cell biology2

Question 1

hypoxia inducible factor 1 (HIF1A)

Answer 1

The alpha subunits of HIF are hydroxylated at conserved proline residues by HIF prolyl-hydroxylases, allowing their recognition and ubiquitination by the VHL E3 ubiquitin ligase, which labels them for rapid degradation by the proteasome. This occurs only in normoxic conditions. In hypoxic conditions, HIF prolyl-hydroxylase is inhibited, since it utilizes oxygen as a cosubstrate. HIF-1, when stabilized by hypoxic conditions, upregulates several genes to promote survival in low-oxygen conditions. These include glycolysis enzymes, which allow ATP synthesis in an oxygen-independent manner, and vascular endothelial growth factor (VEGF), which promotes angiogenesis. HIF is necessary for tumor growth because most cancers demand high metabolic activity and are only supplied by structurally or functionally inadequate vasculature. Activation of HIF allows for enhanced angiogenesis, which in turn allows for increased glucose intake. While HIF is mostly active in hypoxic conditions, VHL-defective renal carcinoma cells show constitutive activation of HIF even in oxygenated environments.

Question 2

Von Hippel–Lindau tumor suppressor

Answer 2

The VHL protein (pVHL) is involved in the regulation of a protein known as hypoxia inducible factor 1α (HIF1α). This is a subunit of a heterodimeric transcription factor that at normal cellular oxygen levels is highly regulated. In normal physiological conditions, pVHL recognises and binds to HIF1α only when oxygen is present due to the post translational hydroxylation of 2 proline residues within the HIF1α protein. pVHL is an E3 ligase that ubiquitinates HIF1α and causes its degradation by the proteasome. In low oxygen conditions or in cases of VHL disease where the VHL gene is mutated, pVHL does not bind to HIF1α. This allows the subunit to dimerise with HIF1β and activate the transcription of a number of genes, including vascular endothelial growth factor, platelet-derived growth factor B, erythropoietin and genes involved in glucose uptake and metabolism.

Question 3

WHY REGULATE MEMBRANE FUSION?

Answer 3

All cellular membranes are composed from a lipid bilayer with charged residues facing the outside and interacting with water molecules. As a consequence, the membrane fusion thermodynamically is highly inefficient event, since it needs to overcome the repulsive ionic forces and dissipate the hydration between to lipid bilayers. Thus, cell need to have the machinery that increases the efficiency of membrane fusion. In addition, cell needs to ensure the specificity of membrane fusion to ensure that the vesicular cargo is delivered to the correct destination.

Question 4

REGULATION OF INTRACELLULAR MEMBRANE FUSION

Answer 4

A transport vesicle fusion involves many coordinated steps. First it needs to be delivered to the fusion site via help of molecular motor proteins that move along actin or microtubule tracks. Once delivered to the appropriate destination it needs to fuse with the target membrane. That is achieved by a help of a set of cellular proteins known as SNAREs (soluble NSF attachment protein receptor).

Question 5

SNARE classes and functions

Answer 5

The primary role of SNARE proteins is to mediate vesicle fusion, that is, the exocytosis of cellular transport vesicles with the cell membrane at the porosome or with a target compartment (such as a lysosome). There are three main classes of SNAREs: syntaxins, VAMPs (vesicle associated membrane protein) and SNAPs (synaptosome associated protein). VAMP is located in transport vesicle, while syntaxin and SNAP are located in target membrane. SNAREs are small, abundant and mostly plasma membrane-bound proteins. Although they vary considerably in structure and size, all share a segment in their cytosolic domain called a SNARE motif that consists of 60-70 amino acids that are capable of reversible assembly into tight, four-helix bundles called “trans”-SNARE complexes. The ‘zipper’ model of SNARE function postulates that the SNARE core complex ‘zips’ from the membrane-dis- tal amino termini to the membrane-proximal carboxyl termini, and the formation of the stable SNARE com- plex overcomes the energy barrier to drive fusion of the lipid bilayers. Many neurotoxins directly affect SNARE complexes. Such toxins as the botulinum and tetanus toxins work by targeting the SNARE components. These toxins prevent proper vesicle recycling and result in poor muscle control, spasms, paralysis, and even death. There are multiple types of each kind of SNARE proteins which leads to its specificity of fusion

Question 6

syntaxins

Answer 6

has transmembrane domain and three helix domain and sits on the targeted plasma membrane. It can form coiled coil complex with its self using the three helix domains. N-sec1 creates a functional structure but keeps it in inactive conformation until it recieves a signal. In this way it regulates when vesicles can fuse when n-sec1 is signaled to leave leading to fusion of membranse

Question 7

VAMPs (vesicle associated membrane protein)

Answer 7

Sits on synaptic vesicles. Has one transmembrane domain and huge helix domain

Question 8

SNAPs (synaptosome associated protein)

Answer 8

has two helix domains and transmembrane domain in between. Also sits of targeted plasma membrane.

Question 9

NSF and αSNAP protein function

Answer 9

Once the fusion occurred, cell needs to recycle the SNARE proteins for another fusion events. Since SNARE complexes are highly stable, cell ahs a set of proteins, namely NSF (N-ethyl-maleimide sensitive fusion protein) and αSNAP (soluble NSF attachment protein) to disassemble SNARE complex. The NSF uses ATP hydrolysis to generate energy needed to disassemble the SNARE complex by unraveling the coiled-coil interactions of the SNARE complex domain

Question 10

Sec1 protein function.

Answer 10

After disassembly of SNARE complex, the SNARE proteins need to be refolded to the active conformation. That is achieved by the Sec1 protein, which binds and changes syntaxin into active conformation.

Question 11

VIRAL MEDIATED MEMBRANE FUSION

Answer 11

The class of so-called enveloped viruses (includes HIV, ebola, influenza viruses) also needs to go through membrane fusion in order to infect cells. Viruses have developed fusion machinery that is remarkably similar to SNARE fusion. This machinery also achieves both objectives of fusion, ensures that it is efficient and specific. The examples discussed in the class are influenza and HIV viruses, although the mechanisms of other enveloped virus fusion are very similar. Viral fusion is achieved by help of specialized viral fusion proteins. This protein contains transmembrane domain at the one end (it is inserted into viral membrane) and highly hydrophobic fusion domain. Normally, fusion domain is folded and hidden within viral protein. Upon receiving specific signal (in case of influenza virus it is change in pH), the fusion domain is exposed and inserted into target cell membrane. That is immediately followed by refolding of fusion protein leading to the fusion of viral and cellular membranes.

Question 12

Why are cells rich in potassium but poor in sodium?

Answer 12

One idea concerns how the oceans grew salty as cells evolved (about a billion years ago). The early oceans were relatively rich in potassium; they have grown more salty with sodium over time. The reason is that, while there are about equal parts potassium and sodium in the earth’s crust, silicates (soils) bind potassium more tightly than they bind sodium. Thus, as the rains fell, they preferentially leached sodium out of the soil, loading up the oceans with sodium. Evidently, it was easier for cells to evolve mechanisms allowing them to retain their primitive potassium-rich intracellular milieu than to evolve enzymes more tolerant of sodium.

Question 13

Intracellular Fluid

Answer 13

Intracellular fluids are high in potassium, magnesium, and polar macromolecules and low in sodium and chloride ions. There are about 27l total in the body.

Question 14

Extracellular fluid

Answer 14

In humans, the normal glucose concentration of extracellular fluid that is regulated by homeostasis is approximately 5 mM/L, and the pH is tightly regulated by buffers around 7.4. The volume of ECF is typically 15 L, of which 12 L is interstitial fluid and 3 L is plasma. Interstitial fluid makes up 16% of human body weight, and blood plasma, 4%. It is high in Na and Cl and low in potassium and polar macromolecules.

