Final

Question 1

Lysosomes contain pump proteins that when activated transport protons against proton concentration gradients. Alpha-Galactosidase A is a soluble enzyme found in the interior of the lysosome. The enzyme catalyzes the hydrolysis of melibiose (which is a disaccharide) into galactose and glucose. Where does the synthesis of the two proteins occur? Justify your answer.

Answer 1

Both proteins are synthesized in the RER. The RER is responsible for synthesizing proteins used outside the cell, transmembrane proteins, and also the: RER, golgi, SER and lysosomes. The pump protein (transmembrane protein) and the enzyme that breaks down sugar (soluble protein) are made in the RER.

Question 2

Integrins mediate the adhesion of cells to their substratum (or to other cells) and the transmission of signals between the external environment and the cell interior. Signaling can occur in an inside-out direction, or in the opposite direction.

a. What is the difference between inside-out and outside-in signaling?
b. what is the role of integrin in each process?
c. Example of either signaling.

Answer 2

a. In inside-out, the signal is initiatied in cytoplasmic region causing activation of the integrin protein that relays the message to extracellular matrix.
For outside-in, the binding of the extracellular domain of integrin to ligand can induce a conformational change at the opposite end. Changes here can alter the way the integrin interacts with many different cytoplasmic proteins, modifying activity.

b. integrin is the mediator in these processes
c. Outside-in signaling: can induce conformational change in talin on the inside of the membrane, initiating a cascade of events leading to the polymerization of actin filaments of the cytoskeleton.

Question 3

The drawing below shows a transmembrane protein. The protein has a group of aspartate and glutamate residues at the C terminal end of the alpha helix, and a group of lysine and arginine in the N terminal end of the alpha helix.

a. Draw the correct orintation.
b. Explain how the orientation of the protein is determined. In your answer, be sure to explain, how does the chemical characteristics of the translocon help to orient the protein and how the protein is finally translocated into the rough ER membrane.

Answer 3

The correct orientation of N to C needs to be labeled with the protein going through the translocon. The explanation in part b needs to include something about opposite charges attracting or how the charges of the translocon orients the protein. A complete answer will have something about how the hydrophobic residues are translocated either in writing or shown via the drawing.

Question 4

Insulin is a peptide hormone that contains 51 amino acids. The islets of Langerhans in the pancreas are responsible for the production and secretion of insulin. Describe the steps that occur between the time a ribosome attaches to an insulin mRNA and the time the protein leaves the RER

Answer 4

Insulin mRNA binds to the ribosome and then a signal sequence of hydrophobic residues is translated and attracts the Signal Recognition Particle (SRP). The SRP brings the ribosome/mRNA complex to the RER membrane and docks the complex at a translocon. Inside the RER, chaperone proteins help fold the protein and the protein is packaged in vesicles to be sent off to the golgi.

Question 5

• Distinguish the structural differences and similarities between a hemidesmosome, desmosome, and an adherens junction. Also explain the function of each of these structures

Answer 5

ALL: involved in cell adhesion, structural integrity and signaling/communication

Both: contain a dense plaque on inner surface of PM with keratin filaments to connect to others. Composed of intermediate filaments

hemidesmosomes: site of interaction between a cell and the underlying basal lamina. Resemble half a desmosome.
desmosomes: site of interaction between two cells. used for mechanical stress

Adherens junction: composed of actin filaments. involved in cell to cell interactions

Question 6

Describe how membrane asymmetry is maintained as the membrane moves from the ER to the plasma membrane. Include in your answer how proteins orient correctly in the correct orientation as it moves through the membrane and how the lipids and the proteins of the membranes are established and maintained.

Answer 6

For proteins membrane asymmetry is initiated in the ER (from where all membranes originate.) The correct position of the integrate proteins is set up when the protein is co-translated-translocated in the translocon and this orientation is maintained through the movement of the protein in the endomembrane system because the chemical characteristics of the lumen of the rough RE, vesicles lumen of the Golgi and the extracellular space is the same.

