Chapter 1: Metabolic Fuels and Dietary ComponentsBook Notes

Mary-Catherine Ciuba

Metabolic requirements (Intro to Unit) Synthesize items for body that arent in diet o Four basic pathways to metabolize various components Fuel oxidative pathways (OIL-RIG) Ex - Calories ingested are converted to energy in order to move, think and reproduce Converts fuels into energy for biosynthetic and mechanical work Fuel storage and mobilization pathways Ex - the body can function during times between meals Biosynthetic pathways Ex amino acids are used to create proteins we naturally cannot make Anabolic build molecules Catabolic break molecules Anabolic + catabolic = metabolic Detoxification/waste disposal pathways Ex remove toxins from the body from cigarette smoke we breathe Xenobiotics - dietary components that have no value and are therefore disposed Protect internal environment from outside environment o Cooperation between tissues and changes in response due to external changes occur through either Transport pathways Intercellular pathways Fuel Metabolism Fuel comes from carbohydrates, fats and proteins Food is o Digested o Absorbed

o Oxidized *oxidative = catabolictherefore breaking down the molecule to release energy o And then produces energy Energy can be stored o Primarily as fat (triacylglycerols) o Glycogen (in muscle and liver) o Protein (in muscle) *these can be oxidized during times of famine and create energy! Require enough energy to live and energy to do PA Must also ingest essential fatty acids, amino acids, vitamins, minerals and water in order to survive Dietary components that are in excessed or already oxidized must be disposed of, as well as toxic chemicals

Dietary Fuels RESPIRATION RXN: Food + O2 -> Waste food + CO2 + heat + energy (ATP) o ATP provides the energy that drives most of the energy-consuming processes in the cell Biosynthetic rxns Muscle contx Active transport across membranes o Once energy is released, ATP -> ADP Carbohydrates o Major carbs Starch poly Sucrose di Lactose di Fructose mono Glucose mono o Produces 4 kcal / 1 g carb o Carb molecules break down quickly when consumed because they are partially oxidized before they enter our bodies Proteins

o o o o o Fats o o o

Amino acids lined in linear chains make up proteins Contain 16% nitrogen, as well as C, H and O Proteins are broken into amino acids and enter the blood RXN: Protein + O2 -> CO2 + H2O + NH4+ + energy Produces 4 kcal / 1 g protein

Triglycerides < lipids < fats Triglyceride is 3 fatty acids combined Fats are more dense and therefore release more energy 9 kcal / 1 g fat o saturated acids = 0 double bonds o monounsaturated acids = 1 double bond o polyunsaturated acids = >1 double bond Alcohol o 1 g Ethanol = 7 kcal Body Fuel Stores Intro o We can fuel our bodies without constantly eating Stores are light weight, virtually limitless and easily oxidized/used o Fat is the primary storage site, found in adipose tissue o Glycogen in muscles and liver is accessible too o Protein is found in muscle Fat o A lipid commonly known among us o Makes up 85% sotred calories o Triglycerides are more efficient as a fuel b/c they can store more energy than carbs or protein can o Triglycerides are only 15% water, when compared to the common 80% Glycogen o Carbs stored in liver and muscles throughout the body o Liver glycogen is used to maintain blood glucose levels at a normal level between meals o Also used to fuel exercise

o Almost all cells have a small glucose supply for emergencies Protein o Not solely a fuel storage o Proteins add in body movement, enzymes and structural components for cells Daily Energy Expenditure Intro o To not gain or lose weight, we must consume the same amount of calories as our daily energy expenditure o DEE = RMR/BMR + PA + DIT Resting metabolic rate (RMR) o RMR measure of the energy required to maintain life (breathing, heart beating, etc) o BMR is defined as energy expenditure of a personal mentally and physically at rest 12 to 18 hours after a meal o RMR is resting metabolic rate RMR and BMR differ very little in value o BMR is typically measured in kcal/day Depends on body size, age, sex and other factors BMR is higher in bigger people and men Hyperthyroidism and hypothyroidism cause BMR to increase or decrease BMR increases with pregnancy and lactation Growing children have higher BMRs than adults and declines as we age 1 kcal/kg/hr is the best estimation we have and multiplying it by a persons weight All of these tend to be estimations anywaystoo many variables Physical Activity o Different EE with different activities (EE depends on difficulty of task) Diet-Induced Thermogenesis o Diet induced thermogenesis and thermic effect of food are interchangeable

o Metabolic rate increases after eating food in order for it to digest, absorb, distribute and store nutrients o 10% of kcals ingested are burned off to morph the food o DIT is usually left out b/c its virtually non-existent Calculations of Daily Energy Expenditure o DEE = RMR + 0.3-2.0 RMR (for PAdepends on active lifestyle) Healthy Body Weight o BMI helps determine if people are in a healthy range for weight o 20-24.9 is overweight o 25-29.9 is obese Weight Gain and Loss o To maintain body weight, must have calories = 0 o Gain, eat more and less to loose lbs o 3500 kcal ~ 1 lb o fad diets move water weight arounddont work Dietary Requirements Intro o Must ingest essential nutrientsbody cant make them if lacking o RDA is the average intake needed to meet the requirement of all healthy individuals in an age and gender group o AI recommends a value when RDA data is unavailable Carbohydrates o Dont completely eliminate carbs, but an excess is not healthy Can convert carbs from proteins but dangerous when low and have to use fats instead Essential fatty acids o Need some dietary lipids for optimal health, like carbs o Essential fatty acids cant be mademust be ingested o Like a-linoleic and a-linolenic acids, EPA and DHA Protein o Intro Need 0.8 g per kg BW 50 g women and 60 g men always look for high quality proteins o Essential amino acids

9 essential amino acids and 11 the body can make lysine, isoleucine, leucine, threonine, valine, tryptophan, phenylalanine, methionine and histidine some are conditionally essential, depending on amount of essential proteins o Nitrogen balance Amino acids are constantly bonding and breaking proteins in order to be used again Can be used again, serve as precursors for synthesis or oxidized for energy Urea is the nitrogen waste from the oxidation reaction In urine mainlybut also in sweat, feces and epithelial cells Balance shows nitrogen in and out of the body (+) means more in than out occurs in growing individuals (-) means more out than in amino acids will be more degraded and bodily functions impaired healthy adults are in nitrogen balance Vitamins o Diverse nutrients needed in small quantities for health, growth and survival Deficiencies cause many diverse symptoms and ultimately death o 2 types: water and fat soluble vitamins o used for synthesis of coenzymes (complex organic molecules that assist enzymes in catalyzing biochemical reactions) some vitamins also act as hormones o deficiencies vitamin C scurvy thiamin beriberi niacin pellagra vitamin D rickets o vitamins cannot be synthesized usually, and if so, in very little amounts

o tolerable intake upper limit fatal to ingest more than this amount of a vitamin most often occur with ingestion of single vitamins, not from foods Minerals o Classifications Electrolytes dissolved into fluids in the body and maintain water balance as well as neutralize charges Sodium Potassium Chloride Minerals required in large quantities Magnesium Phosphorus Sulfur Again, dangerous in excess Ultratrace minerals Water o Makes up to 4/5 BW o Amount required per day depends on metabolism, amount excreted from the body and PA Dietary Guidelines Intro o These recommended food choices on the pyramid can reduce the risk of developing chronic and degenerative diseases o Diet and exercise show a reduced risk in many major diesases (MI, osteoporosis, cancer, hypertensioneverything from EPI) o Recommends types of food and amount of nutrients for everyday eating General recommendations o 30 minutes of moderate activity a day o choose foods represented in the food pyramid o keep food safe to eat with proper refrigerating/ freezing protocol Vegetables, fruits and grains o Eat many veggies, fruits and grains

o o o Fats o o o o o

6+ grains 5+ veggies reduce refined sugar in foods and drinks

Reduce fat intake >30% total calories saturated fat >10% total calories avoid trans fatty acids cholesterol < 300 mg/day < 200 mg/day in atherosclerosis pts Proteins o 0.8 g/kg o high quality proteins (fish, lean poultry and dry beans) o veggies should have adequate amino acids as well Alcohol o One drink per women and 2 for men o Pregnant women should not drink Vitamins and minerals o Decrease sodium intake in most people 3 g/day in salt-sensitive hypertension people the broad diet should cover all other vitamins and minerals o dont supplement vitamins in excess o fluoride should be present either in water or toothpaste Xenobiotics Defined as foods that have no nutritional value o Can be harmful in excess amounts Eat a variety of food to protect against ingestion of a toxic level of xenobiotic Reduce salt-cured, smoked foods that can contribute to cancer Review Questions In the process of respiration, fuels (A) are stored as triacylglycerols. (B) are oxidized to generate ATP.