Question 15

The Structure of the Plasma Membrane

Answer 15

Membranes are mostly lipid molecules, which are impermeable to water and charged substance. Proteins provide pathways for water and solutes to get across. Lipids are also ‘strong’, in an electric sense. That is, they can keep opposite electric charges separated, without collapsing. As we will see, the cytoplasm of nearly all cells is electrically negative, compared to the ECF, because cells contain a few more negative than positive ions. The excess anions create an electrical potential difference between the inside and the outside of the cell, and this membrane potential, which governs some vital cell processes, is wholly dependent on the integrity of the plasma membrane. The ability of the membrane to withstand the imposed electric force (which has a whopping strength of about 100,000 volts/cm=50mV/5nm(thickness of plasma membrane)) is due to its lipid composition. therefore a small change (a few mV) in the electric force can have a large effect on transmembrane proteins. The second physiologically important point about membranes is that they are apparently shot full of holes. That is, many charged and polar molecules can cross membranes, no thanks to the lipids. Their ability to do so depends primarily on the presence of certain proteins inserted in the lipid membrane; these proteins have special abilities to transport particular substances between the ECF and the ICF. From a functional (i.e., our) standpoint, there are two kinds of proteins that mediate the transmembrane movements of charged substances: channels and transporters

Question 16

Channels

Answer 16

most are selective for particular ions. For example, sodium channels will pass Na+ ions, but not K+, or Cl-, or other ions. Some channels permit just about any cation to pass, but not anions; these are called ‘non-selective cation channels’. Second, some channels contain molecular gates. Substances can pass through the channel only when the gate is open. opening and closing of the gates can govern important functions, like action potentials. The gates of channels are controlled by a variety of forces. Some depend on the electric field (membrane potential) across the membrane. The gates in many such voltage-gated channels swing open when the membrane is depolarized (the cell is made less negative inside). Gates in other channels require mechanical stimulation (stretching of the membrane) to open (e.g., hair cells in the cochlea; touch receptors in the skin). Still others require the binding of a particular chemical (synaptic receptors for a neurotransmitter). Others open or close depending on temperature (cutaneous thermal receptors). Some gates in channels depend on more than one force, making for some pretty complicated behavior. For example, a channel gate may require the binding of a particular chemical substance and, simultaneously, a particular value of membrane potential in order to open up. Some channels contain no gates (e.g., water channels, also called aquaporins) , while others contain two gates, and both have to be open to allow ions to pass.

Question 17

patch clamp technique

Answer 17

The Patch Clamp was invented in 1980. A micropipette is lowered onto a cell, but instead of punching through the membrane, it sucks onto it, and a tight seal is formed between the membrane and glass pipette. In this situation, it is possible to see the tiny electric currents produced by the openings and closings of single ion channels. Examples are shown on the right (with the averaged response at the bottom). It is astonishing to see a single protein molecule at work, in real time, in its native membrane environment. The work won a Nobel Prize in 1991 for its German inventors, Bert Sakmann and Erwin Neher. allows one to see individual ion channels at work. A glass pipette touches a cell, and gentle suction draws a membrane bleb into the mouth of the pipette, forming a tight seal with the glass. Then, when a channel gate opens, it is possible to record the electric current through the channel. As channels flicker open and closed, one can describe their behavior in great quantitative detail.

Question 18

Ion channel structure

Answer 18

Channels differ with respect to the ion they let pass (for example, Na+, K+, Cl−), the ways in which they may be regulated, the number of subunits of which they are composed and other aspects of structure. Channels belonging to the largest class, which includes the voltage-gated channels that underlie the nerve impulse, consists of four subunits with six transmembrane helices each. On activation, these helices move about and open the pore. Two of these six helices are separated by a loop that lines the pore and is the primary determinant of ion selectivity and conductance in this channel class and some others.

Question 19

Transporters

Answer 19

Some substances cross membranes by binding to proteins and being escorted (‘carried’) across. Such a mechanism is needed to transport big molecules, like glucose, selectively. Transporters are also needed in order to pump molecules across a membrane, that is, to concentrate them on one side against their energy gradient. Such pumping is called active transport, to distinguish it from passive transport pathways, like channels. Pumping means performing work, and the energy to do the work can come from different sources. If the energy comes directly from metabolism (usually splitting ATP), the transport is called primary active transport. An example is the Na+ pump, which extrudes Na+ from cells, and requires ATP. If the energy comes from other sources, it is called secondary active transport. There are many examples of secondary active transport, and most of them use the energy released when Na+ ions leak into cells, and the energy released is captured and used to pump another ion across the membrane. Ultimately, secondary active transport also depends on metabolism. We are only just beginning to acquire information about the molecular structure of transporters, and that information suggests that they are very similar to channels. For example, by mutating only a few amino acid residues, a transporter of chloride and hydrogen ions, which pumps only about a hundred ions per second, can be turned into a chloride-selective ion channel that allows the passage of tens of thousands of ions per second.

Question 20

Von Hippel–Lindau syndrome

Answer 20

autosomal dominant inherited cancer sysndrome. 1:36,000 live births, mutation in the VHL tumor suppressor gene. Characterized by formation of cystic and highly vascularized tumors in many organs:Cerebellar, spinal cord, and retinal hemangioblastomas, Endolymphatic sac (inner ear) tumors, Bilateral kidney cysts and renal cell carcinomas,Pancreatic cysts and islet cell tumors, Pheochromocytomas, Cystadenomas of the genitourinary tract. Tumors called hemangioblastomas are characteristic of von Hippel-Lindau syndrome. These growths are made of newly formed blood vessels. Although they are typically noncancerous, they can cause serious or life-threatening complications. Hemangioblastomas that develop in the brain and spinal cord can cause headaches, vomiting, weakness, and a loss of muscle coordination (ataxia). Hemangioblastomas can also occur in the light-sensitive tissue that lines the back of the eye (the retina). These tumors, which are also called retinal angiomas, may cause vision loss.Von Hippel-Lindau syndrome is associated with a type of tumor called a pheochromocytoma, which most commonly occurs in the adrenal glands (small hormone-producing glands located on top of each kidney). Pheochromocytomas are usually noncancerous. They may cause no symptoms, but in some cases they are associated with headaches, panic attacks, excess sweating, or dangerously high blood pressure that may not respond to medication. Pheochromocytomas are particularly dangerous if they develop during pregnancy.

Question 21

recommended screening and follow up for VHL patients

Answer 21

Age 5-15: Annual physical examination and neurological assessment by pediatrician, Annual examination by ophthalmologist, Annual test for fractionated metanephrines, Annual abdominal ultrasonography after age 8, Complete audiology assessment by an audiologist every 2-3 years; annually if any hearing loss, tinnitus, or vertigo, MRI with contrast of the internal auditory canal to check for ELST if recurrent ear infections.Age 16 and beyond: Annual eye/ retinal examination by ophthalmologist, Annual physical examination by physician, Annual test for fractionated metanephrines, Annual quality ultrasound, and at least every other year MRI scan of abdomen, Annual abdominal MRI or MIBG scan if biochemical abnormalities found, MRI brain, C/T/L-spine, with thin cuts through the posterior fossa, to rule out hemangioblastomas and ELST; every 2-3 years, Audiology assessment by an audiologist every 2-3 years.

Question 22

VHL type 1

Answer 22

VHL mutation resulting in total or partial VHL loss due to improper folding, this leads to up regulation of HIF. Clinical manifestation: Hemangioblastoma, Renal cell carcinoma, Low risk of pheochromocytoma

Question 23

VHL type 2A

Answer 23

VHL missense mutation leading to upregulation of HIF, inability to stabilize microtubules. Clinical manifestation: Hemangioblastoma, Low risk of renal cell carcinoma, Low risk of pheochromocytoma

Question 24

VHL type 2B

Answer 24

VHL missense mutation which resulats in up regulation of HIF. Clinical manifestations: hemangioblastoma, high risk of renal cell carcinoma, and high risk of pheochromocytoma

Question 25

VHL type 2c

Answer 25

VHL missense mutation. pVHL maintins ability to downregulate HIF. Decreased binding to fibronectin, defective fibronectin matrix assemble. . Clinical manifestations: pheochromocytoma only

Question 26

Clear cell papillary renal cell carcinoma

Answer 26

a rare subtype of renal cell carcinoma (RCC). VHL is most common cause of inherited ccRCC. Familial ccRCC (4%)- typically multifocal, bilateral, early age onset. Sporadic ccRCC (96%)- solitary, unilateral, later age of onset. Patients with familial VHL are at risk of developing of up to 600 tumors per kidney. Goal of surgical management for VHL-associated kidney tumors is to prevent metastasis while preserving renal function. Other genetic alterations in ccRCC are BAP1 and PRBM1which are also located on Chr 3p. Loss of 3p causes heterozygous loss of BAP1 and PRBM1 too. Subsequent PRBM1 or BAP1mutations results in ccRCC with different features and outcomes.