Question 7

It was noted that two different autoimmune diseases can cause severe blistering of the skin. However, one of the two diseases produce antibodies against a component of hemidesmosomes, while the other produces antibodies against a component of desmosomes. Why do you think these two conditions have such similar symptoms?

Answer 7

Both types of adhesive junctions are important in maintaining the integrity of an epithelium. Hemidesmosomes are important in holding the layer to the underlying substrate and desmosomes in holding the cells to one another. Autoimmune disease in in the desmosome or the hemidesmosomes will compromise the integrity of these two structures which lead to the detachment of the cell to the substratum or cell where the hemidesmosome and desmosome must be attached causing epithelial tissue to easily blistering of the tissue with moderated mechanical stress.

Question 8

List the structural characteristics and at least one function of collagen, proteoglycan, fibronectin, and laminin

Answer 8

Collagen: three a helix wound around each other to form rod-like triple helix bound by H+ bonds. These align in rows that stagger to form fibers held by covalent bonds
function: mechanical properties of the fiber, withstanding mechanical stress

Proteoglycans: core protein w/ lots of glycosaminoglycan chains consisting of repeating disaccharide. Then assemble by linking proteins to hyaluronic acid
function: negative charges to attract H2O to regulate extracellular matrix viscocity

Fibronectin: two similar polypeptides joined by disulfide bonds in c-terminal. Each is composed of a linear series of distinct fibronectin domains containing RGB sites to bind collagen or peptidoglycan
function: cell migration and embryonic development

Laminin: consist of three diff polypeptide chains linked via disulfide bonds and organized into cross-like structure
function: involved in migration

Question 9

The pH of the fluid inside the intestinal lumen is extremely low and if it leaks through the intracellular space of the tissue it could destroy the cement created by the adherent junction, which establishes the apical and basal domain of the cells. What structure do epithelial cells use to prevent the destruction of the adherent desmosomes?

Answer 9

desmosomes tight junction must be found in the apical domain of the cell, the barrier of these structure will avoid the leakage in between these cells avoiding contact with the adjacent or desmosome junctions

Question 10

Scurvy is a disease that results from a deficiency of vitamin C (ascorbic acid) and is characterized by inflamed gums and tooth loss, poor wound healing, brittle bones, and the weakening of the lining of blood vessels, causing internal bleeding. It has been found that patients with scurvy have problems in their collagen fiber of the ECM. Describe the structural organization of the collagen fibers and explain how the absence of ascorbic acid affects collagen at the molecular level.

Answer 10

A collagen molecule is composed of three alpha helixes that are wound around each other to form a rod-like molecule triple helix. The alpha helix of the collagen molecule contains many proline and leucine residues that are hydroxylated. The hydroxylation of these residues is used to hold together the three alpha helixes with hydrogen bonds which stabilized the triple helix. Vitamin C (ascorbic acid) is a coenzyme of the enzyme that adds the hydroxyl group to proline and lysine. Therefore, if vitamin C is not present the triple helix of collagen won’t be stable and it will cause the weakness of this extracellular protein.

Question 11

Describe two mechanisms in which energy is utilized to move ions and molecules against their concentration gradient.

Answer 11

1- Active transport- In this mechanism molecules of ATP serve as the source as energy to move molecules using a pump.
2- In the case of co-transport the potential energy of the sodium ions is used as they move from high concentration to lower concentration to transport molecules against their concentration gradient.

Question 12

You are exploring a rather inhospitable planet, which has seas that are somewhat hydrophobic in nature. Surprisingly, there are living organisms in the seas whose cytoplasm is hydrophobic to a similar degree. These organisms have membranes made primarily of phospholipids arranged in a bilayer.

a. What is the most probable orientation of these phospholipids?
b. In these extraterrestrial cells, how would you expect this new condition to affect a transmembrane protein in regards to its extra cellular domain, cytoplasmic domain, and inner membrane?