(C) release energy principally as heat. (D) combine with CO2 and H2O. (E) combine with other dietary components in anabolic pathways. The caloric content per gram of fuel (A) is higher for carbohydrates than triacylglycerols. (B) is higher for protein than for fat. (C) is proportionate to the amount of oxygen in a fuel. (D) is the amount of energy that can be obtained from oxidation of the fuel. (E) is higher for children than adults. The resting metabolic rate is (A) equivalent to the caloric requirement of our major organs and resting muscle. (B) generally higher per kilogram body weight in women than in men. (C) generally lower per kilogram body weight in children than adults. (D) decreased in a cold environment. (E) approximately equivalent to the daily energy expenditure. REVIEW QUESTIONSCHAPTER 1 The RDA is (A) the average amount of a nutrient required each day to maintain normal function in 50% of the U.S. population. (B) the average amount of a nutrient ingested daily by 50% of the U.S. population. (C) the minimum amount of a nutrient ingested daily that prevents deficiency symptoms. (D) a reasonable dietary goal for the intake of a nutrient by a healthy individual. (E) based principally on data obtained with laboratory animals. A 35-year old sedentary male patient weighing 120 kg was experiencing angina (chest pain) and other signs of coronary artery disease. His physician, in consultation with a registered dietician, conducted a 3-day dietary recall. The patient consumed an average of 585 g carbohydrate, 150 g protein, and 95 g fat each day. In addition, he drank 45 g alcohol. The patient

(A) consumed between 2,500 and 3,000 kcal per day. (B) had a fat intake within the range recommended in current dietary guidelines (i.e., year 2000). (C) consumed 50% of his calories as alcohol. (D) was deficient in protein intake. (E) was in negative caloric balance.

The Fed or Absorptive StateIntro

Mary-Catherine Ciuba

nutrients get digested and absorbed during a meal o food is oxidized and meets energy requirements for the body to stay alive the excess is then stored in fuel depots o *a fuel is either oxidized (used immediately) or stored in response to the amount of insulin and glucagon in the blood carbohydrates o always translated into monosaccharides (mainly glucose) o stored as glycogen in liver and muscles glucose is the major biosynthetic precursor in the body o glucose can be converted into fat (triglycerides) o glucose -> very low-density lipoproteins (VLDL) and released into the blood proteins o proteins -> digested amino acids -> new proteins (neurotransmitters and heme) o carbon from proteins can be used to create glucose or energy directly fats o fatty acids -> 2-monoacylglycerols -> triacylglycerols packaged in chylomicrons and secreted into lymph and blood o fatty acids are mainly stored in adipose tissue oxidized to synthesize membrane lipids Digestion and Absorption Intro o Food is absorbed with the aid of enzymes in the mouth, stomach and small intestines o Digestion and absorption ALWAYS occurs in the fed state Carbohydrates o Convert carbs to monosaccharides (glucose primarily) o A-amylase help cleave polysaccharides apart in the border epithelial cells

o Monosaccharides are absorbed by intestinal epithelial cells and released into the hepatic portal vein to end up at the liver Proteins o Proteases cleave apart ingested proteins to form amino acid building blacks o Pepsin in the stomach and proteolytic enzymes from the pancreas act on proteins in the small intestines o Again, amino acids are absorbed by intestinal epithelial cells and released into the hepatic portal vein Fats o Most complex system b/c fat is water insoluble o Triglycerides are emulsified by bile salts Made in the liver and stored in the gallbladder o Pancreatic lipase turns triglycerides into diglycerides (for lack of a better term) Then absorbed by intestinal epithelial cells o Fats must be transported by a protein through the bloodstreamfat is insoluble in water Same goes for cholesterol Changes in hormone levels after a meal Carb meal o Pancreas releases insulin o Glucagon is inhibited o Endocrine glands (like the pancreas) release endocrine hormones o When at target tissue, hormones adjust metabolic pathways After eating glucose is used and stored Fed o Glucagon oxidizes fuel stores and brings energy to body parts that need it! Fate of glucose after a meal Conversion to glycogen, triglycerides, and CO2 in the liver o Once glucose is in the liver, the liver takes some for itself and converts glucose -> ATP Immediate needs for energy are used in these cells

Rest is converted to glycogen and triglycerides Can only hold 200-300 g glycogen Then must use fat Fat is not stored in liverrather it is transported with proteins, phospholipids and cholesterol as VLDLs that are sent out to tissues (esp adipose tissue) o *glucose -> pyruvate -> acetyl CoA -> acetyl group is oxidized in tricarboxylic acid cycle (TCA cycle) -> CO2 o energy from this conversion yields ATP Glucose metabolism in other tissues o Intro Glucose is either stored as glycogen or adipose tissue in systemic areas Muscles have relatively large glycogen stores Insulin stimulates the transport of glucose between the adipose tissue and muscles o Brain and other neural tissues Brain RELIES on glucose for energy needs primarilybig deal! Oxidize glucose through TCA cycle to generate ATP Glucose is a major precursor to MANY NTs Why we become dizzy and lightheaded without it Brain needs 150 g glucose/day o Red blood cells RBCs must use glucose b/c these cells dont have mitochondria to make their own RBCs convert glucose to energy through anaerobic glycolysis Takes place in the cytosol RBCs could not survive without glucose and we couldnt survive without RBCs that transport oxygen o Muscle Muscles convert glucose into lactate, letting the muscles harness their energy through another oxidation reaction Muscles can also use fatty acids as fuel o Adipose tissue

Adipose tissue oxidizes glucose for energy and use it as a source for triglyceride creation/storage Fate of lipoproteins in the fed state Two types of lipoproteins o Chylomicrons produced in fed state Chauffeur triglycerides through blood These triglygerides are formed in intestinal epithelial cells after digestion o VLDL produced in fed state Chauffeur triglycerides through blood These triglycerides are synthesized in the liver o When the lipoprotein complex moves into adipose tissue, the structure degrades into fatty acids and glycerol Triglycerols mix for glycerol moiety (from blood glucose) > fat droplets in adipose tissue Chylomicron remnants are cleared from blood by liver VLDL remnants are either cleared by the liver or LDLs are formed and cleared by liver/remnant cells o Fat storage seems limitless, except when the heart becomes overloaded Fate of amino acids in the fed state Liver uses amino acids to create serum proteins, liver proteins, nonessential amino acids, NTs, hormones and other structures Amino acids may be oxidized to glucose/ketone bodies for body use o Urea wasted nitrogen from these amino acids Amino acids will tend to circulate peripherally to be used by other tissues in protein synthesis Proteins are constantly synthesized and degraded Each individual biosynthetic pathway using an amino acid precursor is found in only a few tissues Summary of the fed (absorptive) state Fuels we eat are oxidized to meet immediate needs Then excess is stored in a variety of ways

o Carbs -> glycogen o Fats -> fatty acids and droplets o Amino acids -> new proteins Review Questions During digestion of a mixed meal, (A) starch and other polysaccharides are transported to the liver. (B) proteins are converted to dipeptides, which enter the blood. (C) dietary triacylglycerols are transported in the portal vein to the liver. (D) monosaccharides are transported to adipose tissue via the lymphatic system. (E) glucose levels increase in the blood. After digestion of a high carbohydrate meal, (A) glucagon is released from the pancreas. (B) insulin stimulates the transport of glucose into the brain. (C) liver and skeletal muscle use glucose as their major fuel. (D) skeletal muscles convert glucose to fatty acids. (E) red blood cells oxidize glucose to CO2. Amino acids derived from digestion of dietary protein (A) provide nitrogen for synthesis of nonessential amino acids in the liver. (B) can be converted to glucose in most tissues. (C) cannot be converted to adipose tissue fat. (D) release nitrogen that is converted to urea in skeletal muscle. (E) are generally converted to body proteins or excreted in the urine. Elevated levels of chylomicrons were measured in the blood of a patient. What dietary therapy would be most helpful in lowering chylomicron levels? (A) Decreased intake of calories (B) Decreased intake of fat (C) Decreased intake of cholesterol (D) Decreased intake of starch (E) Decreased intake of sugar

A male patient exhibited a BMI of 33 kg/m2 and a waist circumference of 47 inches. What dietary therapy would you consider most helpful? (A) Decreased intake of total calories, because all fuels can be converted to adipose tissue triacylglycerols (B) The same amount of total calories, but substitution of carbohydrate calories for fat calories (C) The same amount of total calories, but substitution of protein calories for fat calories (D) A pure-fat diet, because only fatty acids synthesized by the liver can be deposited as adipose triacylglycerols (E) A limited food diet, such as the ice cream and sherry diet