Question 27

pheochromocytoma

Answer 27

Up to 25% of pheochromocytomas may be familial. Mutations of the genes VHL, RET, NF1 (Gene 17 Neurofibromatosis type 1), SDHB and SDHD are all known to cause familial pheochromocytoma, therefore this disease may be accompanied by Von Hippel–Lindau disease, neurofibromatosis,[3] or familial paraganglioma depending on the mutation. Pheochromocytoma is a tumor of the multiple endocrine neoplasia syndrome

Question 28

hemangioblastoma

Answer 28

Hemangioblastomas are composed of endothelial cells, pericytes and stromal cells. In VHL syndrome the von Hippel-Lindau protein (pVHL) is dysfunctional, usually due to mutation and/or gene silencing. In normal circumstances, pVHL is involved in the inhibition of hypoxia-inducible factor 1 α (HIF-1α) by ubiquitin mediated proteosomal degradation. In these dysfunctional cells pVHL cannot degrade HIF-1α, causing it to accumulate. HIF-1α causes the production of vascular endothelial growth factor, platelet derived growth factor B, erythropoietin and transforming growth factor alpha, which act to stimulate growth of cells within the tumour. Tumors of the central nervous system that originate from the vascular system usually during middle-age.

Question 29

fibronectin

Answer 29

a high-molecular weight (~440kDa) glycoprotein of the extracellular matrix that binds to membrane-spanning receptor proteins called integrins. Similar to integrins, fibronectin binds extracellular matrix components such as collagen, fibrin, and heparan sulfate proteoglycans (e.g. syndecans). Fibronectin has numerous functions that ensure the normal functioning of vertebrate organisms. It is involved in cell adhesion, growth, migration, and differentiation. Cellular fibronectin is assembled into the extracellular matrix, an insoluble network that separates and supports the organs and tissues of an organism.

Question 30

Processes regulated by HIF

Answer 30

Angiogenesis, Erythropoiesis, Anaerobic glycolysis, Glucose uptake, Extracellular matrix turnover, pH control, Apoptosis, Mitogenesis. Accumulaton of HIF-α results in over-expression of VEGF, TGF, PDGF

Question 31

VEGF

Answer 31

a signal protein produced by cells that stimulates vasculogenesis and angiogenesis. It is part of the system that restores the oxygen supply to tissues when blood circulation is inadequate.

Question 32

TGFbeta

Answer 32

exists in three known subtypes in humans, TGFβ1, TGFβ2, and TGFβ3. These are upregulated in Marfan’s syndrome and some human cancers, and play crucial roles in tissue regeneration, cell differentiation, embryonic development, and regulation of the immune system. Isoforms of transforming growth factor-beta (TGF-β1) are also thought to be involved in the pathogenesis of pre-eclampsia. TGFβ receptors are single pass serine/threonine kinase receptors.

Question 33

PDGF

Answer 33

one of the numerous growth factors, or proteins that regulate cell growth and division. In particular, it plays a significant role in blood vessel formation (angiogenesis), the growth of blood vessels from already-existing blood vessel tissue. Uncontrolled angiogenesis is a characteristic of cancer.

Question 34

Treatment options in metastatic renal cell carcinoma

Answer 34

immunotherapy, vascular endothelial growth factor inhibitors, mTOR inhibitors

Question 35

mTOR inhibitiors

Answer 35

It regulates cellular metabolism, growth, and proliferation, and therefore is a target for the development of a number of mTOR inhibitors. There is also some cross talk with the VHL pathway. examples include temsirolimus and everolimus, non-infectious pneumonitis with mTOR inhibition (14%) Rule out infectious etiology. Options for treatment: Treatment interruption, Dose reduction, Treatment discontinuation, Corticosteroids.

Question 36

vascular endothelial growth factor inhibitors

Answer 36

examples include sorafenib, sunitinib, bevacizumab, pazopanib, axitinb. Hypertension is a side effect

Question 37

examples of immunotherapy

Answer 37

HD-IL2, IFN

Question 38

high dose interleukin-2

Answer 38

nterleukin-2 (IL-2) is a cytokine produced endogenously by activated T cells and is commercially available as aldesleukin (Proleukin), a human recombinant product. IL-2 is effective in the treatment of a variety of malignancies, including renal cell carcinoma and melanoma, because it has both immune-modulating and antitumor properties. Although high-dose interleukin-2 (IL-2, Proleukin), a highly toxic agent used in the treatment of renal cell carcinoma and melanoma, was initially associated with treatment-related mortality, it can, in the appropriate setting, be administered safely. High-dose IL-2 is associated with significant morbidity; however, the incidence and severity of toxicities have decreased as clinicians have gained experience with this agent and implemented toxicity prevention and management strategies.

Question 39

BAP1

Answer 39

a deubiquitinating enzyme that in humans is encoded by the BAP1 gene. BAP1 functions as the catalytic subunit of the Polycomb repressive deubiquitinase (PR-DUB) complex, which controls homeobox genes by regulating the amount of ubiquitinated Histone H2A in Nucleosomes bound to their promoters. In flies and humans, the PR-DUB complex is formed through the interaction of BAP1 and ASXL1 (Asx in fruit flies) BAP1 has also been shown to associate with other factors involved in chromatin modulation and transcriptional regulation, such as Host cell factor C1, which acts as an adaptor to couple E2F transcription factors to chromatin-modifying complexes during cell cycle progression. BAP1 somatic mutations were identified in a small number of breast and lung cancer cell lines, but BAP1 was first shown to act as a tumor suppressor in cultured cells, where its deubiquitinase (UCH) domain and Nuclear localization sequences were required for BAP1 to suppress cell growth

Question 40

thermodynamic challenge for membrane fusion

Answer 40

you need to strip the water molecules from inbetween the charged polar heads and overcome the repulsive properties. This does not happen spontaneously. Machinary is required

Question 41

structure of alpha helix on SNARE proteins

Answer 41

they have hydrophobic region on one side and hydrophilic on the other this allows them to form coiled- coil domain

Question 42

Coiled coils

Answer 42

a structural motif in proteins in which 2-7 alpha-helices are coiled together like the strands of a rope (dimers and trimers are the most common types). Coiled coils usually contain a repeated pattern, hxxhcxc, of hydrophobic (h) and charged (c) amino-acid residues, referred to as a heptad repeat. The positions in the heptad repeat are usually labeled abcdefg, where a and d are the hydrophobic positions, often being occupied by isoleucine, leucine, or valine. Folding a sequence with this repeating pattern into an alpha-helical secondary structure causes the hydrophobic residues to be presented as a ‘stripe’ that coils gently around the helix in left-handed fashion, forming an amphipathic structure. The most favorable way for two such helices to arrange themselves in the water-filled environment of the cytoplasm is to wrap the hydrophobic strands against each other sandwiched between the hydrophilic amino acids. Thus, it is the burial of hydrophobic surfaces that provides the thermodynamic driving force for the oligomerization. The packing in a coiled-coil interface is exceptionally tight, with almost complete van der Waals contact between the side-chains of the a and d residues.

Question 43

N-ethylmaleimide Sensitive Factor (NSF)

Answer 43

It is a homolog of helicase, is a ATPase, it unwinds the SNARE complex. It functions as a hexamor. It binds to the end of the SNARE and hydrolizes 6 ATPs to twist and unwind SNARE. AlphaSNARE is an adapter.

Question 44

n-sec1

Answer 44

after the NSF does its work, SNARE is denatured. Nsec1 comes in and helps SNAREs to reform their correct conformation.

Question 45

envelope virus

Answer 45

Many viruses (e.g. influenza and many animal viruses) have viral envelopes covering their protective protein capsids. The envelopes typically are derived from portions of the host cell membranes (phospholipids and proteins), but include some viral glycoproteins. Functionally, viral envelopes are essential to entry into host cells. They may help viruses avoid the host immune system. Glycoproteins on the surface of the envelope serve to identify and bind to receptor sites on the host’s membrane. The viral envelope then fuses with the host’s membrane, allowing the capsid and viral genome to enter and infect the host. Usually, the cell from which the virus itself buds will often die or be weakened, and shed more viral particles for an extended period. The lipid bilayer envelope of these viruses is relatively sensitive to desiccation, heat, and detergents, therefore these viruses are easier to sterilize than non-enveloped viruses, have limited survival outside host environments, and typically must transfer directly from host to host. Enveloped viruses possess great adaptability and can change in a short time in order to evade the immune system. Enveloped viruses can cause persistent infections.