Answer 12

A-The polar heads of the phospholipids face toward the middle of the bilayer with the hydrophobic fatty acid tails facing outward.
B-The transmembrane domain of the protein will be hydrophobic and the extracellular and cytoplasmic domain will be hydrophobic

Question 13

2- How might a stealth liposome be prepared so that it could potentially be used in the treatment of a tumor whose cells have a protein called tumor cell antigen (TCA) in their membranes? This protein (TCA) is not found at all in normal cells.

a. Describe how you will design your liposome so it could deliver a toxic drug that will kill the tumor cells without affecting healthy cells. In your design, explain how you will specifically target the tumor cell (be specific) (5 points)
b. How will you protect the stealth liposome from destruction by immune cells. Mention the name of the molecule. (5 points)
c. Explain where the drug will be found in the stealth liposome if the drug is hydrophilic vs hydrophobic. (5 points)

Answer 13

Solution:
1- TCA protein is unique to these tumor cells an antibody against that proteins should be made.
2- Inserted the antibody in the membrane of the liposome.
3- In the toxic drug is hydrophilic then it must be deposited in the aqueous part of the liposome.
4- If the drug is hydrophobic it must be inserted in the inside if the inner membrane.
5- To avoid that the degradation of the liposome by the immune system a coat of polythene glycol will be added over the membrane.

Question 14

3- You determine the transition temperatures for two membranes. The first has a transition temperature of 28°C, the second has a transition temperature of 15°C.

a. What can you conclude about the compositions of these two membranes? (5 points)
b. Explain two mechanisms of membrane remodeling (that does not involve new synthesis of lipids if the temperature decreases from 28°C to 15°C. In your answer mention the enzymes involved in this process. (5 points)

Answer 14

A- At 28oC will have more saturated fatty acids than the 15 oC (alternative correct answer they could answer the following At 28oC will have more longer fatty acids than the 15 oC.)
B- -Desaturases will add double covalent bonds to the fatty acids.
-phospholipases will cut the fatty acids tails and acyl-transferaces will re-shuffle them.

Question 15

4- Parietal cells are responsible for increasing the acidity in the stomach during food digestion using a H+/K+ ATPase pump.

a. Explain the process by which this pump is activated (5 points)
b. How the pump functions and
c. How the initial activation of the pump could be inhibited. (5 points)

Answer 15

A- The pump is activated when histidine bind its receptor causing the vesicles that contains the pumps. Fuse with the cell membrane of the parietal cell inserting the pump in the cell membrane. Once in the membrane the pump start working.
B- The pump moves H+ ions outside of the cell and bring in K+ ion inside the cell.
C- Drugs that bind the histidine receptor will inhibit the initial activation.

Question 16

5- Look at the hydropathy plot of the protein connexin in the figure below and determine.

a. How many transmembrane domains are indicated for the protein connexin in the plot? (2.5 points)
b. How the values plotted in this graph. were calculated (2.5 points)
c. If any transmembrane domains contain charged amino acids, where do you expect to find them within the membrane. (Be specific). (2.5 points)

Answer 16

A- Three transmebrane domains (If a student said 2 I will give it the points because when I reduced the size of the figure it made difficult to clearly distinguish the one of the peak) If a student said 4 that will be incorrect.
B- By making an average of the of the hydrophobicity of group of three amino acids.
C- At the very end of the alpha helix.

Question 17

Note: Use Figure A, Table B, Figure C, and Table D to answer these questions.

a. Explain what changes in the membrane must take place in a lymphocyte that has aged to the point in which it does not function properly and must be targeted for degradation (5 points)
b. Your answer should include: the lipid translocating enzyme that must be activated to accomplish this change. (5 points)

Answer 17

A- Phosphatidylserine must be flip from the inside to the outside.
B- Scrablasses will move them.

Question 18

7- Which of the four 20 amino acid sequences listed below in the single-letter amino acid code is the most likely candidate to form a transmembrane region (alpha helix) of a transmembrane protein. Explain your answer.

