Protein Degradation

 

Information for the Public
Nobel Prize in Chemistry
6 October 2004

 Discovery of Ubiquitin-Mediated Protein Degradation

A human cell contains some hundred thousand different proteins. These have numerous important functions: as accelerators of chemical reactions in the form of enzymes, as signal substances in the form of hormones, as important actors in the immune defense and by being responsible for the cell’s form and structure. This year’s Nobel Laureates in chemistry, Aaron Ciechanover, Avram Hershko and Irwin Rose, have contributed ground-breaking chemical knowledge of how the cell can regulate the presence of a certain protein by marking unwanted proteins with a label consisting of the polypeptide ubiquitin. Proteins so labeled are then broken down – degraded – rapidly in cellular “waste disposers” called proteasomes.

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Through their discovery of this protein-regulating system Aaron Ciechanover, Avram Hershko and Irwin Rose have made it possible to understand at molecular level how the cell controls a number of very important biochemical processes such as the cell cycle, DNA repair, gene transcription and quality control of newly-produced proteins. New knowledge of this form of controlled protein death has also contributed to explaining how the immune defense functions. Defects in the system can lead to various diseases including some types of cancer.

Proteins labeled for destruction

Degradation needs no energy – or does it?

While great attention and much research have been spent on understanding how the cell controls the synthesis of a certain protein – at least five Nobel Prizes have been awarded in this area – the reverse, the degradation of proteins, has long been considered less important. A number of simple protein-degrading enzymes were already known. One example is trypsin, which in the small intestine breaks down proteins in our food to amino acids. Likewise, a type of cell organelle, the lysosome, in which proteins absorbed from outside are broken down, had long been studied. Common to these processes is that they do not require energy in order to function.

Experiments as long ago as the 1950s showed, however, that the breakdown of the cell’s own proteins does require energy. This long puzzled researchers, and it is precisely this paradox that underlies this year’s Nobel Prize in Chemistry: that the breakdown of proteins within the cell requires energy while other protein degradation takes place without added energy. A first step towards an explanation of this energy-dependent protein degradation was taken by Goldberg and his co-workers who in 1977 produced a cell-free extract from immature red blood cells, reticulocytes, which catalyze the breakdown of abnormal proteins in an ATP-dependent manner (ATP = adenosine triphosphate – the cell’s energy currency).

Using such an extract Aaron Ciechanover, Avram Hershko and Irwin Rose, in a series of epoch-making biochemical studies in the late 1970s and early 1980s, succeeded in showing that protein degradation in cells takes place in a series of step-wise reactions that result in the proteins to be destroyed being labeled with the polypeptide ubiquitin. This process enables the cell to break down unwanted proteins with high specificity, and it is this regulation that requires energy. As distinct from reversible protein modifications such as phosphorylation (Nobel Prize in Physiology or Medicine 1992), regulation through polyubiquitination is often irreversible since the target protein is destroyed. Much of the work was done during a series of sabbatical leaves that Avram Hershko and Aaron Ciechanover of the Technion (Israel Institute of Technology) spent with Irwin Rose at the Fox Chase Cancer Center in Philadelphia, USA.

The label is ubiquitin

The molecule that would later prove to be the label that marks out a protein for degradation was isolated as early as 1975. This 76-amino-acid-long polypeptide was isolated from calf sweetbread and was assumed to participate in the maturation of white blood cells. Since the molecule was subsequently found in numerous different tissues and organisms – but not in bacteria – it was given the name ubiquitin (from Latin ubique, “everywhere”) (fig. 1).

Fig 1. Ubiquitin – a common polypeptide that represents the “kiss of death”.

The discovery of ubiquitin-mediated protein degradation

After taking his doctorate, Avram Hershko had studied energy-dependent protein degradation in liver cells, but decided in 1977 to transfer to the reticulocyte extract described above. This extract contained large quantities of hemoglobin, which upset the experiments. In their attempts to remove the hemoglobin using chromatography, Aaron Ciechanover and Avram Hershko discovered that the extract could be divided into two fractions, each inactive on its own. But it turned out that as soon as the two fractions were recombined, the ATP-dependent protein degradation restarted. In 1978 the researchers reported that the active component of one fraction was a heat-stable polypeptide with a molecular weight of only 9000 which they termed APF-1 (active principle in fraction 1). This protein later proved to be ubiquitin.