Fastingdo not read as closelyits not necessary fasting not eating for 2-4 hours after a meal

Mary-Catherine Ciuba

converts building blocks to glycogen and stores extras in muscle, liver and other various pleaces

blood sugar ranges 80-100 mg/dL enter a basal state usually an overnight fast o insulin is low and glucagon rises starvation not eating for 3 days

Chapter 4: Acids, Bases and BuffersObjectives:

Mary-Catherine Ciuba

1. Recognize a hydrogen bond and atoms or groups of atoms that can take part in forming a hydrogen bond. Hydrogen bonds are weak covalent bonds between H (electropositive atom with a weak electron) and an electronegative atom (found on the right side of the periodic table that hold electrons extremely tightly) o Examples hydroxide groups and ammonia groups (OH and NH3) ALWAYS LOOK FOR OXYGEN AND NITROGEN FOR HYDROGEN BONDS!!! Sulfur as well sometimes o These hydrogen bonds are typically soluble in water, making the compound soluble Hydrogens become positive and oxygen becomes more negative Water molecules break and form bonds together extremely quickly and is extremely tough to break those bonds o Hydroelectric energy harnesses this energy -4 kcal/mol released o also has tremendous heat capacity temperature effects bonds in water and energy is easily given off or transferred throughout water Lungs control CO2 in the body o Bicarbonate, phosphate and hemoglobin are buffers for CO2 in the blood Hemoglobin is the principle buffer Not going to cover rxns in fatty membranes, just rxns in water! Water o Fluid compartment in the body 5:3 water balance in and out of the human body cells interstitial fluid and extracellular fluid 60% water in humans and 75% water in children o water in thermal regulation thermal conductivity how fast can heat travel down water? Heat capacity how much heat can you put into a solution?

Heat of vaporization when water evaporates, it cools the body so we cool off when we sweat during exercise o Electrolytes Sodium out of cell Chloride out of cell Inorganic Phosphate in cell Potassium in cell Sodium potassium ATPH pumps sodium back out and potassium back in Water enters the cell and ruptures if pump doesnt work o Osmolality and water movement With a semipermeable membrane, water is allowed through but other compounds are not Each particle in extracellular movement binds with water and they are stuck to that ion Water travels to higher osmotic pressure in order to bind the the excess particles outside the cell Dehydrates the cell If you have an imbalance of ions, the water will flow to those ions Can cause dehydration or cell explosion, both lead to cell death This is why IVs have sodium chloride and not pure water Diabetics must worry about ketoacidosis because dehydration can occur when riding the body of excess glucose 2. Define the ion product for water! Given any hydroxide ion concentration, determine the hydrogen ion concentration and visa versa. pH of water o Remember, water dissociates into H and OH ions and its pH is neutral Acids give hydrogens Concentration of H is 1.0 x 10-7 or pH = 7

Bases accept hydroxides Concentration of OH is 1.0 x 10-7 or pH = 7 pH is 7 for solution because water has equal concentrations of hydrogen and hydroxide atoms o Kw = [acid] [base] Water = 55.5 M Defines the ion product of a solution Kw water = 1x 10-14 ALWAYS o K = [acid] [base] / [compound] Adding hydroxide allows increase in hydroxide and decrease in hydrogen ion o If one goes up, other goes down ALWAYS Strong and weak acids o Weak acids have Ka, the dissociation constant Used in Henderson-Hasselbach equations pKa = -logKa 3. Define pH. Given any hydrogen ion concentration, determine the pH and visa versa. pH = -log [H+] o shown as mol/L solution o determines hydrogen ion concentration o log 1 = 0 o log 2 = 0.3 estimate many logs now! Log 4 = 0.6 because log 4 = log 2x2 = log 2 + log 2 = 0.3 + 0.3 = 0.6 o Log xy = log x + log y o Log x = -log 1/x o Log x/y = -log y/x AND log x log y Log 0.5 = log = -log 2/1 = -log 2 = -0.3 If you decrease the pH by 0.3, you have doubled [H+] o If you increase the pH by 0.3, you have halved the [H+] 4. Define an acid and a base.

Acid BLA something in solution and a proton dissociates o Proton donor o Undissociated acids end in ic Sulfuric acid o Dissociated acids end in ate Sulfate is the dissociated acids and 2 hydrogens o Strong acids water will dissociate those hydrogen atoms o Weak acids water will not dissociates its hydrogen atoms Base BLB something in solution that accepts protons o Proton acceptor Conjugate base ion after losing a proton o Salt of the acid Buffers will form for weak acids, not strong acids 5. Recognize the acids produced by the body during metabolism and be able to state the predominant form at pH = 7.4 (ic or ate, ammonium or ammonia). The predominant form will be covered later in the chapter. Most of the acids produced by the body that we are responsible for are found in Table 4.2. o Hydorchloric acid (HCl) and nitric acid (HNO3) are two more acids that we are responsible for. Strong acids are completely dissociated at any pH. Later, we will use the pH of the solution and the pKa of the weak acid to determine the predomninant form at pH 7.4 and any other pH.

acids will dissociate 50% if pH = pKa 6. What is the difference between a strong acid and a week acid? Can a strong acid serve as a buffer? Weak acids o All acids in above table except for sulfuric acid o These serve as buffers because weak acids only partly dissociate in water o Only weak acids can buffer Strong acids o No, strong acids can NOT serve as a buffer o When added to water, these completely dissociate o H2SO4, HCl and HNO3 7. Given a weak acid, be able to draw the equation for its dissociation and label the conjugate base (salt of the acid). Be able to define the Ka for the acid. Be able to write the Henderson-Hasselbalch equation for an acid.

Ka = [H+] [A-] / [HA] pH = pKa + log [A-] / [HA] o pKa = -log Ka o when [A-] [HA], it = 1 and log 1 = 0 therefore pH = pKa 8. Define the term buffer! Name two factors that determine the effectiveness of the buffer! At what pH are buffers most effective? Buffer a weak acid and a salt (conjugate base) are together in a solution that RESIST changes in pH when another base or acid is added o Effectiveness of a buffer lies upon pKa of the buffer relative to the pH solution the closer the pKa is to the pH, the stronger the buffer must be within +/- 1 pH concentration of buffer present must have enough buffer to recombine or dissociate with hydrogen atoms found in solution accordingly o when acid is added to a buffered solution, the salt combines with the added acid and prevents the pH from changing o when a base is added to a buffered solution, the weak acid combines with the added base to prevent a change in pH when the pH < pKa, there is very little salt present to combine with acid o this is why buffers are only effective within 1 unit of the pKa of the weak acid when the pH > pKa, there is very little acid present to combine with the base o this also is why buffers are only effective +/- 1 unit pKa value 9. What is the most important reason for maintaining a physiological pH? The body requires a certain pH to maintain protein structures among other things o Proteins have a narrow pH range that they require in order to function properly o Changing pH changes the structure and therefore function o A change in function alters enzyme production among other biological pathways

10. What are the two most important buffers in blood? Bicarbonate buffer system Hemoglobin buffer system o Blood pH = 7.3 In the intracellular fluid, proteins and organic phosphates are the most important buffers o pH = 7.1 buffers must work on a constant, regular basis 11. Be able to draw the equations that show how the bicarbonate buffer system works in blood! What is the respiratory compensation when the blood pH drops to 7.3? What is the respiratory compensation when the blood pH rises to 7.5? bicarbonate buffer system: o CO2 + H2O H2CO3 (carbonic acid) HCO3- + H+ (bicarbonate) Catalyzed by carbonic anhydrase (one of the fastest and most abundant enzymes known) o CO2 is constantly produced in the tissues by metabolism, carried by the blood, and is exhaled by the lungs. CO2 pressure in the tissues is higher than in the blood and the pressure (concentration) in the blood is usually higher in the blood than in the lung sacs (alveoli). Normally the lungs regulate the concentration of CO2 in the blood. Multiply by 0.03 40 mm Hg x 0.03 = 1.2 mm carbonic acid o The synthesis of HCO3- by the kidney is much slower than the synthesis of carbonic acid. It takes several hours to replace lost bicarbonate. Henderson-Hasselbalch Equation for bicarbonate buffer system: o pH = 6.1 + log([HCO3-]/[H2CO3]) Even though the pKa of 6.1 for the bicarbonate buffering system is more than one pH unit from the normal pH of blood (7.4), the system works because the lungs regulate the concentration of CO2.

