and CO2 (carbondioxide), which then undergo anaerobic respiration.

The products are CO2 (carbon dioxide) andH2O (water), which then become the reactants forphotosynthesis.

The products are O2 (oxygen) and H2O(water), which are then used for growth and to keep the organismhydrated.

Question 3

If a researcher added a quantity of phospholipid molecules to abeaker of water, what would be most likely to happen to theorientation of the phospholipid molecules?.l.14.2>

They would break apart, and the remaining lipid tails wouldfloat up and form a layer on the surface of the water.

Their lipid tails would align within a spherical double layerwith their phosphate heads in the water inside and outside thesphere.

They would form a single-layered sphere with the phosphate headspointing outward and the lipid tails pointing inward.

Their lipid tails would adhere to the glass of the beaker andtheir phosphate heads would float in the water.

Question 4

While looking at a cell under a microscope, a scientist is ableto see a biological molecule. This molecule is a nucleic acid witha ladder structure and contains the nitrogenous bases adenine,thymine, cytosine, and guanine. Based on this information, whatconclusion can the scientist draw regarding this nucleicacid?.l.18.1>

The nucleic acid is RNA and would be found in the cell’snucleus.

The nucleic acid is RNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cell’snucleus.

Question 5


Which of the following is needed for cellular respiration tooccur?.l.18.8>

Sodium

Oxygen

Water

Nitrogen

Question 6


What would be the most likely effect if the side chain of an aminoacid group was unable to vary its structure?.l.18.4>

The organism would not be able to make and store starches.

The organism would not be able to process glucose moleculeseffectively.

The organism would not be able to regulate the creation ofspecialized cells.

The organism would not be able to make more than one type ofprotein.

Question 7


During photosynthesis, unused oxygen atoms form oxygen gas(O2) which is released as a waste product. How did theseoxygen atoms originally enter the plant?.l.18.7>

They were part of the ATP molecules gathered by thechloroplasts.

They were part of the sugar molecules(C6H12O6) the plant brought in forfood.

They were part of the carbon dioxide (CO2) moleculesthat entered through the leaves.

They were part of the water molecules (H2O) pulled inthrough the roots.

Question 8


Which of the following organelles houses DNA in eukaryoticcells?.l.14.3>

Endoplasmic reticulum

Golgi apparatus

Nucleus

Mitochondrium

Question 9


M. J. Schleiden shifted the scientific community's attention tocellular processes with his.l.14.1>

attention to disproving spontaneous generation

publication of ideas on cell generation

theories on the division of cells

claim that cells were the basic unit of all plant matter

Question 10


While the average human is able to hold his or her breath forapproximately one minute, a whale can dive for over 30 minuteswithout returning to the surface. Which of the following correctlydescribes this difference?.l.18.10>

Whales need less energy than humans.

Whales gather energy from their environment better thanhumans.

Whales are able to tolerate the products of cellular respirationbetter than humans.

Whales have a slower cellular respiration metabolism rate thanhumans.

Question 11


What can you conclude about the advantage of the cell membranehaving hydrophobic ends in the phospholipid bilayer?.l.18.3>

The cell requires the ability to digest and eliminateenzymes.

The cell requires a barrier between itself and itsenvironment.

The cell requires a means to store starches and glucose.

The cell requires a structure for eliminating waste.

Question 12


What would be the most likely outcome if the ribosomes of a cellwere damaged and stopped functioning?.l.14.3>

The cell would lose structure and shape.

Cell growth and repair would slow or stop.

Cell waste would no longer be processed.

The cell would produce more mitochondria.


Tasmanian devils (Sarcophilus harrisii), the world’s largest carnivorous marsupials, are facing extinction. Not from hunting, or habitat loss, but from within; a transmissible cancerous derivative of Schwann cells, responsible for devil facial tumour disease (DFTD). First seen in 1996, the disease has spread widely though the population, and is now estimated to be present in 80% of the population. With the grim prognosis of ~100% mortality in 12-18 months, animals that once went through multiple breeding cycles are now lucky to reproduce once before death, resulting in a 70% decline in their population over the past decade. The cancer is transmitted by biting (a common ritual used by these animals to establish and maintain social position), with cancer cells from one animal (the biter) being implanted in the wound inflicted on the other animal. Equally common in both sexes, the disease itself is more prevalent in animals higher on the social scale, presumably reflecting higher levels of social interaction (biting). The cancer is usually first evident as facial tumors, from which metastasis to cervical lymph nodes occurs, often followed by dissemination to other organ systems. Cause of death is by organ failure or by malnutrition in those with facial tumors large enough to prevent feeding. Based on what we have covered regarding the Major Histocompatibility Complex, you would expect the passage of a cancer from one individual to another* to be a rare event (to see just how rare, read the accompanying popular press piece from the New York Times). Here are some things that you already know: *Yes, some viruses can cause cancer (example: Hepatitis B virus and liver cancer), and you can transmit a virus from one individual to another; but that is not the same thing as passing the cancer itself. v Almost all nucleated cells express high levels of class I MHC proteins on their surface. v Class I MHC proteins collect peptides produced from endogenous proteins and display them at the cell surface. v Different Class I MHC proteins bind and display different peptides. v Developing cytotoxic T-cells that recognize self-antigens are eliminated. v Mature cytotoxic T-cells conduct surveillance by scanning the peptides displayed by class I MHC proteins. v Mature cytotoxic T-cells that recognize an antigen on a cell will kill that cell (recall that cytotoxic T cell activity is important in rejection of tissue transplants).

Before the questions:
v Tasmanian Devils are not immunosuppressed.
v Tasmanian Devil cytotoxic T-cells are not broken, they can kill cells.
So neither of those should be part of your answers.
This is an ongoing research topic with large gaps in our knowledge; we really don’t fully understand how this works
and so there is no “correct” answer. Your job is to provide a logical argument, with some research and details to
support your contention. Note that in questions 1 & 2 class I MHC should be part of your answer, but not in
question 3.

Questions:
1.) Acquiring DFTD is basically like receiving a transplant that is not rejected. How can that be? Suggest a possible
way this could occur. Be sure to explain why you think this would work, and then explain how you would test
that model to determine if it were correct. (10 point question).
2.) Canine Transmissible Venereal Tumor (CTVT) is the other know transmissible cancer in mammals. Unlike
DFTD, CTVT does not usually result in death. Do a little research on this disease and summarize how it is
different from DFTD. Provide an explanation as to why one is lethal and the other is generally not. (10 point
question).
3.) Recently, researchers at Washington State University determined that some Tasmanian devils had developed
resistance to this cancer. By scanning partial genome sequences of resistant and susceptible animals (a
complete devil genome has yet to be assembled) they identified a set of genes where mutations appeared to
enhance resistance. Speculate as to the function of the proteins these genes encode, and explain why that
might provide for resistance (note: they were NOT MHC genes).

Mainly need guidance/ideas for #1's 1 and 3!

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