The decisive breakthrough in the research was reported in two works that Ciechanover, Hershko and Rose published in 1980. Until that time the function of APF-1 was entirely unknown. In the first work it was shown that APF-1 was bound covalently, i.e. with a very stable chemical bond, to various proteins in the extract.

In the second work it was further shown that many APF-1 molecules could be bound to the same target protein; the latter phenomenon was termed polyubiquitination. We now know that this polyubiquitination of substrate proteins is the triggering signal that leads to degradation of the protein in the proteasome. It is this reaction that constitutes the actual labeling, the “kiss of death” if you will.

At a stroke, these entirely unanticipated discoveries changed the conditions for future work: it now became possible to concentrate on identifying the enzyme system that binds ubiquitin to its target proteins. Since ubiquitin occurs so generally in various tissues and organisms, it was quickly realized that ubiquitin-mediated protein degradation must be of general significance for the cell. In addition, the researchers guessed that the energy requirement in the form of ATP enabled the cell to control the specificity of the process.

The field was now open and between 1981 and 1983 Ciechanover, Hershko, Rose and their post docs and students developed “the multistep ubiquitin-tagging hypothesis” based on three newly-discovered enzyme activities they termed E1, E2 and E3 (fig. 2). We now know that a typical mammalian cell contains one or a few different E1 enzymes, some tens of E2 enzymes and several hundred different E3 enzymes. It is the specificity of the E3 enzyme that determines which proteins in the cell are to be marked for destruction in the proteasomes.

Fig 2. Ubiquitin-mediated protein degradation

 

  1. The E1 enzyme activates the ubiquitin molecule. This reaction requires energy in the form of ATP.
  2. The ubiquitin molecule is transferred to a different enzyme, E2.
  3. The E3 enzyme can recognize the protein target which is to be destroyed. The E2-ubiquitin complex binds so near to the protein target that the actual ubiquitin label can be transferred from E2 to the target.
  4. The E3 enzyme now releases the ubiquitin-labeled protein.
  5. This last step is repeated until the protein has a short chain of ubiquitin molecules attached to itself.
  6. This ubiquitin chain is recognized in the opening of the proteasome. The ubiquitin label is disconnected and the protein is admitted and chopped into small pieces.

 

All the studies up to this point had been done in cell-free systems. To be able to study the physiological function of ubiquitin-mediated protein degradation as well, Avram Hershko and his co-workers developed an immunochemical method. By using antibodies to ubiquitin, ubiquitin-protein-conjugate could be isolated from cells where the cell proteins had been pulse-labeled with a radioactive amino acid not present in ubiquitin. The results showed that cells really break down faulty proteins using the ubiquitin system, and we now know that up to 30% of the newly-synthesized proteins in a cell are broken down via the proteasomes since they do not pass the cell’s rigorous quality control.

The proteasome – the cell’s waste disposer

What is a proteasome? A human cell contains about 30,000 proteasomes: these barrel-formed structures can break down practically all proteins to 7-9-amino-acid-long peptides. The active surface of the proteasome is within the barrel where it is shielded from the rest of the cell. The only way in to the active surface is via the “lock”, which recognizes polyubiquitinated proteins, denatures them with ATP energy and admits them to the barrel for disassembly once the ubiquitin label has been removed. The peptides formed are released from the other end of the proteasome. Thus the proteasome itself cannot choose proteins; it is chiefly the E3 enzyme that does this by ubiquitin-labeling the right protein for breakdown (fig. 3).

Fig 3. The cell’s waste disposer, the proteasome. The black spots indicate active, protein-degrading surfaces.

 

More recent research

While the biochemical mechanisms underlying ubiquitin-labeled protein degradation were laid bare around 1983 its physiological significance had not yet been fully understood. That it is of importance in destroying defective intracellular proteins was known but, to proceed, a mutated cell was needed in the ubiquitin system. By studying in detail how the mutated cell differs from a normal cell under various growth conditions, it was hoped to gain a better idea of what reactions in the cell depend on the ubiquitin system.