Question 46

capsid proteins

Answer 46

the protein shell of a virus. It consists of several oligomeric structural subunits made of protein called protomers. The observable 3-dimensional morphological subunits, which may or may not correspond to individual proteins, are called capsomeres. The capsid encloses the genetic material of the virus. The functions of the virion are to protect the genome, deliver the genome and interact with the host. The virion must assemble a stable, protective protein shell to protect the genome from lethal chemical and physical agents. These include forms of natural radiation, extremes of pH or temperature and proteolytic and nucleolytic enzymes. Delivery of the genome is also important by specific binding to external receptors of the host cell, transmission of specific signals that induce uncoating of the genome, and induction of fusion with host cell membranes.

Question 47

envelope protein

Answer 47

Env is a viral protein that serves to form the viral envelope. The expression of the env gene enables retroviruses to target and attach to specific cell types, and to infiltrate the target cell membrane. In HIV, the env gene codes for gp160; gp160 is later processed by a host cell protease to form the cleavage products gp120 and gp41.

Question 48

gp41

Answer 48

a subunit of the envelope protein complex of retroviruses, including Human immunodeficiency virus (HIV). Gp41 is a transmembrane protein that contains several sites within its ectodomain that are required for infection of host cells. In a free virion, the fusion peptides at the amino termini of gp41 are buried within the envelope complex in an inactive non-fusogengic state that is stabilized by a non-covalent bond with gp120. Gp120 binds to a CD4 and a co-receptor (CCR5 or CXCR4), found on susceptible cells such as Helper T cells and macrophages. As a result, a cascade of conformational changes occurs in the gp120 and gp41 proteins. The core of gp41 folds into a six helical bundle structure exposing the previously hidden gp41 fusion peptides which then assist in the fusion with the host cell. The activation process occurs readily which suggest that the inactive state of gp41 is metastable and the conformational changes allow gp41 to achieve its more stable active state. The interaction of gp41 fusion peptides with the target cell causes a formation of an intermediate, pre-hairpin structure which bridges and fuses the viral and host membranes together. The pre-hairpin structure has a relatively long half-life which makes it a target for therapeutic intervention and inhibitory peptides. Enfuvirtide (also known as T-20) is a fusion inhibitor drug that binds to the pre-hairpin structure and prevents membrane fusion and HIV-1 entry to the cell. The vulnerability of this structure has initiated development towards a whole spectrum of fusion preventing drugs. A variety of naturally-occurring molecules have also been shown to bind gp41 and prevent HIV-1 entry.

Question 49

fusion of influenza virus

Answer 49

Fusion is triggered by low pH, which induces conformational changes in the protein, leading to insertion of a hydrophobic ‘fusion peptide’ into the viral membrane and the target membrane for fusion. The cell membrane then engulfs the virus and the portion of the membrane that encloses it pinches off to form a new membrane-bound compartment within the cell called an endosome, which contains the engulfed virus. The cell then attempts to begin digesting the contents of the endosome by acidifying its interior and transforming it into a lysosome. However, as soon as the pH within the endosome drops to about 6.0, the original folded structure of the HA molecule becomes unstable, causing it to partially unfold and release a very hydrophobic portion of its peptide chain that was previously hidden within the protein. This so-called “fusion peptide” acts like a molecular grappling hook by inserting itself into the endosomal membrane and locking on. Then, when the rest of the HA molecule refolds into a new structure (which is more stable at the lower pH), it “retracts the grappling hook” and pulls the endosomal membrane right up next to the virus particle’s own membrane, causing the two to fuse together. Once this has happened, the contents of the virus, including its RNA genome, are free to pour out into the cell’s cytoplasm.

Question 50

HIV trans infection and dendritic cells

Answer 50

Dendritic cells initiate and sustain immune responses by migrating to sites of pathogenic insult, transporting antigens to lymphoid tissues and signaling immune specific activation of T cells through the formation of the immunological synapse. Dendritic cells can also transfer intact, infectious HIV-1 to CD4 T cells through an analogous structure, the infectious synapse. This replication independent mode of HIV-1 transmission, known as trans-infection, greatly increases T cell infection in vitro and is thought to contribute to viral dissemination in vivo. DCs can also “hide” the HIV virus. anti-retroviral drugs can wipe out all viruses in the blood however some will be in the DCs where they are not replicating. Later they can emerge and attack the CD4 T cells

Question 51

diffusion

Answer 51

the net movement of a substance (e.g., an atom, ion or molecule) from a region of high concentration to a region of low concentration. This is also referred to as the movement of a substance down a concentration gradient. A gradient is the change in the value of a quantity (e.g., concentration, pressure, temperature) with the change in another variable (e.g., distance). For example, a change in concentration over a distance is called a concentration gradient, a change in pressure over a distance is called a pressure gradient, and a change in temperature over a distance is a called a temperature gradient.

Question 52

osmosis

Answer 52

the spontaneous net movement of solvent molecules through a partially permeable membrane into a region of higher solute concentration, in the direction that tends to equalize the solute concentrations on the two sides.[1][2][3] It may also be used to describe a physical process in which any solvent moves across a semipermeable membrane (permeable to the solvent, but not the solute) separating two solutions of different concentrations.[4][5] Osmosis can be made to do work

Question 53

body fleuids composition

Answer 53

It turns out that about 99% of all the chemical particles in the 45 liters of body fluids are water molecules. Of the remaining 1%, about 5 out of every 6 are simple inorganic ions, principally Na+, K+, and Cl-. All of the rest – all carbohydrates, proteins, nucleic acids, lipids, and so forth – constitute a very small fraction of the total. changes in these composition can make you sick

Question 54

membrane permiability

Answer 54

it is permiable to water, Na (except there is a Na pump), K, and Cl. However, ICF is negative, hence the higher concentration of K+. The ECF is positive, hence the higher concentration of Cl-.

Question 55

hyperkalemia

Answer 55

Normal serum potassium levels are generally considered to be between 3.5 and 5.0 mEq/L. Levels above 5.0 mEq/L indicate hyperkalemia, and those below 3.5 mgEq/L indicate hypokalemia. Both hyperkalemia and hypokalemia are potentially life-threatening. For example, recent data evaluating the relationship between mortality and levels of potassium in patients with chronic kidney disease who were not on dialysis, found that lower serum potassium levels (i.e., 5.5 mmol/L), on the other hand, were associated with cardiovascular events (or the composite endpoint of cardiovascular events or death). These observations may suggest the need for reevaluation of what constitutes a “normal range” for serum potassium in certain patient populations.[9] Potassium is the most abundant intracellular cation and about 98% of the body’s potassium is found inside cells, with the remainder in the extracellular fluid including the blood. Membrane potential is maintained principally by the concentration gradient and membrane permeability to potassium with some contribution from the Na+/K+ pump. The potassium gradient is critically important for many physiological processes, including maintenance of cellular membrane potential, homeostasis of cell volume, and transmission of action potentials in nerve cells. Hyperkalemia develops when there is excessive production (oral intake, tissue breakdown) or ineffective elimination of potassium. Ineffective elimination can be hormonal (in aldosterone deficiency) or due to causes in the renal parenchyma that impair excretion. Increased extracellular potassium levels result in depolarization of the membrane potentials of cells due to the increase in the equilibrium potential of potassium. This depolarization opens some voltage-gated sodium channels, but also increases the inactivation at the same time. Since depolarization due to concentration change is slow, it never generates an action potential by itself instead, it results in accommodation. Above a certain level of potassium the depolarization inactivates sodium channels, opens potassium channels, thus the cells become refractory. This leads to the impairment of neuromuscular, cardiac, and gastrointestinal organ systems. Of most concern is the impairment of cardiac conduction which can result in ventricular fibrillation or asystole.

Question 56

cell volume regulation

Answer 56

all volume changes are produced by the movement of water, or, stated in reverse, only water movement can change the volume of a cell.

Question 57

van hoft equcation

Answer 57

The quantitative relationship between the osmotic suction and the pressure one has to exert in order to balance it was given last century by a Dutch physicist named Van’t Hoff. He found that the amount of pressure you have to apply is proportional to the difference in solute concentration on the two sides of the (semi-permeable) membrane: p = RT.DC, where p = osmotic pressure. The constants of proportionality are the gas constant (R) and the temperature (T, in degrees Kelvin).