Respiratory Compensation in response to metabolic acidosis/alkalosis, compensation corrects this cycle through respiratory acidosis/alkalosis o Respiratory Acidosis person breaths slower and/or shallower than normal so [H2CO3] (CO2) increases in the blood. If the denominator in the Henderson-Hasselbalch Equation increases, the pH must drop. o Respiratory Alkalosis person breaths faster and/or deeper than normal so [H2CO3] (CO2) decreases in the blood. If the denominator in the Henderson-Hasselbalch Equation decreases, the pH must rise. Denominator of this equation is determined by the lungs Respiratory Compensation if the pH of the blood drops to 7.3: o Assume that the drop in pH is due to metabolic acidosis. Too much lactate Lactate and hydrogen and cotransported out of the cell The protons produced by the lactic acid are buffered by the bicarbonate buffer system out in the blood stream so the amount of bicarbonate has dropped The excess carbonic acid produced has been exhaled by the lungs. This initially resulted in a drop in the bicarbonate concentration and no change in the carbonic acid level. You can see from the Henderson-Hasselbalch Equation that the pH had to decrease. Now for compensation. The body senses that there is a decrease in pH and begins compensatory respiratory alkalosis. This decreases the carbonic acid term of the equation and raises the pH. Respiratory Compensation if the pH of the blood raises to 7.5: o Assume that the rise in pH is due to metabolic alkalosis. Due to vomit or other trauma, there is too much bicarb Bicarb rises and no change in carbonic acid You can see from the Henderson-Hasselbalch Equation that the pH had to increase. Now for compensation. The body senses that there is an increase in pH and begins compensatory respiratory acidosis. This increase in the carbonic

acid term of the equation and lowers the pH back towards normal. To get over this and return to normal, The body must stop producing more bicarbonate than can normally be handled and the kidney must get rid of the excess bicarbonate. 12. Draw the dissociation of ammonium ion. If the pKa =9.3, what form is found at pH = 7? Ammonium ion dissociation o NH4+ NH3 + H+ pH = pKa + log [salt]/[acid] o 7 = 9.3 + log[NH3]/[NH4+] o -2.3 = log[NH3]/[NH4+] o 2.3 = log[NH4+]/[NH3] o (2 + 0.3) = log[NH4+]/[NH3] o [NH4+]/[NH3] = 10(2 + 0.3)= (102) x (100.3) = :100 x 2 = 200 o 200=[NH4+]/[NH3] Shortcut If the pH of the solution is below the pKa, then the protonated acid form on the left of the above equation predominates. o pH > pKa, the dissociated base (otherwise known as the salt or conjugate base) predominates. Because the pH (7) in this instance is far below the pKa (9.3), Ammonium ion predominates in this solution. Another shortcut At step 5 above, enter 2.3 into your calculator and depress the 10X button 13. Regarding Di Abietes: Define Type 1 diabetes. With out insulin, what happens to the blood glucagon concentration? What happens to her blood glucose and ketone body levels (concentrations)? Type 1 diabetes blood sugar is extremely high because the pancreas fails to produce insulin o Complete loss of beta cells o Formerly called insulin dependent DM/juvenile diabetes o Complete lack of insulin because autoimmune disease destroys these beta cells

Without insulin, they die (insulin is a major regulator of glucagon) Have an indirect relationship o Patient is in a hyperglycemic state (euglycemia) with a blood sugar level ridiculously high at 600 mg/dL Insulin inhibits and glucagon increases to turn fatty acids into ketone bodies o Without insulin, ketone bodies are synthesized very quickly and can send a type 1 diabetic into metabolic acidosis 14. Regarding Di Abietes: Explain why an increase in a metabolic acid would cause the changes seen in PaCO2 and serum bicarbonate. What would happen to the concentration of CO2 and serum bicarbonate after the insulin injection? Diabetic metabolic acidosis b/c too much metabolic acid in the form of ketone bodies or lactic acid o pH drops b/c of excess metabolic acid due to drop in pH, diabetic breathes deeper or more rapidly to bring up to normal insulin injection o will stop making excess metabolic acid o kidney will replace the bicarbonate over several hours, allowing the lungs to readjust and go back to normal confusing: [carbonic acid] never changes o lungs control [CO2], so carbonic acid concentration does not need to change! Metabolic Acidosis: pH decreases; HCO3- decreases H2CO3 compensatory drop. o Metabolic acidosis is a condition in which there is too much acid in the body fluids (accumulation of metabolic acids, i.e. lactic acid, ketone bodies, etc.). Metabolic Alkalosis:' pH-Increases; HCO3- increases; H2CO3 compensatory increase. Respiratory Acidosis: H2CO3 increases; pH-Decreases (think brown paper bag). o Kussmaul breathing is an abnormal respiratory pattern (hyperventilation) characterized by rapid and/or deep breathing,

often seen in people with metabolic acidosis, where the respiratory system is attempting to rid the body of CO2 (an acid) to increase the pH. Respiratory Alkalosis: H2CO3 decreases; pH-Increase (think hyperventilation) 15. Regarding any person suspected of having diabetes: What levels of fasting plasma glucose or levels of random plasma glucose would you expect to measure? Fasting plasma glucose level of suspected diabetic: 126 mg/dL (milligram/deciliter)or higher. o A person is confirmed as diabetic when they have two or more fasting blood glucose tests of 126 mg/dL or above. Normal random (any time) plasma glucose levels should not be over 200mg/dL. A level over 200 mg/dL defines "overt" diabetes mellitus. Other: o impaired fasting glucose greater than or equal to 100 and less than 126 for at least two tests (often termed "prediabetes") o Normal fasting plasma glucose level- can range from 70-100 mg/dL 16. Concerning Dennis Veere: If the pKa for acetylsalicylic acid to acetylsalicylate is 3.5, is aspirin a weak or strong acid? Which form is prevalent in the stomach at a pH of 1? Which form is prevalent in blood at pH of 7.4? Prove it using the Henderson-Hasselbalch equation. Aspirin is a weak acid. Any compound with a pKa between 1 and 14 is a weak acid. o In the stomach''', where the pH is 1, acetylsalicylic acid (acid) is prevalent relative to acetylsalicylate (salt). o pH = pKa + log [salt/acid] 1 = 3.5 + log [salt/acid] -2.5 = log [salt/acid] 2.5 = log [acid/salt] Gluconate + H2O2 Glucose oxidase will ONLY react with glucose and no other sugar o Glucose and oxygen are converted to gluconate and hydrogen peroxide. Glucose is oxidized and oxygen is reduced. o The Hydrogen peroxide then reacts with a dye to produce a color; the intensity of the color is directly proportional to the concentration of glucose in the patient's blood. This test is specific for glucose: o Because only glucose can serve as a substrate of this particular enzyme. o In general, all enzyme reactions are more specific that nonenzymatic chemical tests because enzymes are very specific for their substrates.

o For example, the copper reducing test above reacted with glucose, mannose, fructose and many more carbohydrates. This test only reacts with glucose. No other sugars will react with the enzyme. Objective 29. Concerning Lotta Topaigne, Given that the pKa for uric acid is 5.7 and that uric acid is 20 times less soluble than sodium urate, explain which is more likely to precipitate in the toe (pH = 7.4) and which is more likely to precipitate in the kidney tubule (pH = 6.3). This problem is worked out in Chapter 5, Other Help. The equation is: uric acid = urate + H+ o In the toe: Use a pH of 7.4 and the Hendersen-Hasselbach equation determine that the ratio of [Sodium Urate]/[Uric Acid] is 50. Therefore, even though uric acid is 20 times less soluble than sodium urate, sodium urate will precipitate. o In urine: Use a pH of 6.3 (Urine pH range is 4.5-8, average is 6.3) and the Hendersen-Hasselbach equation determine that the ratio of [Sodium Urate]/[Uric Acid] is 4. Therefore, uric acid will precipitate. When it turns into a salt, these are the formation of the kidney stones

Chapter 5 Questions

Mary-Catherine Ciuba

1. Di Beatty has a metabolic acidosis resulting from an increased hepatic production of ketone bodies. For a patient with this disorder a screening test using the Nitroprusside reaction can test for the presence of ketones in the blood. Which of the following molecular structures is a ketone? R-CH2-OH RCHO (RCH2-)2C=O RCOOH RCH2-O-CH3 2. All of the following statements regarding oxidation-reduction reactions are true EXCEPT? KNOW Carbon atoms in organic molecules have different oxidation states depending on whether they are involved in single or double bonds or whether they are part of an alcohol, aldehyde, ketone, or carboxylic acid The oxidation of a molecule involves the loss of electrons The reduction of a molecule involves the addition of electrons When CH3-CH2-OH is converted to CH3-CHO (ethanol converted to acetaldehyde), ethanol is oxidized. When CH3-CH2-OH is converted to CH3-CHO (ethanol converted to acetaldehyde), ethanol is reduced. 3. All of the following are characteristics of the reduction of a biological molecules EXCEPT KNOW Gains hydrogen atoms Gains oxygen atom Looses oxygen atoms Gains electrons It gains hydrogens/electrons but loses oxygen and leaves electrons behind 4. Which of the following two substrates undergo a condensation reaction to form an ester? Alcohol + Alcohol Acid + Acid