A mutated mouse cell had been isolated in 1980 by a research group in Tokyo. Their mouse-cell mutant contained a protein that, because of the mutation, was sensitive to temperature. At lower temperatures the protein functioned as it should, but not at higher. Cells cultured at the higher temperature stopped growing. In addition, they showed defective DNA synthesis and other erroneous functions at the higher temperature. Researchers in Boston quickly showed that the heat-sensitive protein in the mutant mouse cell was the ubiquitin-activating enzyme E1. Obviously, ubiquitin activation was necessary for the cell to function and reproduce itself at all. Controlled protein breakdown was not only important for degrading incorrect proteins in the cell but it probably also took part in control of the cell cycle, DNA replication and chromosome structure.

Since the late 1980s a number of physiologically important substrates for ubiquitin-mediated protein breakdown have been identified. Only a few of the most important will be mentioned here.

Prevention of self-pollination in plants

Most plants are bisexual, hermaphroditic. Self-pollination leads to a gradual decline in genetic diversity which in the long run can cause the whole species to die out. To prevent this, plants use ubiquitin-mediated degradation to reject “own” pollen. The exact mechanism has not yet been clarified but the E3 enzyme has been encountered and when proteasome inhibitors have been introduced, the rejection has been impaired.

Regulation of the cell cycle

When a cell is to make a copy of itself, many chemical reactions are involved. In a human being, six thousand million base pairs must be duplicated in DNA. These are gathered in 23 chromosome pairs that must be copied. Ordinary cell division, mitosis, and the formation of sex cells, meiosis, have many points of contact with the subjects of this year’s Nobel Prize. The E3 enzyme responsible, a protein complex termed the “anaphase-promoting complex” (APC) checks that the cell goes out of mitosis. This enzyme complex has also proved to play an important role in the separation of the chromosomes during mitosis and meiosis. A different protein complex acts like a rope around the chromosome pair, holding it together. At a given signal, the APC labels an inhibitor of a certain protein-degrading enzyme, whereupon the inhibitor is carried to the proteasome and destroyed. The enzyme is released, is activated and cuts the rope around the chromosome pair. Once the rope is gone, the chromosome pair can be separated. Incorrect chromosome division during meiosis is the commonest cause of spontaneous miscarriage during pregnancy, and an extra chromosome 21 in humans leads to Down’s syndrome. Most malignant tumors have cells with changed numbers of chromosomes as a result of incorrect chromosome division during mitosis.

 

 DNA repair, cancer and programmed cell death

Protein p53 has been dubbed “the guardian of the genome” and it is a tumor-suppressor gene. This means that as long as a cell can produce p53 the development of cancer is hampered. Sure enough, the protein is mutated in at least 50% of all human cancer. The amount of protein p53 in a normal cell is low in consequence of continual production and breakdown. The breakdown is regulated through ubiquitination and the E3 enzyme responsible forms a complex with protein p53. Following DNA injury, protein p53 is phosphorylated and can no longer bind to its E3 enzyme. The breakdown stops and the quantity of p53 in the cell rises rapidly. Protein p53 acts as a transcription factor, i.e. a protein that controls the expression of a certain gene. Protein p53 binds to and controls genes that regulate DNA repair and programmed cell death. Raised levels of protein p53 lead first to interruption of the cell cycle to allow time for repair of DNA damage. If the damage is too extensive the cell triggers programmed cell death and “commits suicide”.

Infection with human papilloma virus correlates strongly to the occurrence of cervical cancer. The virus avoids the protein p53 control function through one of its proteins activating and changing the recognition pattern of a certain cellular E3 enzyme, E6-AP, which is tricked into ubiquitinating the protein p53, which is totally destroyed. In consequence of this the infected cell can no longer repair DNA damage in a normal manner or trigger programmed cell death. The DNA mutations increase in number and this can ultimately lead to the development of cancer.

Immune and inflammatory reactions

A certain transcription factor regulates many of the genes in the cell that are important for immune defense and inflammatory reactions. This protein, the transcription factor, occurs bound to an inhibitor protein in the cytoplasm of the cell, and the bound form of the transcription factor lacks activity. When cells are exposed to bacteria or various signal substances, the inhibitor protein is phosphorylated, and this results in its being ubiquitinated and broken down in the proteasome. The released transcription factor is transported to the cell nucleus where it binds to, and activates the expression of, specific genes.

The ubiquitin-proteasome system also produces the peptides that are presented by the immune defense on the surface of a virus-infected cell by breaking down virus proteins to suitable sizes. T lymphocytes recognize these peptides and attack the cell as an important part of our defense against virus infections.