A. ITEIYFGRMAGVIGTDLLIS
B. ITLIYFGNMSSVTQTILLIS
C. LLKKFFRDMAAVHETILEES
D. LLLIFFGVMALVIVVILLIA

Answer 18

D is the correct answer because all the amino acids are hydrophobic with only

Question 19

Aquaporins are channel proteins that facilitate the movement of water molecules across the plasma membrane. It is important that for this protein to be effective, it must be extremely selective for water. Explain what would be the physiological consequence to the cell and proteins if residues 203 AND 68 are mutated on aquaporin.

Answer 19

if the amino acid is mutated for a another positively charge amino acid it will not cause changes in the function of this channel proteins

Question 20

9- KcsA is a prokaryotic potassium channel. If a mutation is introduced in the sequence of the portion of DNA that encodes the region for the selectivity filter.
A) Predict what will be the consequence on the function of this channel if this mutation causes a decrease in the width of the selective filter from 0.134 to 0.096 nM.
B) Explain your answer.
NOTE. The width of both ions are K+  0.133nM and NA+  0.095nM.

Answer 20

Reducing the width of the channel will allow now NA+ to interact with the 4 oxygens of the channel which will allow them to cross the membrane.

Question 21

9- Explain how cells deal with small tonicity (osmolarity) changes in their environment. (5 points)

Answer 21

By Allowing ions toc cross the membrane to counteract these changes in osmolarity.

Question 22

10- Assume that you are measuring the diffusion rate of a group of membrane proteins by exposing the cell to a cancer drug to determine whether the cell will experience a dramatic change in the movement of proteins after treatment with the drug.

a. What technique you will use (1 point)
b. Explain what steps need to be done to perform this study (4 points)

Answer 22

A- The technique is FRAP.

B- 1) Proteins are labeled with florescence dye. 2) photobleach spots with laser beam 3) measure the rate of recovery in the region that was photobleached.

Question 23

11- Describe the characteristics of lipid rafts, how they have been observed, and why some researchers are concerned that they don’t exist in cells in vivo.

Answer 23

It is a region with specific composition of proteins and lipids that is more gelated and highly ordered than surrounded regions.

The vast majority of these regions have been produced and study in in-vitro after unnatural processes have been used to create the region while attempts to study in living cells are usually unsuccessful.

Question 24

12- You are studying a protein that you believe works as a secondary active transport in cells lining the small intestine. Previous data suggests that the protein assists glucose in entering these cells against its concentration gradient using sodium ions as a source of energy.

A- Design an experiment to isolate this protein. What type of detergent you will use and explain why. (5 Points)
B- Describe how you will create a liposome, so you can test this protein. (5 points)
C- Explain how you will obtain the lipids that you need to create the liposome. Explain, why this is important. (5 points)
D- How would you maintain the flow of Na+ ions going through the cell. Explain why this step is important. (5 points)
E- Glucose is moving against its concentration gradient, explain where the energy for this movement will come from. (5 points)

Answer 24

1- The cell is expose to an non-ionic detergent do avoid the denaturalization of the protein.
2- The complexed of proteins and detergents are used to purify the protein of interest
3- The micelles of detergent and membrane lipids are saved because the specific lipid composition could be required for the proper function of the protein.
4- Micelles of detergent/lipid are added to the purified protein of interest and the detergent is remove. At this point the function protein could be tested.
___________________________________________________________

D- A high concentration of NA+ is created in the environment

E- Secondary active transport requires ions to move in favor of concentration gradient to provide the energy that will allow glucose to move against its concentration gradient

Question 25

Describe the quality control steps that take place in the RER to ensure that glycoproteins are properly folded.

Answer 25

1- The two outermost glucose molecules that are part of the oligopeptide that are attached to the now folded protein are cleave leaving only one glucose molecule.
2- Oligoprotein binds to calnexin and their glucose is cleave.
3- In protein is correctly folded it will continue to final target.
4- If protein is misfolded it will be bind to the conformation enzyme (UGGT) which add a glucose to the oligosaccharide.
5- Protein binds again to Calnexin. To recheck 3D structure again
6- After several attempts if protein does not get correct shape it will be send for degradation.