Question 58

How do cells counter osmotic force

Answer 58

1. A simple solution to the problem would be to make the cell membrane impermeable to water, as well as to the internal solute molecules. However, this would create many new problems for the cell; growing cells, for example, must have a way of accumulating water as they grow (remember, cells are mostly water by volume), and therefore they must have surface membranes permeable to water. There are some special kinds of cell membranes that do exhibit extremely low water permeability (sweat glands, some cells in the kidney). In general, however, plasma membranes are quite highly permeable to water (although diffusion of water, and all substances we will consider, through even the most permeable membrane, is still thousands of times slower than diffusion in free solution). 2. Another solution (Fig. 2, middle) is to build a strong wall around the cell, and so keep the cell from swelling by brute force (that is, apply a hydrostatic force to counter the osmotic force drawing water into the cell). Plant cells, fungal cells, and bacteria do just this, by building tough cell walls outside their plasma membranes. But it’s an expensive way to go, in the sense that the osmotic force is not trivial (as shown on the right, a small container of pure water would support a column of a 1 M solution, from which it is separated by a semi-permeable membrane, that is 900 feet high), so that cell walls require considerable metabolic resources (building materials, energy for synthesis, etc.), and greatly limit cell shape (try to imagine a neuron in the cerebral cortex, with all of its delicate dendrites surrounded by a cell wall!). 3. The solution in animal cells to the problem of volume control is to fight fire with fire, and balance the osmotic force osmotically, by having solute molecules in the ECF, in order to balance those in the ICF. Thus, the concentration (or, more accurately, the activity) of water is exactly the same in the ICF and ECF. How simple physiology can be. There are two further important points about osmotic balance: First, the chemistry doesn’t matter: the ECF solutes can be different from the ICF solutes (and they are: mainly NaCl in the ECF, and K+ and big anions inside the cell). What matters is that the total concentration of solute particles, that is, the osmolarity of the two solutions is the same. Second, the solutes in the ECF must be nonpermeating (to match those in the ICF); putting a permeating solute, like glycerol, in the ECF would do no good at all in the long run, as the example problems below illustrate.

Question 59

Osmolarity

Answer 59

Osmolarity is the total concentration of solute particles: for example, a 1 M solution of CaCl2 gives a 3 osM solution (3 solute particles/molecule dissolved). If the whole system is at equilibrium, then each individual species must be at equilibrium. This means that if a substance can cross the membrane, then the concentration on the two sides must be the same at equilibrium. Thus, if glycerol is present, [Gly]i must equal [Gly]o at equilibrium. Because water can always cross the membrane, [H2O]i= [H2O]o at equilibrium. Now ordinarily, we don’t talk about the “concentration of water”. An equivalent statement is that the osmolarities are the same inside and out: osMi= osMo.

Question 60

Time course of volume changes

Answer 60

example F (600 mosM glycerol) and consider what happens to cell volume immediately after the cell is exposed to the solution. The test solution is hyperosmotic to the cell, so initially water will leave the cell. Glycerol of course will enter the cell. Cell volume depends only on the movement of water, so the cell will shrink at first. But then, as glycerol continues to enter, the internal osmolarity will rise, and at some point the movement of water will reverse direction, and begin to enter the cell. From then on, glycerol and water will both enter, swell beyond about 140% of their resting volume). The point is that the rate of change of volume can tell us something about the way solutes cross the membrane. The exact time course of volume change will depend on how easily (relative to water) the solute can cross the membrane. If it crosses slowly, the cell will stay shrunken for a longer period, although eventually it must swell and burst.

Question 61

Reflection coefficient

Answer 61

Another consequence of a permeating solute is that it will not exert as large an osmotic force as a nonpermeating solute at the same concentration. How much the osmotic pressure is diminished depends again on how easily the molecule can cross the membrane, relative to water. For example, consider a mixture of H2O and tritiated water, THO (one H is replaced with one T). The THO can be considered as a solute. Of course, it exerts no osmotic pressure, because THO and H2O cross membranes with equal ease. A molecule that crosses half as easily as water will exert half of the ideal osmotic pressure of a non-permeating solute. The Van’t Hoff equation, which relates osmotic pressure to hydrostatic pressure, is modified to account for this non-ideal behavior by multiplying by a factor called the reflection coefficient, which has a range from zero (for THO) to one (for nonpermeating solute). It’s a measure of how well the membrane ‘reflects’ the solute. The modified equation is: π=σRTΔC where σ = reflection coefficient, R=gas constant, T=temperature, and ΔC = difference in solute concentration across the membrane.

Question 62

cerebral edema in DKA

Answer 62

Often, when children first present with diabetes mellitus (Type I, or childhood diabetes), their body fluids are hyperosmotic due to the high concentration of glucose in the ECF (lack of insulin prevents glucose uptake by most cells). As insulin is administered in the clinic, the glucose is taken up by cells, and cellular metabolism is rescued. Glucose concentration in the ECF falls. The potential danger concerns the brain. Capillaries in the brain have a much lower permeability to nearly all substances than do systemic capillaries (you will study this ‘blood-brain barrier’ in detail later). Thus, if plasma osmolarity falls too quickly (insulin given too quickly), the glucose stranded in the brain ECF will create an osmotic gradient across the brain capillaries (the plasma will be hypo-osmotic) and the glucose in the brain ECF will osmotically suck water out of the capillaries. Anywhere else in the body, this would produce swelling (edema), but the brain is encased in rigid bone, and cannot swell very much. Instead, the pressure in the brain rises, which can seriously disrupt brain function, sometimes with fatal consequences. This condition is called cerebral edema. The danger is minimized by giving insulin slowly. The take-home message is simple: avoid rapid changes in plasma osmolarity.

Question 63

clysis

Answer 63

the nonoral insertion or injection of a fluid into tissue spaces, the rectum, or the abdominal cavity, such as the administration of an enema. It is used when IV access is not possible. sticking an infant’s vein is not easy, and in decades past, before special IV kits were developed (and even today, in underserved regions), the fluid (~300 mosM glucose in water) was give subcutaneously (‘clysis’; usually in the lower back where skin is loose), whence it would shortly be absorbed into the blood via capillaries located there. Surprisingly, the ‘blister’ of injected fluid at first increased in size. Fluid was being sucked out of the dehydrated infant’s circulation, exactly the opposite of the intended treatment! (and risking circulatory collapse and death) Eventually, however, the blister decreased in size and disappeared as the injected fluid entered the circulation. NaCl crosses capillaries more easily than glucose does.While the injected fluid (isotonic glucose) had the same osmolarity as plasma, the solutes were very different – glucose on one side, NaCl on the other (blood) side. Because the capillaries were more permeable to NaCl than to glucose, it (NaCl) rapidly diffused out of the blood into the injected fluid (which contained no NaCl). This of course raised the osmolarity of the fluid in the blister , which caused water to move into the blister from the blood. To say it another way: the reflection coefficient across capillaries is higher for glucose than for NaCl. Thus a 300 mM solution of glucose exerts a higher osmotic suction than 300 mosM NaCl, so that water moves into the glucose solution. Eventually, as the sucrose diffuses into the blood, the blister will shrink.

Question 64

The Nernst equation

Answer 64

In electrochemistry, the Nernst equation is an equation that relates the reduction potential of a half-cell (or the total voltage (electromotive force) of the full cell) at any point in time to the standard electrode potential, temperature, activity, and reaction quotient of the underlying reactions and species used. When the reaction quotient is equal to the equilibrium constant of the reaction for a given temperature, i.e. when the concentration of species are at their equilibrium values, the Nernst equation gives the equilibrium voltage of the half-cell (or the full cell), which is zero; at equilibrium, Q=K, ΔG=0, and therefore, E=0. E = (60/z) log10 (Co/Ci), where E= equilibrium potential; Co= outside concentration of the ion; Ci= inside concentration; R= gas constant; T= temperature; z= valence of the ion in question; F= Faraday constant. E is expressed in millivolts, mV. Please note that we arbitrarily define E as the potential of the inside of the cell with respect to the outside. Thus, if E= -40 mV, we say that the inside must be 40 mV negative to the outside in order for the ion in question to be at equilibrium.

Question 65

Electric Force

Answer 65

opposite charges attract and like charges repel each other. These forces are very strong, much, much stronger in fact than osmotic forces.

Question 66

electrochemical gradient.

Answer 66

1. The passive movement of an ion across the membrane is governed by two forces: its concentration difference and the electrical potential difference across the membrane (the membrane potential). The two forces, combined, make an electrochemical gradient. 2. Membrane potentials are produced by only one thing, namely an imbalance in the number of cations and anions inside a cell. 3. The electric force is very much more powerful than the diffusional force produced by a concentration difference, which means that relatively few excess ions are needed to counter large concentration differences. In other words, it is very safe and appropriate to consider that the concentration of chloride in the cell does not change, or more generally, bulk solutions are always electrically neutral. It is called the equilibrium potential for the ion in question.