Cystic fibrosis (CF)

The hereditary disease cystic fibrosis, CF, is caused by a non-functioning plasma membrane chloride channel called CFTR, the “cystic fibrosis transmembrane conductance regulator”. Most CF patients have one and the same genetic damage, loss of the amino acid phenylalanine in the CFTR protein. The mutation causes faulty folding of the protein and this in turn leads to the protein being retained in the cell’s control system for protein quality. This system ensures that the incorrectly folded protein is destroyed through ubiquitin-mediated protein breakdown instead of being transported out to the cell wall. A cell with no functioning chloride channel can no longer transport chloride ions through its wall. This affects secretion in, among other organs, the lungs and leads to the accretion of thick phlegm in the lungs which impairs their function, greatly increasing the risk of infection.

The ubiquitin system has become an interesting area of research for medicines against various diseases. Such preparations can be aimed at components of the ubiquitin-mediated breakdown system to prevent the degradation of specific proteins. They can also be designed to cause the system to destroy unwanted proteins. A medicine already being tested clinically is the proteasome inhibitor Velcade (PS341) which is used against multiple myeloma, a cancer disease that affects the body’s antigen-producing cells.

This year’s Laureates have explained the molecular background to a protein regulation system of great importance for all higher cells. New cell functions controlled by ubiquitin-mediated protein degradation are being discovered all the time and this research is being conducted in numerous laboratories all over the world.

The Laureates
Aaron Ciechanover

Technion (Israel Institute of
Technology)
Rappaport Institute
1 Efron Street
P.O. Box 9697
Haifa 31096
Israel

 

Israeli citizen. Born 1947 (57 years) in Haifa, Israel. Doctor’s degree in medicine in 1975 at Hebrew University of Jerusalem, and in biology in 1982 at the Technion (Israel Institute of Technology), Haifa. Distinguished Professor at the Center for Cancer and Vascular Biology, the Rappaport Faculty of Medicine and Research Institute at the Technion, Haifa, Israel.

Aaron Ciechanover

 

Avram Hershko

Technion (Israel Institute of Technology)
Rappaport Institute
1 Efron Street
P.O. Box 9697
Haifa 31096
Israel

 

Israeli citizen. Born 1937 (67 years) in Karcag, Hungary. Doctor’s degree in medicine in 1969 at the Hadassah and the Hebrew University Medical School, Jerusalem. Distinguished Professor at the Rappaport Family Institute for Research in Medical Sciences at the Technion, Haifa, Israel.

Avram Hershko
Irwin Rose

Dept. of Physiology and Biophysics
College of Medicine
University of California, Irvine
Irvine, CA 92697
USA

 

American citizen. Born 1926 (78 years) in New York, USA. Doctor’s degree in in 1952 at the University of Chicago, USA. Specialist at the Department of Physiology and Biophysics, College of Medicine, University of California, Irvine, USA.

Irwin Rose

Illustrations: Typoform

Source: http://nobelprize.org/nobel_prizes/chemistry/laureates/2004/press.html

 

Potato Osmosis Bi Lab

 

Potato Osmosis

 

Introduction:

A shipwrecked sailor is stranded on a small desert island with no fresh water to drink. She knows she could last without food for up to a month, but if she didn’t have water to drink she would be dead within a week. Hoping to postpone the inevitable, her thirst drove her to drink the salty seawater. She was dead in two days. Why do you think drinking seawater killed the sailor faster than not drinking any water at all? Today we explore the cause of the sailor’s death. We’ll prepare solutions of salt water to represent the sea, and we’ll cut up slices of potato to represent the sailor. Potatoes are made of cells, as is the sailor!

Objective:

The concentration of solute in a solution will affect the movement of water across potato cell membranes.

Materials:

potato, corer, 3 plastic cups, marker, salt, sugar, distilled water, paper, pencil, electronic balance, clock with second hand or timer, metric ruler, small ziplock plastic bag, foil or plastic wrap

Procedure:

Day 1

  1. Use a knife to square off the ends of your potato. Your potato’s cells will act like the sailor’s cells.
  2. Stand your potato on end & use your cork borers to bore 3 vertical holes.