Question 26

• Describe and compare the molecular composition of microtubules and microfilaments. Be sure to include in your response critical characteristics of the monomers that comprise each one of them. Also, mention one function for each of these structures.

Answer 26

Microtubules:

The structural units of the microtubules are alpha tubulin and beta tubulin that are arrange in longitudinal rows known as protofilaments. Each microtubule is a hollow structure with 13 microfilament that are kept together by non-covalent interactions.

Microfilaments:

The structural units of the microtubules is the acting protein which form long longitudinal strands to come together as a dimer.

Question 27

Explain the process of polymerization of the microtubules. In your answer include
how the plus and minus ends are established, the process of nucleation, and the role of dynamic instability in the process of elongation. (include in your answer the molecules and structures are in the minus end.)

Answer 27

The nucleation process occurs in the microtubule organization center where the gamma rings are located. The gamma tubulin subunits bind to the Gamma tubulin rings and then the alpha tubulin binds to the gamma tubulin. The minus end faces the gamma tubuling. Dimers of alpha and tubulin keep adding themselves to the plus end, alpha facing the gamma rings and the beta-tubulin facing the plus side. Elongation takes place as the subunits add themselves the structure does rapidly forming an open structure. But as the open structure start to form the seam the beta-tubulins get dephosphorylated forming GDP. The GDP form of the beta-tubulins slightly bends and once enough beta-tubulin GDP are found in the plus end it will initiate catastrophe. Eventually catastrophe stop and the rescue (grows) begging to take place and this cycle is repeated.

Question 28

• Explain the process of polymerization of the microfilaments. In your answer include how the plus and minus ends are established, and how the process of nucleation and elongation take place. (5 points)

Answer 28

• All of the monomers within an actin filament are pointed in the same direction, resulting in a polar filament with so-called “barbed” and “pointed” ends. Before it is incorporated into a filament, an actin monomer binds a molecule of ATP. Actin is an ATPase, just as tubulin is a GTPase. The ATP associated with the actin monomer is hydrolyzed to ADP at some time after it is incorporated into the end of a growing actin filament. The initial nucleation event in filament formation occurs slowly in vitro, whereas the subsequent stage of filament elongation occurs much more rapidly. The barbed and pointed ends require different minimal concentrations of ATP-actin monomers in order to elongate, a measure known as the critical concentration. The critical concentration of the barbed end is much lower than the pointed end, meaning that the barbed end can continue to elongate at lower ATP-actin concentrations than the pointed end can.

Question 29

Explain the process of polymerization in intermediate filaments.

Answer 29

• Intermediate Filaments polypeptides have diverse amino acid sequences, yet all share a similar structural organization and form similar-looking filaments.
• Polypeptides of IFs all contain a central, rod-shaped, α-helical domain of similar length.
• The central fibrous domain is flanked on each side by globular domains of variable size.
• Two polypeptides interact as their α-helical rods wrap around each other to form a ropelike dimer.
• The two polypeptides are aligned parallel to one another so the dimer has polarity, with one end defined by the C-termini of the polypeptides and the opposite end by their N-termini.
• The basic building block of IF assembly is thought to be a rodlike tetramer.
• Eight tetramers associate with one another in a lateral arrangement to form a filament that is one unit in length (about 60 nm).
• Unit lengths of filaments associate with one another in an end-to-end fashion to form the highly elongated intermediate filament.
• None of these assembly steps require the direct involvement of either ATP or GTP.
• The tetrameric building blocks lack polarity as does the assembled filament, which distinguishes IFs from other cytoskeletal elements.