Question 67

equilibrium potential

Answer 67

the membrane potential at which there is no net (overall) flow of that particular ion from one side of the membrane to the other. It relates a concentration gradient to an electrical force. Specifically, it is the electrical potential difference across the membrane that must exist if the ion is to be at equilibrium at the given concentrations. The Nernst equation can be applied to any ion, and thus there are as many equilibrium potentials as there are ion species. The equilibrium potential for chloride is ECl; for sodium it’s ENa; and so on – EK, ECa, EHCO3, EH, etc.Suppose [K+]i = 140 mM and [K ]o = 4 mM. Will EK be positive or negative? Well, to keep potassium from diffusing down its concentration gradient out of the cell, we must make the inside of the cell negative to hold on the positively-charged potassium ions. Nernst tells us that EK = -92.6 mV. Suppose an ion is at equal concentrations across a membrane. What will be its equilibrium potential? (Answer: zero). Equilibrium potentials are not real voltages - that’s why they’re written with an E (for “electromotive force”) not a V. The real voltage is called the membrane potential, Vm, and can be measured by impaling the cell with a microelectrode. If E and Vm are the same, then we can say that Cl- is distributed at its electrochemical equilibrium. We do not have to postulate the existence of a Cl- “pump” to maintain it in a state away from equilibrium.

Question 68

What if Vm is not the same as ECl?

Answer 68

Then we can say that either a) the membrane is impermeable to Cl-, or b) Cl- must be pumped across the membrane, because it is not distributed at equilibrium. In this case, Cl- ions will perpetually diffuse in one direction, ‘down’ their electrochemical energy gradient, and an exactly equal number of Cl- ions will be pumped in the opposite direction. So the Cl- concentration in the cell won’t change so long as the pump works.

Question 69

The Principle of Electrical Neutrality

Answer 69

bulk solutions (inside and out) have to be electrically neutral: the total cation concentration in the external solution must equal the total anion concentration in the external solution. The same holds for the internal solution: [cations]i = [anions]i. However, this clearly contradicts something said earlier, namely that membrane potentials arise from an excess of one type of ion in the cell; a cell with Vm= -50 mV contains more anions than cations. Well, both are correct. The key is once again to note that the electric force is so awesomely powerful that, in a cell, the excess number of anions is very, very small compared to the total number of ions present in the cell. So electrical neutrality is only an approximation, but it is a very good one.

Question 70

The Donnan rule

Answer 70

the behavior of charged particles near a semi-permeable membrane that sometimes fail to distribute evenly across the two sides of the membrane.[1] The usual cause is the presence of a different charged substance that is unable to pass through the membrane and thus creates an uneven electrical charge. The problem is, can both permeating ions, K+ and Cl-, be at equilibrium simultaneously? Well, if they can work things out, then there must be an arrangement whereby ECl = EK = Vm, because a cell can have only one value of membrane potential. Then we can write: EK = 60 log ([K+]o/[K+]i)= ECl= -60 log ([Cl-]o/[Cl-]i). This simplifies to the Donnan Rule: [K+]o.[Cl-]o = [K+]i.[Cl-]i

Question 71

Hyponatremia

Answer 71

Hypervolemic hyponatremia —Both sodium & water content increase: Increase in sodium content leads to hypervolemia and water content to hyponatremia. Total body water and sodium are regulated independently. Euvolemic hyponatremia — there is volume expansion in the body, no edema, but hyponatremia occurs. Hypovolemic hyponatremia — The hypovolemia (extracellular volume loss) is due to total body sodium loss. The hyponatremia is caused by a relatively smaller loss in total body water.

Question 72

The Law of Mass Action

Answer 72

The rate, vf, of the forward reaction is proportional to product of the concentrations [A] and [B], Vf=Kv[A][B], The rate, vr, of the reverse reaction is proportional to product of [C] and [D], The equilibrium constant Keq = kf / kr = [C][D] / [A][B]

Question 73

Water is in equilibrium with its component ions H+ and OH-

Answer 73

Keq = 1.8 X 10-16 = [H+] [OH-]/[H2O], The concentration of water is 55.5Molar so: 55.5 X 1.8 X 10-16 = [H+] [OH-] = 1.0 X 10^-14, Expressed as logarithms: log [H+] + log [OH-] = -14 or pH +pOH = 14, A solution is neutral when [H+] = [OH-], in other words pH = pOH = 7

Question 74

acids and bases

Answer 74

An acid is a compound or chemical group that acts as a proton donor, An base is a compound or chemical group that acts as a proton acceptor, Strong acids like HCl and bases like NaOH are completely dissociated in water whereas weak acids and bases are reversibly protonated and deprotonated. Many drugs, amino acids, peptides, proteins and nucleic acids are weak acids or bases.

Question 75

pKa

Answer 75

pKa = -log Ka = - log ([H+] [A-]/[HA]), the lower the pKa, the stronger the acid
the higher the pKa, the stronger the base

Question 76

acetoacetic acid

Answer 76

the organic compound with the formula CH3COCH2COOH. It is the simplest beta-keto acid group and like other members of this class is unstable. When ketone bodies are measured by way of urine concentration, acetoacetic acid, along with beta-hydroxybutyric acid (BHB), and acetone, is what is detected.

Question 77

Examples of medically important weak bases

Answer 77

purines, pyrimidines, amphetamines, procainamide, nortriptylene, local anesthetics

Question 78

procainamide

Answer 78

It is a sodium channel blocker which blocks open sodium channels and prolongs the cardiac action potential (outward potassium (K+) currents may be blocked). This results in slowed conduction, and ultimately the decreased rate of rise of the action potential, which may result in widening of QRS on electrocardiogram. This drug is used for both supraventricular and ventricular arrhythmias. For example, it can be used to convert new-onset atrial fibrillation, though it is suboptimal for this purpose. It can also be used to treat Wolff-Parkinson-White syndrome by prolonging the refractory period of the accessory pathway.

Question 79

Nortriptyline

Answer 79

Like other tricyclic antidepressants, nortriptyline also blocks sodium channels, possibly accounting in part for its analgesic action

Question 80

Helicobacter pylori

Answer 80

lives in the stomach, creates a neutral pH environment using erease

Question 81

biological importance of pH

Answer 81

it can strongly affect enzyme activity (eg pepsin, trypsin, and alkaline phosphatase). It also affects drug availability (eg for drugs with ionizable groups, a critical factor governing uptake and excretion will be the pH of the local environment. Asprin can be protonated in the stomach allowing it to be taken up, in the duodenum pH is higher)

Question 82

alkaseltcer

Answer 82

NaHCO3, HCO3 is taken up into the blood stream. Therefore, kidney disease is a contraindication.

Question 83

The Henderson-Hasselbalch equation

Answer 83

This equation is a log-based expression of relationship between an acid HA and its dissociation products in water . Ka = [H+] [A-]/[HA], since pKa is –logKa and pH is – log[H+] then we can rewrite the expression as pKa =pH - log([A-]/[HA]), pH = pKa + log([A-]/[HA]). When an acid or base is 50% deprotonated and 50 % protonated then pH=pKa. At less than half dissociated pH < pKa. if pH>pKa the acid is predominately deprotonated.

Question 84

buffer

Answer 84

an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small or moderate amount of strong acid or base is added to it and thus it is used to prevent changes in the pH of a solution. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. Many life forms thrive only in a relatively small pH range so they utilize a buffer solution to maintain a constant pH. One example of a buffer solution found in nature is blood. Buffer solutions achieve their resistance to pH change because of the presence of an equilibrium between the acid HA and its conjugate base A-. When some strong acid is added to an equilibrium mixture of the weak acid and its conjugate base, the equilibrium is shifted to the left, in accordance with Le Chatelier’s principle. Because of this, the hydrogen ion concentration increases by less than the amount expected for the quantity of strong acid added. Similarly, if strong alkali is added to the mixture the hydrogen ion concentration decreases by less than the amount expected for the quantity of alkali added. The effect is illustrated by the simulated titration of a weak acid with pKa = 4.7. The relative concentration of undissociated acid is shown in blue and of its conjugate base in red. The pH changes relatively slowly in the buffer region, pH = pKa ± 1 ([A-]/[HA] = 0.1 to 10 ), centered at pH = 4.7 where [HA] = [A-]. The hydrogen ion concentration decreases by less than the amount expected because most of the added hydroxide ion is consumed in the reaction