  1. Remove the potato cylinders from the cork borer & measure their length in centimeters.
  2. Cut the 3 potato cylinders to the same length (about 4 -5 centimeters long).
  3. Record the length & turgidity of the potato cylinders in your data table (day 1).
  4. Place the 3 potato cylinders in a small ziplock bag to prevent them from dehydrating before they’re used.
  5. Take 3 plastic cups and label them with the solution that will be placed in each one — sugar, salt, distilled water.
  6. Prepare a saturated solution of salt by mixing as much salt as you can with water.
  7. Repeat this step by making a saturated sugar solution.
  8. Now fill each cup 2/3’s full of the correct solution —- sugar water, salt water, or distilled water.
  9. Mass each of the potato cylinders & record this mass in grams on your data table.
  10. Place one of your potato cylinders into each cup and cover the top of the cup with foil or plastic.
  11. Leave the potato cylinders in the solution for 24 hours.

Day 2

  1. Carefully remove the potato cylinder from the distilled water solution & pat it dry on a paper towel.
  2. Measure the length of the potato cylinder & record this length & the appearance of the cylinder on your data table. (day 2)
  3. Measure & record the mass of this cylinder.
  4. Repeat steps 13-15 for the potato cylinders in the salt solution & the sugar solution.
  5. Clean up your equipment & area and return materials to their proper place.

Data:

 

Results of Osmosis in Potato Cells
Solution Initial length
cm
(day1)
Final length
cm
(day2)
Change in length
cm
Initial Mass
g
(day1)
Final Mass
g
(day2)
Change in mass
g
Initial Turgidity
(flaccid or crisp)
Final Turgidity
(flaccid or crisp)
Tonicity of Solution
(iso-, hypo-, or hpertonic)
Distilled water
Salt Solution
Sugar Solution

 

Results & Conclusions:

1. Did any of the potato cylinders change in their turgidity (flexibility), and if so, which ones changed?

 

2. Explain why the flexibility of the potato slices changed.

 

3. Define isotonic, hypotonic, & hypertonic solutions.

 

4. If potato slices changed in length or turgidity, what process was responsible for this?

 

5. Make a sketch of your potato cylinder in the distilled water and use arrows to show the direction of water movement across the potato cell membranes.

 

 

6. What type of solutions were the salt & sugar solutions. Explain how you know this.

 

7. Which solution served as the control for this experiment & why?

 

8. In which solutions was their a greater solute concentration outside of the cells?

 

9. In which direction did water move through these cell membranes?

 

10. In what type of solution do plant cells do best & why?

 

11. Using the information you’ve discovered from this experiment, explain why the sailor died that drank saltwater.

 


BACK

Preap Cell Study Guide

 

Cell Structure & Function  Review   

 

1. The first Person to describe microscopic organisms and living cells was
________________________________.

2. The maximum size to which a cell may grow is limited mainly by the cell’s ___________________________  ____________________________.

3. Short, hair-like organelles that can move and may cover a unicellular organism or line the respiratory tract are called ______________________________________.

4. Some Ribosomes are free in the cytoplasm, while others line the membrane of the
_________________  __________________  __________________________.

5. Everything between the cell membrane and the nucleus, is the cell’s
____________________________________.

6. All cells, from all organisms, are surrounded by a _______________   _____________________.

7. Membranes are _______________________ and have the consistency of vegetable oil.

8. The organelle that stores DNA and synthesizes RNA _________________________.

9. The organelle that processes and packages substances produced by the cell ______________________  _________________________.

10. The ____________________________ is the control center of the cell.

11. The DNA in the form of a long strand is called ______________________________.

12. Cytoplasm consists of two main components:  ____________________________ and
______________________________.

13. The cell membrane functions like a ______________________, controlling what
__________________ and _______________________ the cell.

14. A lipid is a simple form of ________________________________.

15. There are many kinds of ______________________ in cell membranes; they help to move material into and out of the cell.

16. Scientist call the modern view of the cell membrane structure the
______________________________ ____________________ _________________.

17. The nucleus is surrounded by a double layer membrane called the
__________________________  _________________________________.

18. During cell division, _________________________ strands coil and condense into thick structures called _____________________________________.