Acid + Alcohol Phosphoric acid + Phosphoric acid Acid + Amine 5. Hydrolysis of a thioester produces a Carboxylic acid + alcohol Phosphoric acid + alcohol Carboxylic acid + sulfhydryl compound Carboxylic acid + carboxylic acid Carboxylic acid + amino group 6. If a monosaccharide has 6 carbon atoms and the carbonyl group is on the number 3 carbon atom, the sugar is called a aldohexose ketohexose aldopentose ketopentose aldotriose 7. If a monosaccharide has 5 carbon atoms and the carbonyl group is on the number 1 carbon atom, the sugar is called a ketohexose aldohexose aldopentose ketopentose aldotriose 8. A carbon atom is said to be asymmetric If it has four different chemical groups attached to it If it has both single and double bonds If it forms an achiral center If it contains less than four bonds None of the above describes an asymmetric carbon

9. You are presented with two compounds and they both have the formula C6H12O6. They differ only in their orientation about the number 3 asymmetric carbon atom. They are epimers sterioisomers carbohydrates isomers All of the above But not enantiomers 10. Which of the following characteristics would allow you to distinguish between a stereoisomer and an epimer? Epimers have the same chemical formula but differ in the position of the hydroxyl groups on one or more of their asymmetric carbons Stereoisomers are like one another with respect to which atoms are joined to which other atoms but have different chemical formulas Stereoisomers have the same chemical formula but differ in the position of the hydroxyl groups on one or more of their non-asymmetric carbons Epimers are stereoisomers that differ in the orientation of the hydroxyl group about a single asymmetric carbon atom Stereoisomers are epimers that differ in the orientation of the hydroxyl group about number 3 asymmetric carbon atom 11. D-Galactose and D-Mannose are aldohexoses that differ in the orientation of hydroxyl groups at two different carbon atoms. They are Enantiomers of each other Isomers of each other Anomers of each other Epimers of each other Mirror images of each other Not enantiomers or epimers because they differ on two different carbons 12. When glucose is dissolved in water, the straight chain form can produce either an -hemiacetal or a -hemiacetal. All of the following are characteristics associated with -D-Glucopyranose and b-D-Glucopyranose, EXCEPT? KNOW

Conversion of one isomer to the other is referred to as mutarotation Once the anomeric carbon atom forms a bond with another molecule it is locked in either the alpha or beta position -D-Glucopyranose can not be converted into b-D-Glucopyranose Enzymes differentiate between -D-Glucopyranose and b-D Glucopyranose, and are specific for only one of them -D-Glucopyranose and b-D-Glucopyranose are two different configurations of the same molecule it can be converted? By breaking apart again and flipping I suppose? 13. When the number 6 carbon atom of glucose is oxidized to a carboxyl group at pH = 7, the new compound is called KNOW Fructose Glucuronate Gluconate Sorbitol Ribose 14. Which of the following monosaccharides is a result of the reduction of the glucose aldehyde group? Sorbitol Gluconic acid Glucuronic acid Galactitol Ribose 15. Which of the following statements is true regarding the formation of a glycosidic bond? A glycosidic bond is formed when the OH or NH of the anomeric carbon reacts with the hydroxyl group of another molecule The anomeric carbon of a glycosidic bond is always in the beta position A glycosidic bond is formed when the hydroxyl group of the anomeric carbon reacts with the OH or NH group of another molecule The anomeric carbon of a glycosidic bond is always in the alpha position.

A glycosidic bond is formed between glycose and sidose. 16. Eicosanoids are a group of hormone-like compounds produced by many cells in the body. They include the prostaglandins, thromboxanes and leukotrienes. They are synthesized from polyunsaturated fatty acids containing 20 carbon atoms and 3, 4, or 5 double bonds. Which of the following fatty acids is the precursor to the eicosanoids? Palmitic acid Palmitoleic acid Oleic acid Arachidonic acid Stearic acid 17. Which of the following is an example of an 18:0 saturated fat? Palmitoleic acid Arachidonic acid Stearic acid Oleic acid Palmitic acid 18. Using delta numeric nomenclature, which of the following would be used to describe a fatty acid with 18 carbons and 1 double bond, which is located at position 9 of the carbon chain? 18:91. 18:91 18:19 18:19 18:99 19. Arachidonic acid has 20 carbon atoms and four double bonds. It may be described as -6 or as 20:4,5,8,11,14 The -6 indicates that there are no double bonds between carbon atoms 20 and 15 1 and 6 20 and 14

20 and 5 5 and 14 20. Which of the following characteristics regarding the length and saturation of the fatty acids are true? The melting point of a fatty acid decreases with chain length and increases with the degree of unsaturation The melting point of a fatty acid increases with chain length and decreases with the degree of unsaturation The fatty acid composition of membrane phospholipids determines the fluidity of membranes at body temperature A and C are correct B and C are correct 21. Unsaturated fats such as those in vegetable oils are liquids at room temperature and thus, contribute to the fluidity of our cellular membranes due to their Having lower melting points Having many double bonds Having higher melting points Both A & B Both B & C 22. Which of the following is true of a triacylglycerol? A triacylglycerol is formed when a fatty acid reacts with the hydroxyl group attached to the 3rd carbon of the glycerol moiety A triacylglycerol rarely contains the same fatty acid at all 3 positions A triacylglycerol is formed when 3 fatty acids reacts with the 3 hydroxyl groups of the glycerol moiety A triacylglycerol always contains the same fatty acid at all 3 positions B and C are both correct 23. The three OH groups on glycerol can react with one, two, or three fatty acids to form KNOW Amide groups Anhydride groups

Ester groups Carboxyl groups 24. The components of phosphatidic acid are? 1 Phosphate, 1 glycerol, and 1 choline 1 Phosphate, 1 glycerol, and 1 lecithin 1 Phosphate, 1 glycerol, and 1 fatty acid 1 Phosphate, 1 glycerol, and 2 fatty acids 1 Phosphate, 1 glycerol, and 3 fatty acids 25. Phospholipids such as phosphatidylcholine contain fatty acids esterified at which position of the glycerol molecule KNOW had no idea Position 1 only Position 3 only Position 1 and 2 Both position 1 and 3 Both position 2 and 3 26. Lysolecithin is produced by KNOW Addition of an acyl group to the glycerol molecule of lecithin Addition of a fatty acid to the phosphate group of lecithin Addition of choline to the phosphate group of lecithin Removal of a fatty acid (acyl) group from lecithin Removal of phosphate group from lecithin 27. One difference between sphingomyelin and lecithin is that Sphingomyelin contains choline and lecithin does not Sphingomyelin contains an acyl (fatty acyl) group and lecithin does not Sphingomyelin contains phosphate and lecithin does not Sphingomyelin contains a ceramide and lecithin does not Sphingomyelin contains a glycerol and lecithin does not 28. All of the following are characteristics of cholesterol, EXCEPT: Cholesterol contains 4 rings known as the steroid nucleus Cholesterol is synthesized in animals but not plants

Cholesterol is the parent compound from which other steroids are produced including bile salts, bile acids, adrenocortical steroids and sex hormones Cholesterol contains one hydroxide group Cholesterol is very water-soluble 29. By definition, amphipathic molecules contain both hydrophobic and hydrophilic regions. Which of the following statements correctly identifies the hydrophobic and hydrophilic regions of cholesterol? The steroid nucleus of cholesterol is hydrophilic and the carboxyl and hydroxyl groups are hydrophobic The carboxyl group of cholesterol is hydrophobic and the steroid nucleus and hydroxyl groups are hydrophilic The steroid nucleus of cholesterol is hydrophobic and the hydroxyl group is hydrophilic The hydroxyl group of cholesterol is hydrophobic and the steroid nucleus is hydrophilic The carboxyl and hydroxyl groups of cholesterol and the steroid nucleus are all hydrophilic 30. Which characteristic of bile acids helps them act as emulsifying agents in the intestine, helping to prepare dietary triacylglycerol and other complex lipids for degradation by pancreatic digestive enzymes? They have two or three hydroxyl groups and a carboxyl group They have steroid nucleus that is hydrophobic The molecules have both a polar and a nonpolar face They are amphipathic All of the above 31. Which of the following statements correctly describes the -carbon of most amino acids? It contains a hydroxyl group, an amino group and a side chain It contains a carboxyl group, an amino group and a side chain It contains a sulfhydryl group, an amino acid and a side chain. It is covalently bonded to 4 other carbons

All of the constituents vary from one amino acid to another 32. Which of the following contains a pyrimidine base? Thymine Adenine Guanine A and C B and C 33. Adenosine Triphosphate (ATP) is: KNOW Nucleotide Nucleoside Nucleoside triphosphate A and C A and B 34. Free radicals are atoms that have an unpaired electron in the outer orbital which makes them: Highly reactive and able to initiate chain reactions by extracting electrons from neighboring molecules Very stable Less destructive and potent catalysts Unreactive Very useful for stabilizing lipid membranes 35. Which of the following is NOT a free radical? Hydroxide radical Superoxide radical Nitrogen dioxide Oxygen 36. A semiquantitative reagent strip (ketostrip) or Acetest tablets contain nitroprusside that is used to estimate the level of ketone groups in someone who is ketoacidotic. Which of the following ketone bodies is responsible for most of the color developed in with nitroprusside?