Question 85

bicarbonate buffering system

Answer 85

an important buffer system in the acid-base homeostasis of living things, including humans. As a buffer, it tends to maintain a relatively constant plasma pH and counteract any force that would alter. In this system, carbon dioxide (CO2) combines with water (H2O) to form carbonic acid (H2CO3), which in turn rapidly dissociates to form hydrogen ions (H+) and bicarbonate (HCO3- ). The carbon dioxide - carbonic acid equilibrium is catalyzed by the enzyme carbonic anhydrase; the carbonic acid - bicarbonate equilibrium is simple proton dissociation/association and needs no catalyst.Any disturbance of the system will be compensated by a shift in the chemical equilibrium according to Le Chatelier’s principle. For example, if one attempted to acidify the blood by dumping in an excess of hydrogen ions (acidemia), some of those hydrogen ions will associate with bicarbonate, forming carbonic acid, resulting in a smaller net increase of acidity than otherwise. This buffering system becomes an even more powerful regulator of acidity when it is coupled with the body’s capacity for respiratory compensation, in which breathing is altered to modify the amount of CO2 in circulation. In the above example, increased ventilation would increase the loss of CO2 to the atmosphere, driving the equilibria above to the left. The bicarbonate buffer system is OPEN modulated by regulation of CO2 loss in the lung and H+ /HCO3- excretion in the kidney allowing the system to be an effective buffer 1.3 units above its pKa. pH = 6.1 + log ([HCO3-] / .03Pco2),

Question 86

the three reactions of the bicarbonate buffer system

Answer 86

reaction 1: pKa = 3.8, pH = 3.8 + log([HCO3-]/[H2CO3]), H-H, Normal [HCO3-] = 24 mM, Reaction 2 (catalyzed by carbonic anhydrase): K2 = [CO2]d / [H2CO3] = 214, [H2CO3]= [CO2]d / 214, [CO2]d (mM0= 0.03PCO2 mmHg, normal Pco2=40mmHg, normal [CO2]d = 1.2 mM. [H2CO3]= 0.03Pco2 / 214, therefore reaction 3: pH = 3.8 + log 214 ([HCO3-] / .03Pco2), or pH = 6.1 + log [HCO3-]mM/ .03Pco2mmHg

Question 87

respiratory acidosis

Answer 87

a medical emergency in which decreased ventilation (hypoventilation) causes increased blood carbon dioxide concentration and decreased pH (a condition generally called acidosis). Metabolism rapidly generates a large quantity of volatile acid (H2CO3) and nonvolatile acid. The metabolism of fats and carbohydrates leads to the formation of a large amount of CO2. The CO2 combines with H2O to form carbonic acid (H2CO3). The lungs normally excrete the volatile fraction through ventilation, and acid accumulation does not occur. A significant alteration in ventilation that affects elimination of CO2 can cause a respiratory acid-base disorder. The PaCO2 is maintained within a range of 35–45 mm Hg in normal states. Alveolar ventilation is under the control of the central respiratory centers, which are located in the pons and the medulla. Ventilation is influenced and regulated by chemoreceptors for PaCO2, PaO2, and pH located in the brainstem,and in the aortic and carotid bodies as well as by neural impulses from lung stretch receptors and impulses from the cerebral cortex. Failure of ventilation quickly increases the PaCO2. In acute respiratory acidosis, compensation occurs in 2 steps. 1) The initial response is cellular buffering that occurs over minutes to hours. Cellular buffering elevates plasma bicarbonate (HCO3−) only slightly, approximately 1 mEq/L for each 10-mm Hg increase in PaCO2. 2)
The second step is renal compensation that occurs over 3–5 days. With renal compensation, renal excretion of carbonic acid is increased and bicarbonate reabsorption is increased. For instance, PEPCK is upregulated in renal proximal tubule brush border cells, in order to secrete more NH3 and thus to produce more HCO3−. Respiratory acidosis does not have a great effect on electrolyte levels. Some small effects occur on calcium and potassium levels. Acidosis decreases binding of calcium to albumin and tends to increase serum ionized calcium levels. In addition, acidemia causes an extracellular shift of potassium, but respiratory acidosis rarely causes clinically significant hyperkalemia.

Question 88

respiratory alkalosis

Answer 88

a medical condition in which increased respiration elevates the blood pH (a condition generally called alkalosis). It is one of four basic categories of disruption of acid-base homeostasis. This condition is commonly associated with a decrease in PaCO2 (hyperventilation). Respiratory alkalosis generally occurs when some stimulus (see “Causes” below) makes a person hyperventilate. The increased breathing produces increased alveolar respiration, expelling CO2 from the circulation. This alters the dynamic chemical equilibrium of carbon dioxide in the circulatory system, and the system reacts according to Le Chatelier’s principle. Circulating hydrogen ions and bicarbonate are shifted through the carbonic acid (H2CO3) intermediate to make more CO2 via the enzyme carbonic anhydrase. The net result of this is decreased circulating hydrogen ion concentration, and thus increased pH (alkalosis). There is also a decrease in ionized blood calcium concentration.

Question 89

causes of hypokalemia

Answer 89

Normal plasma potassium levels are between 3.5 and 5.0 meq/l; about 98% of the body’s potassium is found inside cells, with the remainder in the extracellular fluid including the blood. A special case of potassium loss occurs with diabetic ketoacidosis. Hypokalemia is observed with low total body potassium and a low intracellular concentration of potassium. In addition to urinary losses from polyuria and volume contraction, also an obligate loss of potassium from kidney tubules occurs as a cationic partner to the negatively charged ketone, β-hydroxybutyrate. An increase in the pH of the blood can cause temporary hypokalemia by two mechanisms. First, the alkalosis causes a shift of potassium from the plasma and interstitial fluids into cells, perhaps mediated by stimulation of Na+-H+ exchange and a subsequent activation of Na+/K+ pump activity. Second, an acute rise of plasma HCO3- concentration (caused by vomiting, for example) will exceed the capacity of the renal proximal tubule to reabsorb this anion, and potassium will be excreted as an obligate cation partner to the bicarbonate. Metabolic alkalosis is often present in states of volume depletion, so potassium is also lost via aldosterone-mediated mechanisms.Disease states that lead to abnormally high aldosterone levels can cause hypertension and excessive urinary losses of potassium. A more common cause is excessive loss of potassium, often associated with heavy fluid losses that “flush” potassium out of the body. Typically, this is a consequence of diarrhea, excessive perspiration, or losses associated with surgical procedures. Vomiting can also cause hypokalemia, although not much potassium is lost from the vomitus. Rather, heavy urinary losses of K+ in the setting of postemetic bicarbonaturia force urinary potassium excretion

Question 90

anomalous rectifier

Answer 90

A channel that is “inwardly-rectifying” is one that passes current (positive charge) more easily in the inward direction (into the cell) than in the outward direction (out of the cell). It is thought that this current may play an important role in regulating neuronal activity, by helping to establish the resting membrane potential of the cell. By convention, inward current is displayed in voltage clamp as a downward deflection, while an outward current (positive charge moving out of the cell) is shown as an upward deflection. At membrane potentials negative to potassium’s reversal potential, inwardly rectifying K+ channels support the flow of positively charged K+ ions into the cell, pushing the membrane potential back to the resting potential. This can be seen in figure 1: when the membrane potential is clamped negative to the channel’s resting potential (e.g. -60 mV), inward current flows (i.e. positive charge flows into the cell). However, when the membrane potential is set positive to the channel’s resting potential (e.g. +60 mV), these channels pass very little charge out of the cell. Simply put, this channel passes much more current in the inward direction than the outward one. Note that these channels are not perfect rectifiers, as they can pass some outward current in the voltage range up to about 30 mV above resting potential.

Question 91

Osmole (mosM)

Answer 91

In chemistry, the osmole (Osm or osmol) is a non-SI unit of measurement that defines the number of moles of solute that contribute to the osmotic pressure of a solution. The term comes from the phenomenon of osmosis, and is typically used for osmotically active solutions. For example, a solution of 1 mol/L NaCl corresponds to an osmolarity of 2 osmol/L. The NaCl salt particle dissociates fully in water to become two separate particles: an Na+ ion and a Cl- ion. Therefore, each mole of NaCl becomes two osmoles in solution, one mole of Na+ and one mole of Cl-. Similarly, a solution of 1 mol/L CaCl2, gives a solution of 3 osmol/L (Ca2+ and 2 Cl-).

Question 92

Intracellular “big anions”

Answer 92

In real cells, the internal impermeant species (mostly proteins, SO4=, and HPO4=) carry a net negative charge. Assume that each A molecule carries n negative charges (in real cells, n = -1 to - 2).