19. The nucleoli make ___________________________. Which in turn build proteins.

20. Membranes are made mostly of ___________________  and  ______________________.

21. The _________________ is the smallest unit that can carry out all of the processes of life.  The basic unit of life.

22. The maximum size to which a cell may grow is limited mainly by the cell’s ___________________  ____________________.

23. The discovery of cells is linked most directly the development of the __________________________.

24. Organisms whose cells never contain a membrane bound nucleus are called _____________________________________.

24. Suspended in the cell’s cytosol are tiny ___________________________________.

25. Cell membranes consist of two phospholipid layers called a ___________________.

26. The chromosomes in the nucleus contain coded _____________________ that control all cellular activity.

27. When a cell prepares to reproduce the _______________________ disappears.

28. Cytosol is a jelylike mixture that consists mostly of _____________________.

29. The nucleus is one ______________________________.

30. In Eukaryotic cells, most organelles are surrounded by a _____________________.

31. Organisms whose cells always or usually contain a nucleus or nuclei are called
____________________________________.

32.  ________________________ are structures that carry out specific functions in the cell.

33. Most cells have a single ______________________; some cells have more than one.

34. Unicellular organisms such as bacteria and their relatives are ___________________________.

35. The Fluid Mosaic Model presents the modern view of a
__________________  ___________________________.

36. The “Blueprints” in a Cell that controls all its activity are the ___________________.

37. Where are poisons and waste detoxified in a cell? _________________________ _________________________________.

38. A cell synthesizes protein by using organelles called _______________________________.

39. The Mitochondria of a cell contain an inner membrane called _____________________________.

40. What are the membrane-bound sacs that package and secrete cell products?
___________________________ ___________________________.

41. Unlike animal cells, plant cells have ______________  ________________.

42. A Chloroplast can convert _________________, __________________________, and ____________________________ into ________________________.

43. What are Flagella? ___________________________________________________.

44. In animal cells, the Cytoskeleton maintains three-dimensional structure and helps the cell ___________________________.

45. The organelle that digest molecules, old organelles, and foreign substances in the cell   _______________________________________.

46. A pigment that absorbs energy in sunlight ________________________________.

47. The organelle that prepares proteins for export and synthesizes steroids is  ________________________  ________________________.

48. Ribosomes differ from most organelles because they have no ___________________________.

49. What type of cells would you expect to find large numbers of mitochondria?  _______________________  _________________.

50. The “Powerhouse” of the cell _______________________________.

51. Short, hairlike organelles that can move and may cover a unicellular organism or line the respiratory tract are called _______________________________.

52. The first cells on Earth were likely _______________________ that did __________ make their own _________________.

53. Microfilaments and microtubules function in cell _______________________ and ____________________________.

54. What is the correct order of structures in living things, from simplest to the most complex? ______________________, __________________________, ______________________________, ______________________________.

55. The is the organelle that transfers energy in ATP _______________________________.

56. What word means “Water Fearing”? ____________________________.

57.  What word means “Water Loving”? _____________________________.

58. What is cell specialization? Give an example.
59. Distinguish between the structure of rough ER and that of smooth ER.

60. Explain how ribosomes, endoplasmic reticulum, Golgi apparatus function together in protein synthesis.

61. Explain the difference between a tissue and an organ.

62.  Why is the cell membrane said to be selectively permeable?

63.  If a cell has a high energy requirement, would you expect it to have many or few mitochondria? Explain your answer.

64. Describe TWO differences between prokaryotic cells and eukaryotic cells.

65. How can you determine whether a unicellular organism is a prokaryote or a eukaryote?

66. Plant cells have cell walls, but animal cells do not. Why do you think that is so?

67. What are the THREE Parts of the Cell Theory?

68. Describe three differences between plant and animal cells.

69. Name the TWO different kinds of animal cells, and describe how their shape is related to their function.

70. What is the difference between chromatin and chromosomes?

71. What are the major roles of the nucleus, and what parts of the nucleus carry out these roles?

72. What is a colonial organism, and what does it have in common with multicellular organisms?

 

Preap Cellular Respiration Study Guide

 

Cellular Respiration Review  

 

1. Most eukaryotic cells produce only about ___________  ATP Molecules per Glucose Molecule.

2. What is the process by which glucose is converted to pyruvic acid? ________________________________________

3. At the beginning of aerobic respiration, pyruvic acid bonds to a molecule called ______________________________________ to form Acetyl CoA.

4. The breakdown of pyruvic acid in the presence of oxygen is called ______________________________  _______________________.