Acetoacetate Acetone Beta-hydroxybutyrate 37. Given a basic solution containing adenosine monophosphate, glucose, fructose and galactose, which of these compounds will NOT undergo a positive reducing sugar test? KNOW AMP Glucose Galactose Fructose 38. Which functional group is oxidized as it donates electrons to copper in the reducing sugar test? Alcohol group Hydroxyl group Aldehyde group Ketone group 39. The glucose oxidase test is _______ for glucose and oxidizes glucose to ________: Specific ; Glucorunate Specific ; Gluconate Nonspecific : Hydrogen peroxide Nonspecific ; Glucuronate Nonspecific ; Gluconate 40. The pKa for uric acid is 5.7. At pH = 4 in the urine, most of the compound is in the form of Uric acid Sodium Urate 41. A 47 year old woman presents to her physicians office complaining of a severe throbbing pain in her right big toe that began 8 hours earlier. She is

diagnosed with acute gouty arthritis (podagra). Which compound is responsible for the severe pain experienced by this patient? Uric acid Sodium urate crystals Calcium pyrophosphate crystals All of the above 42. Refer to the structures in Figure 1 below. Which column contains the structure for an aldotriose? Column A Column B Column C Column D Column E Figure 1.

43. Refer to the structures in Figure 1 above. Which column contains glucuronic acid (glucuronate) KNOW Column A Column B Column C Column D Column E 44. Refer to the structures in Figure 2 below. Which column contains the structure for pyrimidine? KNOW Column A Column B Column C Column D

Figure 2.

45. Refer to the structures in Figure 3 below. Which row contains the structure for palmitic acid (palmitate)? Column A Column B Column C Column D Column E

Figure 3.

46. Three ketone bodies, ________, ________, and ________ are released into the blood from the ________ Acetaldehyde, acetone, beta-hydroxybutyrate; muscle Acetoacetate, provolone, and beta-hydroxybutyrate; liver Acetoacetate, acetone, and alpha-hydroxybutyrate; brain Acetoacetate, acetone, and beta-hydroxybutyrate; liver Acetoacetate, acetone, and beta-hydroxybutyrate; kidney

Chapter 6: Amino Acids in ProteinsObjectives:

Mary-Catherine Ciuba

1. What is the approximate pKa for all alpha-amino groups and for all alphacarboxyl groups? Are they charged at pH=7? The pKa for alpha-amino groups is approximately 9.5 (8.8 - 11.0). o they are protonated at pH of 7 with a positive charge. The pKa for alpha-carboxyl groups is approximately 2 (1.8 - 2.4). o they are dissociated at pH of 7 with a negative charge. Alpha amino carbon have a o Carboxyl group o Amino group o Hydrogen o And R group, which makes 20 different amino acids possible! 2. Which stereoisomer is most prevalent for any amino acid, the D- or the L- form? Alpha carbon is usually asymmetrical o Except lycine* The amino acid glycine cannot be in the D nor L form because the alpha-carbon is not an asymmetric carbon due to the atom having two hydrogen atoms Proteins in humans are made from L amino acids. Some bacteria make D amino acids (examples:. Antibiotics and cell walls) . o Most sugars have a D conformation. R group is the most important substituent in an amino acid 3. Be able to recognize a peptide or amide bond and be able to tell the difference. See Figures 6.3 and 5.2. o Peptide bonds are amide bonds between the carboxyl group on one amino acid and the amino group of the next amino acid. Peptide bonds are made between 2 amino acids o Amide bonds are between the carboxyl group on one compound (not necessarily an amino acid) and the amino group of another compound (not necessarily an amino acid).

o A peptide bond is always an amide bond, but an amide bond is only a peptide bond when it is formed between two amino acids. Other information: o The peptide backbone is made of amino acids linked together by peptide (amide) bonds. o It will appear as -N-C-C-N-C-C-N-C-C Proteins are commonly formed from hundreds of amino acids. Peptide and amide bonds are covalent bonds. : 4. Be able to recognize the three-letter abbreviations for the 20 amino acids! Name Alanine Arginine Asparagine Aspartate Cysteine Glutamate Glutamine Glycine Histidine Isoleucine Leucine Lysine Abbrev. Ala Arg Asn Asp Cys Glu Gln Gly His Ile Leu Lys

Methionine

Met

Phenylalanine Phe Proline Serine Threonine Tryptophan Tyrosine Valine Pro Ser Thr Trp Tyr Val

Note: Only 4 of the 20 amino acid abbreviations differ from the first 3 letters: o Asparagine (Asn) o Glutamine (Gln) o Isoleucine (Ile) o Tryptophan (Trp) MAKE FLASHCARDS! 5. Be able to name the structures of all 20 amino acids. Be able to say which contain R-groups that are nonpolar, polar, or charged! Be able to predict which R groups should be soluble in water and which should not. Name the structure: o Reference Fig 6.4 for structures Here is a helpful tool to learn the structures and to become comfortable with the 3-letter abbreviations: http://www.biology.arizona.edu/biochemistry/problem_sets/ aa/AAid/ID.html Which contain R-groups that are nonpolar, polar, charged? o Instead of memorizing this for all twenty AAs, use this approximation: If an R-group does not contain oxygen or nitrogen, it is not polar and not charged. If an R-group contains an oxygen or a nitrogen, it is polar and may or not be charged. . If it is the R-group of asp, glu, arg, lys, or his, it is charged over a pH range. Ignore the sulfur containing R-groups. Ignore the possible charge on tyr, ser, thr, met, and cys. Which R-groups are soluble in water? o The polar R-groups (see above) and gly. o Dont worry about solubility!! Hydrogen bonding o Hydroxide groups in amino acids can be Serine, theronine and another can form a hydrogen bond Uses NH or OH on carbonyl group

6. Be able to recognize cysteine, cystine, a sulfhydryl group, and a disulfide bond. Which are more oxidized and which are more reduced? (See Figure 6.6 Cystine is formed when two molecules of cysteine or two cysteine residues (R-groups) in a protein combine, forming a covalent disulfide bond between them. o Cysteine R group CH2 - SH Sulfhydryl Group: R-SH Disulfide Bond: R-S-S-R o A disulfide bond in cystine is more oxidized; a sulfhydryl group in cysteine is more reduced. Must remove hydrogen atoms to let the redox reaction play out o Remember that when you add hydrogen to a compound, you reduce the compound. Cysteine -> cystine is reduction because we add hydrogens o When you remove hydrogen from a compound, you oxidize the compound. 2 cystines -> cysteine is oxidation because we take away hydrogens 7. Be able to recognize a salt bond (ionic bond). (See Figure 6.7) A salt bond between two amino acids is when a positively charged R group of one amino acid interacts with a negatively charged R group of another amino acid. o Negative R and positive R o There are only 5 amino acids that can form a full negative or full positive bond (because they are charged amino acids): aspartic acid negative glutamic acid negative histidine positive lysine positive arginine positive Note: Electrostatic bonds on the R-groups of these amino acids can only form at specific pH levels corresponding to their pKas. o For instance, histidine loses the positive charge on its R-group at around pH=6

o in aqueous solutions with a pH > 6 the histidine R-group cannot form electrostatic bonds (although the pKa of an R-group can change if it is not in a pure aqueous environment). Since it takes the attraction of a positive charge and a negative charge, either aspartate or glutamate (negative charge) must be paired with histidine, lysine, or arginine (positive charge). o The -amino and carboxyl groups were not included because most of them have been used to make peptide bonds in the protein. Other information: o Salt bonds are also called ionic bonds because they must be formed between two oppositely charged ions. o An ion is an atom or a group of atoms that has acquired a net positive or negative electric charge by gaining or losing one or more electrons. How to memorize pKas o Throw out cysteine and tyrosine o 2 = alpha carboxyl o 4 = aspartate and glutamate o 6 = histidine o 8 = THROW OUT o 10 = alpha amino and lysine o 12 = arginine 8. Given any two amino acids, predict whether their R-groups could form a salt bond at pH = 7.4. (Seed Figure 6.9) At a pH of 7.4, aspartate and glutamate have negative charges on their Rgroups and arginine and lysine have positive charges on their R-groups. o Therefore, either arginine or lysine can form a salt bond with either aspartate or glutamate. At pH of 7.4, histidine does not have a charge on its R-group and cannot form a salt bond. There are 5 amino acids with R-groups that can be charged: o Aspartate: The R-group pKa is 4, so at any pH greater than 4, a negative charge predominates and aspartate can form a salt bond with any amino acid R-group having a positive charge.