Question 30

Analysis of a miscarried embryo revealed that termination of the pregnancy was the result of an aneuploidy event that occurred during meiosis I. Explain the following: 1. What molecular abnormality occurred during meiosis that caused aneuploidy in the embryo? 2. Which of the two cells (sperm or egg) was the primary contributor to this defect and explain why? In your explanation include why the other cell did not contribute to aneuploidy 3. Why do some embryos with aneuploidy survive and reach full pregnancy while others do not. In this answer use the terms trisomy and monosomy. 4. Explain why an older father increases the chances of having an autistic child. (15 points)

Answer 30

1. Aneuploidy occurs when the homologous chromosomes fail to separate during meiosis 1 or the sister chromatids fail to separate during meiosis II. Two possible explanations could explain this aneuploidy; 1) oocytes of older women have remained arrested in meiosis I for a long period within the ovary. According to one hypothesis meiotic spindles of older oocutes are less able to hold together weakly constructed bivalents than those of younger, increasing the likelihood that homologous chromosomes will missegregate at anaphase I 2) is that sister chromosome cohesion , which prevent the chiasmata from sliding off the end of the chromosomes, it is not fully maintained over an extended period, allowing homologues to separate prematurely.
   2- Egg are the prematurely contributors to the aneuploid defects.
   3- The embryos with aneuploidy with a monosomy conditions do not survive because they miss
   the copy of an entire chromosome and therefore their genes. In the case of trisomy most embryos will die with exceptions of trisomy in smaller chromosomes (like chromosome 21) which carry a low number of genes.
   4- In contrast, because males are constantly regenerating germ cells through mitosis the number of mutation that the father pass to the offspring increases as a man ages which is linked to a higher risk of autism.

Question 31

A mutation in fission yeast strain results in two smaller cells when compared to normal fission yeast strains. 1. Explain what are the respective roles of CAK, Wee1, and Cdc25 in controlling cdc2 activity in fission yeast cells? In your answer, be specific as to which molecule is a kinase or phosphatase. In the case of the kinases, what specific residue is being phosphorylated and the consequence/purpose of adding the phosphate to the substrate. 2. Explain the reasons why the mutant has smaller cells?

Answer 31

When a cyclin reaches a sufficient concentration in the cell, it binds to the catalytic subunit of a cdc2.
 When cyclin binds to cdc2 and enzyme called CAK phosphorylate the threonine 161 residue which is necessary for the activation of cdc2.
 However, another protein kinase called Wee1 phosphorylate tyrosine 15 residue of cdc2 inactivating cdc2.
 This will prevent cdc2 from enter prematurely in the M phase.
 At the end of G2 a phosphate called Cdc25 will remove the phosphate from the tyrosine 15 residue from cdc2 activating it and the cell will enter in mitosis.

Question 32

When normal cells are subjected to treatments that damage DNA, such as ionizing radiation or DNA-altering drugs, their progress through the cell cycle stops while the damage gets repaired. This surveillance mechanism is known as check points. There are two well-studied pathways available to mammalian cells to arrest their cell cycle in response to DNA damage. Select one of the two pathways and describe how it leads to cell cycle arrest as a response to DNA damage. Make it clear which one you have chosen.

Answer 32

UV light radiation:
1- ATR kinase is activated and the cell arrests in G2. When this molecule is recruited to sites of protein-coated, single-stranded DNA.
2- ATR phosphorylates and activates a checkpoint kinase, called Chk1
3- Chk1 phosphorylates Cdc25 on a particular serine residue, making the Cdc25 molecule a target for a special adaptor protein that binds to Cdc25 in the cytoplasm.
4- Cdc25 normally plays a key role in the G2/M transition by removing inhibitory phosphates from Cdk1.
5- Thus, the absence of Cdc25 from the nucleus leaves the Cdk in an inactive state and the cell arrested in G2.
Ionizing radiation-
1- Damage to DNA also leads to the synthesis of proteins that directly inhibit the cyclin–Cdk complex that drives the cell cycle.
2- ATM is involved in this checkpoint mechanism. In this particular DNA damage response, the breaks in DNA that are caused by ionizing radiation serve as sites for the recruitment of a protein complex termed MRN.
3- MRN can be considered as a sensor of DNA breaks.
4- MRN recruits and activates ATM, which phosphorylates and activates another checkpoint kinase called Chk2.
5- Chk2 in turn phosphorylates a transcription factor (p53), which leads to the transcription and translation of the p21 gene and subsequent inhibition of Cdk.