5. With every completion of the Krebs Cycle, how many ATP Molecules are made? ________________

6. What is the waste product of the Krebs Cycle? _____________________________________________.

7. The conversion of pyruvic acid to carbon dioxide and ethanol is called ___________________________________   _____________________________________________.

8. The release of energy from food molecules in the absence of oxygen is ______________________________________     _________________________________________________________.

9. What is the byproduct of the electron transport Chain?_______________________________________________.

10. How efficient is Anaerobic Respiration? __________%  Aerobic Respiration? ____________%

11. What is the first pathway of cellular respiration called? ________________________________________________

12.What is the location of Glycolysis? _______________________________________________________

13. What is the scientific unit of Energy? ________________________________________________

14. What do you call cellular respiration in the presence of oxygen? _______________________________________  _________________________________________________________.

15. Yeast produces ______________________________ and _______________________________ in the process known as ____________________________________  ___________________________________________.

16. In cellular respiration, glycolysis proceeds the _______________________________  ___________________________.

17. In cellular respiration, more energy is transferred in the ___________________________  ________________________  _________________________________ than in any other step.

18. Glucose molecules are converted into _______________________________  _______________________ molecules in the process of glycolysis.

19. What is the location of the electron transport chain in prokaryotes? ________________  _______________________.

20. The processes of glycolysis and the anaerobic pathways is called ___________________________________.

21. What is the product of acetyl CoA and oxaloacetic acid? _________________  ___________________

22. What molecule is the electron acceptor of glycolysis? _________________________________________

23. The breakdown of organic compounds to produce ATP is known as ____________________________________  ________________________-_______________________________.

24. Glycolysis begins with glucose and produces ______________________________  _________________________.

25. An important molecule generated by both lactic acid and alcoholic fermentation is ______________________________.

26.  In the first step of aerobic respiration, pyruvic acid from glycolysis produces CO2, NADH, H+, and _________________________________  _____________________________________.

27. The electron transport chain is driven by two products of the Krebs Cycle – ______________________  and  ___________________________.

28. What happens to electrons as they are transported along the electron transport chain? _________________________________________________________________

29. The energy efficiency of aerobic respiration (including glycolysis) is approximately ______________  __________________________________________________.

30. Where in the mitochondria do the reactions of the Krebs cycle occur? _____________________________   ___________________________________________________________

31. Where in the mitochondria is the electron transport chain located? _____________________________          __________________________________________________

32. In alcoholic fermentation, ethyl alcohol is produced from _______________________________  ______________________________________.

33.  ____________________________________, and _______________________________ supply electrons and protons to the electron transport chain.

34. Cellular respiration takes place in Two Stages: _______________________________________, then ________________________________________  ________________________________.

35. Water is an end product in the ________________________________________________________________
___________________________________________________________________.

36. In cellular respiration, a two-carbon molecule combines with a four-carbon molecule to form citric acid as part of the _____________________________________________________________________________________.

37. When glycolysis occurs, a molecule of glucose is ___________________________________________.

38. The name of the process that takes place when organic compounds are broken down in the absence of oxygen is _____________________________________________ or _______________________________________.

39. Energetic electrons that provide the energy for the production of most of a cell’s ATP are carried to the electron transport chain by _______________________________ and __________________________________________.

40. _______________________________________ is a biochemical pathway of cellular respiration that is anaerobic.

41. Glucose is split into smaller molecules during the biochemical pathway called __________________________________.

42. In the absence of oxygen, instead of oxidative respiration following glycolysis, glycolysis is followed by ______________________________________________________.

43. During fermentation, either ethyl alcohol and carbon dioxide or _______________________________________ is formed.

DIRECTIONS: Answer the questions below as completely and as thoroughly as possible. Answer the question in essay form (not outline form), using complete sentences. You may use diagrams to supplement your answers, but a diagram alone without appropriate discussion is inadequate.

1. How does aerobic respiration ultimately depend on photosynthesis?

2. Explain the role of oxaloacetic acid with respect to the cyclical nature of the Krebs cycle.

3. Glycolysis produces only 3.5% of the energy that would be produced if an equal quantity of glucose were completely oxidized.  What has happened to the remaining energy in the glucose?

4. Why do most cells produce fewer than 38 ATP molecules for every glucose molecule that is oxidized through aerobic respiration?