o Glutamate: The R-group pKa is 4, so at any pH greater than 4, a negative charge predominates and glutamate can form a salt bond with any amino acid R-group having a positive charge. o Lysine: The R-group pKa is 10.5, so at any pH less than 10.5, a positive charge predominates and lysine can form a salt bond with any amino acid R-group having a negative charge. o Arginine: The R group pKa is 12.5, so at any pH below that, a positive charge predominates and arginine can form a salt bond with any amino acid R-group having a negative charge. o Histidine: The R group pKa is 6, so at any pH below that, a positive charge predominates and at any pH above 6, the R-group is neutral. Histidine can NOT form a salt bond with any amino acid R-group having a negative charge unless the pH is below 6. Note! When it came to problems of ionic bonds, we were told to ignore all the amino acids except aspartate, glutamate, histidine, lysine and arginine. 9. What are the pKa's for the R-groups of aspartic acid, glutamic acid, histidine, lysine, and arginine? Aspartic acid - 3.9 Glutamic acid - 4.1 Histidine - 6.0 Lysine - 10.5 Arginine - 12.5 Other Information: o Reminders from Objective 1: The pKa for the alpha carboxyl group of all amino acids is 2 (1.8 - 2.4), The pKa for the alpha-amino group for all amino acids is 9.5 (8.8 - 11.0), The pKa for any molecule will change depending upon the amount and types of other chemical compounds in the solution, so one can usually round off the last digit. That is, 10.5 is 10 and 12.5 is 12. p

10. Given the pH, predict whether the R-groups of the amino acids aspartic acid, glutamic acid, histidine, lysine, and arginine would be neutral or would carry a net negative or net positive charge. (See Figure 6.9) Aspartic acid pKa = 3.9 o At any pH below 3.9, neutral (no charge) charge predominates. o When pH is equal to pKa, the negative charge is equal to the neutral charge. o At any pH above 3.9, the negative charge predominates. Glutamic acid pKa = 4.1 o At any pH below 4.1, neutral (no charge) charge predominates. o When pH is equal to pKa, the negative charge is equal to the neutral charge. o At any pH above 4.1, the negative charge predominates. Histidine pKa = 6.0 o At any pH below 6.0, the positive charge predominates. o When pH is equal to pKa, the positive charge is equal to the neutral charge. o At any pH above 4.1, the neutral charge predominates. Lysine pKa = 10.5 o At any pH below 10.5, the positive charge predominates. o When pH is equal to pKa, the positive charge is equal to the neutral charge. o At any pH above 10.5, the neutral charge predominates. Arginine pKa = 12.5 o At any pH below 12.5, the positive charge predominates. o When pH is equal to pKa, the positive charge is equal to the neutral charge. o At any pH above 12.5, the neutral charge predominates. 11. Given the pH, predict whether the amino acids aspartic acid, glutamic acid, histidine, lysine, and arginine would be neutral or would carry a net negative or net positive charge. The charge on any amino acid would be the sum of the charge on the alpha-amino group, the charge on the alpha-carboxyl group, and, if there is one, the charge on the R-group.

The charge on the alpha-carboxyl group is neutral if the pH of the solution is below 2 and is negative if the pH of the solution is above 2. The charge on the alpha-amino group is positive if the pH of the solution is below 9.5 and is neutral if the pH of the solution is above 9.5. The charge on the R-group carboxyl of aspartate or glutamate is neutral if the pH of the solution is below 4 and is negative if the pH of the solution is above 4. The charge on the basic R-groups is positive if the pH of the solution is below their respective pKa and is neutral if the pH of the solution is above their respective pKa. Just add up the charges and you have the net charge: o Charge on amino acid = alpha-amino charge + alpha-carboxyl charge + R-group charge o If in a peptide bondthe R group alone would dictate the net charge of the amino acids o In a protein, there could be 150-200 charges Isoelectric form of the protein where the entire molecule has a 0 net charge pI = pH of isoelectric form (found on dissociation curve) 12. Define the pI (the isoelectric point) for an amino acid. (See Chapter 6, Other Help) The isoelectric point (pI) is the pH at which a particular molecule carries no net electrical charge. o Meaning that it carries just as many positive as negative charges. o For instance, if the substance were placed in solution between + (anode) and -(cathode) poles, it would not migrate either way if the pH of the solution was equal to the pI. Lowering the pH below the pI would cause the net charge on the molecule to become more positive and it would migrate towards the cathode. Raising the pH above the pI would cause the net charge on the molecule to become more negative and it would migrate towards the anode. This should make sense. If you increase the hydrogen ion concentration (lower pH) more protons will go onto the carboxyl groups making them

more neutral (less negative), and more protons will go onto amino groups making them more positive. 13. Given any amino acid except cysteine serine, threonine and tyrosine, be able to predict the isoelectric point! If you know the charged groups on an amino acid and you know the pKa of the charged groups, you can determine the pI. Lecture tells us that there are only 4 groups of pIs that we need to calculate: o All amino acids without a charged R-group Glutamate and aspartate Arginine and Lysine Histidine Each amino acid has: o a net positive charge when the pH of the solution is below the pI. o no net charge when the pH of the solution is at the pI. o a net negative charge when the pH of the solution is above the pI. Calculating the pIs: o For an amino acid with only one amino and one carboxyl group (most other amino acids): pI = (pKa1 + pKa2)/2 = (pKa alpha-amino 9.5 + pKa alpha-carboxyl 2)/2 = 5.75 o For the two negative acidic amino acids (Aspartate, Glutamate): pI = (pKa alpha-carboxyl 2 + pKa R-group-carboxyl 4)/2 = 3 average of pKa 1 and pKa 2 o For Lysine: pI = (pKa alpha-amino 9.5 + pKa R-group 10.5)/2 = 10 o For Arginine: pI = (pKa alpha-amino 9.5 + pKa R-group 12.5)/2 = 11 o For Histidine: pI = (pKa alpha-amino 9.5 + pKa R-group 6 )/2 = 7.75 14. Is the substitution of a valine for a glutamate in sickle cell hemoglobin a conservative replacement? What about the substitution of an aspartate for a glutamate?

Hemoglobin - each of the 4 heme groups on 4 protein chains o 2 Alpha globin genes o 2 Beta globin genes o carries oxygen on the heme groups through the blood o also acts as a buffer in the blood o quaternary structure more than 1 polypeptide chains form a structure Conservative replacement: Replaces one amino acid with another of a similar structure and charge (meaning a positive for a positive, negative for a negative, or a neutral for a neutral). o The substitution of a valine for glutamate (as in sickle cell anemia) is not a conservative replacement. Glutamate is a negatively charged amino acid and it is replaced by a valine which is a hydrophobic, branched-chain aliphatic (neutral) amino acid. o Histidine MUST attach to the heme group. It is not a conservative replacement not be substituted o The substitution of an aspartate for a glutamate is a conservative replacement because the two amino acids have the same charge and are nearly the same size. o Invariant regions cannot be changed o Variant regions can use conservative replacement here Sickle cell anemia is caused by a point mutation in DNA that changes the 6th amino acid in the beta-globin chain of hemoglobin from (normal) glutamate to (abnormal) valine. This substitution results in a hydrophobic knob. Isozymes enzymes that come from the same primordial gene that can help spur on the same reactions o Used in many different tissues 15. Given the written sequence for a protein, be able to identify the aminoterminus, the carboxyl-terminus, and the R-groups for each residue. (See Figure 6.10) For a protein, the amino acid sequence is written left to right, from the amino terminal to the carboxyl terminal. The amino terminus will be the first amino acid in the chain (on the left) and will have the only free alpha-amino group (-NH3+).