Question 33

Explain how the sister chromatids are held after replication and during prophase. Include in your answer 1. the specific role of cohesin and condensing proteins (in other words, what do they hold?) 2. The role of the Polo-like A and Aurora-B plays 3. how the centromeres of the system chromatids are held together and what prevents them from premature detachment.

Answer 33

1- After replication, the DNA helices of a pair of sister chromatids would be held in association by cohesin molecules that encircled the sister DNA helices.
2- As the cell entered mitosis, the compaction process would begin, aided by condensin molecules
3- In this model, condensin brings about chromosome compaction by forming a ring around supercoiled loops of DNA within chromatin.
4- Cohesin molecules would continue to hold the DNA of sister chromatids together.
5- Most of the are removed during prophase when Polo-like kinase and Aurora B kinase phosphorylate cohesin, and the protein WAPL then removes it.
6- However, the centromeres ae kept attached because cohesin persists due to a phosphatase that removes any phosphate groups added by kinases.

Question 34

Describe the roles of astral, kinetochore, and polar microtubules in the mitotic spindle. (10 points)

Answer 34

(1) Astral microtubules radiate outward from the centrosome and outside of the body of the
spindle. These microtubules are thought to help position the spindle within the dividing cell.
(2) Kinetochore (or chromosomal) microtubules extend between the centrosome and the
kinetochore of the chromosomes and are responsible for maintaining the position of the
chromosomes at the metaphase plate.
(3) Polar microtubules extend from the centrosome past the chromosome, overlapping with their
counterparts from the opposite centrosome. This interaction allows the spindle to form a football-like structure and provides mechanical integrity to the spindle.

Question 35

Describe the role of the anaphase promoting factors during the M phase. In your answer include the mechanisms underlying the transition between metaphase and anaphase and the transition between anaphase and the G1 phase. Your answer must include the adaptor molecules as well as the substrate that is targeted for degradation

Answer 35

APC are multisubunit complexes that ubiquitinate substrates, leading to their destruction.
 APC (anaphase promoting complex) is active during mitosis and G1.
 Two different pathways of APC are active in the M phase differ in containing either a Cdc20 or a Cdh1 adaptor protein, which alters the substrates recognized by the APC.
 APCCdc20 is active early in mitosis, at a time when Cdh1 is inhibited by Cdk1-mediated phosphorylation.
 As Cdk1 activity drops sharply in late mitosis, Cdh1 is activated, leading to the activation of APCCdh1.
 APCCdc20 is responsible for destroying proteins, such as securin, that inhibit anaphase. Destruction of these substrates promotes the metaphase–anaphase transition.
 APCCdh1 is responsible for ubiquitinating proteins, such as mitotic cyclins, that inhibit exit from mitosis. Destruction of these substrates promotes the mitosis–G1 transition.

Question 36

What is the role of DNA strand breaks in genetic recombination?

Answer 36

Recombination involves the physical breakage of individual DNA molecules and the ligation of the split ends from one DNA duplex with the split ends of the duplex from the homologous chromosome.
To occur so faithfully, recombination depends on the complementary base sequences that exist between a single strand from one chromosome and the homologous strand of another chromosome.
During this process, two DNA duplexes that are about to recombine become aligned next to one another as the result of some type of homology search in which homologous DNA molecules associate with one another in preparation for recombination.
Once they are aligned introduces a double-stranded break into one of the duplexes. The gap is subsequently widened (resected).
Resection may occur by the action of a 5’ —> 3’ exonuclease
the broken strands possess exposed single-stranded tails, each bearing a 3’ OH terminus.
One of the single-stranded tails leaves its own duplex and invades the DNA molecule of a non-sister chromatid.
The single-stranded DNA to search for and invade an homologous double helix. Eukaryotic cells Strand invasion activates a DNA repair activity that fills the gaps.
As a result of the reciprocal exchange of DNA strands, the two duplexes are covalently linked to one another to form a joint molecule (or heteroduplex) that contains a pair of DNA crossovers, or Holliday junctions.