5. What happens to electrons that accumulate at the end of the electron transport chain?

6. What role does chemiosmosis play in aerobic respiration?

7. What condition must exist in a cell for the cell to engage in fermentation?

8. How is the synthesis of ATP in the electron transport chain of mitochondria similar to the synthesis of ATP in chloroplasts?

9. The fourth step of glycolysis yields four ATP molecules, but the net yield is only two ATP molecules.  Explain this discrepancy.

10. Under what conditions would cells in your body undergo lactic-acid fermentation?

11. What role does oxygen play in aerobic respiration? What molecule does oxygen become a part of as a result of aerobic respiration?

12. Where in the mitochondrion do protons accumulate, and what is the source of the protons?

pH in Living Systems

 

 

pH and Living Systems

 

Introduction:

Scientists use something called the pH scale to measure how acidic or basic a liquid is. The scale goes from 0 to 14. Distilled water is neutral and has a pH of 7. Acids are found between 0 and 7. Bases are from 7 to 14. Most of the liquids you find every day have a pH near 7. They are either a little below or a little above that mark. When you start looking at the pH of chemicals, the numbers go to the extremes. Substances with the highest pH (strong bases) and the lowest pH (strong acids) are very dangerous chemicals. Molecules that make up or are produced by living organisms usually only function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). Many biological solutions, such as blood, have a pH near neutral.

The biological molecule used in this lab is a protein found in milk. Proteins are used to build cells and do most of the cell’s work. They also act as enzymes. For proteins to work, they must maintain their globular shape. Changing the shape of a protein denatures and the protein will no longer work.

Materials:

Small squares of wide-range pH paper, pH color chart, paper towels, 4 dropper bottles, ammonia, lemon juice, skim milk, distilled water, forceps, 50 ml beakers, small squares of narrow-range pH paper, 2 stirring rods

Procedure (part A): Testing the pH of Substances

  1. Line up 4 squares of wide-range pH paper about 1 cm apart on a paper towel.
  2. Put one drop of distilled water on the pH square.
  3. Compare the color of the pH paper to the color chart and record the pH in data table 1.
  4. Repeat this procedure for the ammonia, lemon juice, and skim milk.

Questions (Part A): Determining the pH of Solutions

  1. Which substance was the most acidic?
  2. Which substance was the most basic?
  3. Did any of the substances have a pH close to neutral? Name them.

Procedure (part B): Showing the Effect of pH on a Biological Molecule (Milk Proteins)

  1. Place 100 drops of skim milk in a 50 ml beaker.
  2. Pick up a piece of narrow-range pH paper with forceps.
  3. Touch the pH paper to the milk and remove it.
  4. Compare the color of the pH paper to the pH color chart.
  5. Record the initial pH in data table 2.
  6. Add a drop of lemon juice to the milk in the cup & stir with a stirring rod. Keep track of how many drops you add to the milk!
  7. Measure and record the pH of the solution with the narrow-range pH paper.
  8. Repeat step 7 until you notice an obvious change in the appearance of the milk. record this final pH and appearance of the milk in your data table.
  9. Repeat steps 1-8 using a clean 50 ml beaker and fresh milk, and substitute ammonia for the lemon juice.
  10. Add drops of ammonia to the milk until the change in pH of the milk is equal to the change in pH you measured in step 8. Be sure to keep track of the number of drops added. HINT: If the pH changed by 2 units with the lemon juice, then add ammonia until you also get 2 pH units of change!

Data:

Table 1

 

Substance Tested pH Acid Base Neutral

 

Table 2

Substance Tested Substance used to Produce Change Starting pH of Milk Final pH of Milk Original Appearance of Milk Final Appearance of Milk Total Number of drops added to Produce the change
100 drops Skim Milk Lemon Juice
100 drops Skim Milk Ammonia

Questions:

1. Which substance tested from table 1 was the most acidic?

2. Which substance was most basic?

3. Did any substance from table 1 have a neutral, or near neutral pH? If so, which substance was neutral?

4. Why did you use narrow-range pH paper to measure the milk’s change in pH?

 

5. Describe the change in appearance of the milk as more lemon juice was added. Explain why this change occurred.

 

 

6. How much did the pH of milk change when lemon juice was added?

7. Why do you think lemon juice “curdled”  (precipitated out the proteins) from the milk?

 

8. Did you get the same change when ammonia was used? Why or why not?