Question 93

sodium-potassium pump

Answer 93

pumps sodium out of cells, while pumping potassium into cells. Active transport is responsible for cells’ containing relatively high concentrations of potassium ions but low concentrations of sodium ions. n fact, both Na+ and K+ ions must be present simultaneously, or the pump won’t work. For example, the pump can’t extrude Na+ from the cell very well if the potassium concentration in the ECF is low. Also, cells with a poisoned pump lose K+ as they gain Na+ (they still swell, because they also gain Cl- as Na+ leaks in). Thus, the sodium pump is really an obligatorily coupled sodium-potassium exchange pump. The Na/K pump is “saturable”, that is, it has an easily demonstrable maximum rate of activity of only about 100 cycles per second. In general, such saturation is characteristic of carrier mediated transport, but not of most fluxes through ion channels, which can transport millions of ions per second. Another property of the pump is that it is not a 1:1 pump; rather, the Na/K ratio is 3 Na+ for 2 K+. Thus, the pump is electrogenic, and not electro-neutral. This has the effect of making Vm a little bit more negative than it would otherwise be. The effect in most cases is very small, so that for our purposes we can ignore its electrogenicity. It does, however, become important in certain neurons, which you will study next year. A simple way to envision the Na/K pump is like a channel with 2 gates, one at the outside surface of the membrane, and one at the inside (see cartoon below). Only one gate can be open at a time; both gates are never open simultaneously. In addition, the ion binding affinity of the channel switches between Na and K, depending on which gate is open. ATP provides the energy for the transitions (gates swinging and affinity changing). The pump contains a large alpha subunit, and a smaller beta subunit, but 4the exact (atomic scale) structure is unknown (it has not yet been possible to make crystals suitable for x-ray diffraction studies).

Question 94

The Na/K pump cycle.

Answer 94

A. Both gates are closed, 2 K+ ions are inside. B. ATP binds, the inner gate opens, and affinity changes from K+ to Na+. So K+ leaves and Na+ enters. C. `ATP is split, leaving the pump phosphorylated. The inner gate closes. D. Spontaneously, the outer gate opens and affinity changes from Na+ to K+. D to A. The pump loses its phosphate group, and the outer gate closes, completing the cycle.

Question 95

What will determine the value of the measured membrane potential, Vm? In real cell biology

Answer 95

We have made a very fundamental change in going from the model cell to real biology. The cell is no longer in a state of equilibrium; it requires a constant input of energy (from metabolism) to keep its ICF composition from changing. It is in a steady state, which means that, like the model cell at equilibrium, the ion concentrations aren’t changing over time, but unlike the model cell, a constant input of energy is needed in the real cell (in the form of ATP, to drive the Na/K pump). Thus, in real cells, Na+ is constantly leaking into the cell, trying to get to equilibrium. What is equilibrium for Na+? It simply means getting Vm to equal ENa. For potassium, equilibrium is achieved when Vm is equal to EK. So a continuous struggle ensues between the two ions - Na+ leaking in, trying to pull Vm up to ENa, and K+ leaking out, trying to pull Vm down to EK. So, which one wins, Na or K? It’s different for different cells. If we were to take a micropipette and impale different cells all over the body, we would find a large range in resting membrane potentials. They are all negative, but range from nearly zero to nearly EK. What are the differences due to? The answer is: relative permeability. A cell with many more K+ channels than Na+ channels will have a membrane potential close to EK. Nerve and muscle cells are examples; Vm is -70 to -90 mV. Conversely, a cell with relatively more Na+ channels will have a membrane potential closer to ENa. Glial cells are nearly perfect ‘potassium electrodes’ (permeable only to K), while red blood cells are about equally permeable to Na and K. A second determinant of Vm is ion concentration. But in normal resting cells, this isn’t important, because all cells have similar intracellular compositions (rich in K) and are bathed in similar ECF (rich in Na). The most important exception to this, of great medical importance, is potassium. A small change in ECF potassium concentration has a big effect on EK and (in nerve, muscle, and other cells highly permeable to potassium) a big effect on Vm.

Question 96

Ohm’s Law

Answer 96

V=I*R where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. Here, V is the ‘driving force’ on the ion, I is the current carried by the ion, and R is the membrane resistance to the ion. It’s easier to think of conductance, G, which is the reciprocal of resistance (G=1/R), which is close to membrane permeability (the higher the permeability, the higher the conductance). Thus, Ohm’s law says, I=(Driving Force) times G. This says simply that the current carried by an ion (say, Na+) through the membrane is given approximately by the product of [driving force on the Na+ ion] times [sodium conductance] (conductance is proportional to the number of sodium channels that are conducting). It makes sense: more channels, more current, and more driving force, more current. What is the driving force on sodium? It is not just the value of membrane potential, Vm. Think of it this way: At what membrane potential will the net movement of Na+ be zero? That will occur when Vm= ENa (Na+ will then be at equilibrium). In other words, when Vm= ENa, the driving force is zero. Thus, the driving force at any instant is the difference between Vm and ENa. So the way we write Ohm’s Law for sodium is as follows: INa= GNa.(Vm-ENa). The same holds for K+: IK= GK.(Vm-EK). When these two currents are equal and opposite (INa = -IK) there will be no net movement of charge across the membrane (we are ignoring other ions), so Vm will be at its resting value. If we set INa = -IK and substitute, we get GNa. (Vm-ENa) = - GK.(Vm-EK). Solving this for Vm gives: Vm = (GKEK + GNaENa) / (GK + GNa).

Question 97

relative conductance

Answer 97

When the conductance for a substitute ion relative to that of a standard ion is determined. Remember that membrane potential is determined by the relative conductances or permeabilities of the membrane to various ions, not the actual values of conductances or permeabilities. So, when the sodium conductance becomes very large relative to the other conductances, the membrane potential approaches the sodium Nernst potential, VNa. Vm=(Ek+Gr*Ena)/(Gr+1). GR=GNa/GK.. GR is the ratio of sodium to potassium conductance. Notice that besides EK and ENa (which differ little from cell to cell), the only variable is GR, the relative conductance of the membrane to sodium and potassium. In summary, for all practical purposes (that is, under normal physiological conditions), membrane potential is determined by relative conductance. We can illustrate this with a simple example. Consider two cells. One (leaky membrane) has 100 K channels and 10 Na channels. The other (tight membrane) has only a tenth as many channels (10 K and 1 Na). Their membrane potentials will be the same (in both cases, GR = 0.1). But the cells are clearly different – how? The leaky cell will have ten times more sodium leaking in, and ten times more potassium leaking out, compared to the tight cell, and so will have to expend 10 times more energy to keep the Na/K pump working. Thus, the tight cell is ten times more efficient, energetically speaking.

Question 98

Goldman equation

Answer 98

is used in cell membrane physiology to determine the reversal potential across a cell’s membrane, taking into account all of the ions that are permeant through that membrane.

Question 99

The sodium/potassium pump and membrane potential.

Answer 99

While the Na/K pump is absolutely necessary over the long run, in many cells it plays only a negligible short term role. For example, if the Na/K pump is completely blocked by drugs, nothing much happens initially (in many cells). Gradually, of course., the cell will fill up with Na, and lose its K. As these changes occur, both ENa and EK move towards zero, and cells will depolarize. How long it takes depends simply on how big and how leaky the cell is. In a large, spherical cell (which has a low surface-to-volume ratio), with few ion channels, it might take many hours for these changes to occur. On the other hand, in a cell with a relatively large surface area (like many neurons in the brain, which have long, slender processes), and a high density of channels, blocking the sodium pump can cause them to fill up with Na and lose K in a matter of seconds.

Question 100

normal plasma value of sodium

Answer 100

140 mM

Question 101

normal intercellular value of potassium

Answer 101

140 mM

Question 102

normal plasma value of H2O

Answer 102

55M

Question 103

normal plasma value of potassium

Answer 103

4mM

Question 104

normal plasma value of chloride

Answer 104

115mM

Question 105

normal plasma value of bicarbonate (HCO3)

Answer 105

24mM

Question 106

normal plasma value of Calcium

Answer 106

1mM

Question 107

normal intercellular level of calcium

Answer 107

0.1microM

Question 108

normal intercellular level of hydrogen

Answer 108

0.1microM

Question 109

normal plasma value of hydrogen

Answer 109

40 nM

Question 110

normal plasma value of urine

Answer 110

1200mosM

Question 111

normal plasma osmolarity

Answer 111

300mosM