The carboxyl terminus will be the last amino acid in the chain (on the right) and will have the only alpha-carboxyl group (-COO-). All of the other alpha-amino and alpha-carboxyl groups were used up to make the peptide bonds and are no longer present in the protein. Only the amino-terminus, the carboxy-terminus, and the charged Rgroups have charges. Determining the charge on a polypeptide or protein: o If you had a protein made up of several amino acids, you would determine the charge of each of the amino acids at the given pH and sum them to determine the overall charge of the protein at that pH. o Example: What is the charge of the following protein at physiological pH?: Lys-Pro-Val-Asp Answer: At physiological pH (7.4) the charge on the alpha-amino group will be +1 the charge on the R-group of Lysine will be +1, the charge on the R-group of Proline will be 0, the charge on the R-group of Valine will be 0, the charge on the R-group of Aspartate will be -1. the charge on the alpha-carboxy group will be -1 +1+1+0+0-1-1 = 0, the protein will have no net charge. Other information: The numbers at the top indicate the number of amino acids from the amino-terminus (#1). 16. Concerning Michael Sichel: To the extent covered in this chapter, explain his disease in terms of amino acids. In the African American population, is the Sickle Cell allele a polymorphism. Pathology: o Will Sichel has two identical alleles for the sickle variant of the beta-globin gene that resulted from the substitution of valine (abnormal) for glutamate (normal) at the sixth position of the beta-globin chain. Putting in a valine forms a hydrophobic patch This patch goes into hydrophobic interaction, resulting in the polymerization of hemoglobin molecules

This polymerization creates an artificial cytoskeleton that sickles the normally disc-shaped red blood cell. When oxygen pressure is low, the sickle variant of the betaglobin can change its conformation so that a polymer is formed. Sickling usually occurs in the capillaries since tension of O2 is very low there. It seems that anytime there is a lack of oxygen reaching tissue, pain results. Have severe pain when sickling occurs "Note": the persistence of the sickle cell allele is most likely attributable to the selective pressure for the heterozygous mutant phenotype, which has been found to offer some protection against malaria. The sickle cell allele is a polymorphism. o Alleles are alternative versions of a gene at a given locus (specific position, or location) on a chromosome. Get 1 allele from mom, one from dad o For each locus, we have two alleles. If the alleles are identical, the individual is homozygous for this gene. If the alleles are different, the individual is heterozygous for this gene. Then the alleles matter From a practical point of view, a polymorphism exist when the locus in question is polymorphic. o A locus is polymorphic if there are two allelic forms at that locus and the least common is present at least one percent of the time in a given population. o This means that at least one in 50 people will be carriers of the variant allele. o The sickle cell allele makes up more than 1% of the total alleles in the black population and, therefore constitutes a polymorphism. o Polymorphism allele represents a small percentage of the population and it causes an abnormality (like sicklecell) Must be in 1% of human population Often these genes are healthy

Other Information: o Sickle Cell Disease: homozygosity for the mutation that causes HbS. o Sickle Cell Trait: the locus has one sickle gene and one normal adult hemoglobin gene. Many more trait carriers than disease carriers o Sickle-Cell anemia results from the destruction of HbS. o Furthermore, the destruction of HbS, specifically heme, can be detected via blood work as an elevated bilirubin. Bilirubin is the reduced form of biliverdin, is produced as the liver breaks down old or sickled HbS, and is found in the bile. 17. Concerning Cal Kullis: To the extent covered in this chapter, explain his disease in terms of amino acids. This patient has cystinurea, an inherited autosomal recessive disease that is characterized by the formation of cystine stones in the kidneys, ureter, and bladder. The mutation results in a defect in the reabsorption of cystine (also arginine, and lysine) into the blood from the kidney filtrate. Because cystine is not reabsorbed, it rises in concentration until its solubility product is surpassed and it forms stones. o Cystine is not as soluble as other amino acids so it is much more likely to form stones than other amino acids like arginine and lysine. Other information about treatment: o Because cystine is produced by oxidation of cysteine, conservative treatment of cystinuria includes decreasing the amount of cysteine within the blood and the amount of cystine eventually filtered by the kidneys. o Reduction of cysteine levels is accompanied by restricting dietary methionine, which contributes its sulfur to the pathway for cysteine biosynthesis. Increased daily fluid volume, to decrease the overall concentration of cystine that remains in solution. Urine can be chronically alkalized to prevent crystallization of cystine. Drugs may be used to enhance the conversion of urinary cystine to more soluble compounds.

Stones may be surgically removed by a technique that involves sonic fracture of the stones. 18. Concerning Di Abetes and in terms of amino acids, explain the difference between pork insulin, lispro, and synthetic human insulin. What are the possible advantages of each. (See Figure 6.12) Pork Insulin: o Difference: only one amino acid is different from human (at the 30th position of the beta chain). o Possible advantage: Along with beef, was one of the sources of mass produced insulin for over 50 years. o Possible disadvantage: Antigenic (causes an immune reaction). Beef Insulin: o Difference: only three amino acids are different from human (one at the 30th position of the beta chain and two in the alpha chain). o Possible advantage: Along with pork, was one of the sources of mass produced insulin for over 50 years. o Possible disadvantage: Antigenic (causes an immune reaction). Synthetic Human Insulin (Humulin): o Difference: Identical amino acid sequence to human. o Possible advantages: Non-antigenic and lower cost than older methods. It is synthesized in a laboratory from a strain of E. coli bacteria or yeast genome which has been genetically altered with recombinant DNA to produce biosynthetic human insulin. o Possible disadvantages: Dissolves slowly and can form a precipitate with Zinc that can clog insulin pumps. Lispro (also known as Humalog): o Difference: Identical to human except that the lysine (think LISpro), at position 29, and proline (think lisPRO), at position 28, of the beta-chain are reversed. This prevents formation of the hexamer precipitate that is normally formed with human insulin (a smaller dimer is formed instead). o Possible advantages: Much more soluble (think faster acting) because it does not precipitate with Zinc. Therefore, it does not clog insulin pumps. Non-antigenic. Allows patients to time their

injections minutes before consumption of carbohydrates rather than have to remember to give an injection one hour prior to a meal. o Possible disadvantages: None mentioned. 19. Concerning Ann Jeina, describe the isozymes of creatine kinase found in her body, their release following injury, and their use in diagnosing a myocardial infarction. Why do they react differently as antigens and why do they move differently in an electric field? (See Figure 6.15) Angina pectoris crushing and compressing pain in the chest and down your left arm o Sx blood clot occludes capillaries and lack of oxygen causes cell death and tremendous pain o Enzymes in the heart flow into the bloodstream o About 4 hours later CK increases in the blood The isozymes of creatine kinase (CK) are CK-MM (found in skeletal and cardiac muscle), CK-BB (found in brain tissue), and CK-MB (specific to cardiac muscle). Isozyme is a contraction of isoenzyme. As cardiac muscle cells die, CK-MB is released, and its level in the blood rises. A ratio of CK-MB to total CK can be measured, and if >5% of total CK is found to be CK-MB, a myocardial infarction is likely to be present. Unfortunately, levels of CK and CK-MB in the blood do not peak until 12-36 hours after a myocardial infarction, so it is possible to miss a myocardial infarction with this test if the test is performed within an hour or two of the beginning of the infarction. o Serial enzyme tests are usually done to ensure a myocardial infarction is not missed. If damage has occurred to muscles other than the heart, total CK will rise without significant elevation of the level of CK-MB. (i.e. CK-MB ratio will be L+P, This is the dissociation reaction o L+P __> LP, This is the association reaction o Kd = ([L][P])/ [LP] = 1/Ka o Ka = [LP]/([L][P]) = 1/Kd [LP] is the concentration of the ligand-protein complex. [L] is the concentration of the ligand

[P] is the concentrations of the protein. In practical terms, the stronger the bond between the ligand and the protein, the higher the Ka will be, and the lower the Kd will be. Ka is useful for describing the affinity a particular drug has for the protein it affects. 14. Using the terms ferrous iron, heme, hydrophobic pocket, histidine, alpha-helix, alpha-turns, salt bonds, hydrophobic interactions, hydrogen bonds, oxygen, and subunits, describe the myoglobin and hemoglobin molecule. Myoglobin and Hemoglobin (Hb): o Homologous proteins, therefore part of the same fold family o Myoglobin and Hemoglobin are both oxygen binding proteins Hemoglobin travels in the blood inside a red blood cell to deliver oxygen to tissues Myoglobin remains in the heart and skeletal muscle cells to bind oxygen released by hemoglobin o The single globin chain of myoglobin and the four globin chains of hemoglobin are all homologous proteins Each polypeptide chain folds into similar tertiary structures o Like all proteins, the tertiary and quaternary structures are stabilized by hydrophobic interaction, hydrogen bonds, and salt bonds (no disulfide bonds in these proteins). Myoglobin: o is a monomer has 8 alpha-helices linked together by alpha-turns (No beta sheets/strands) o Has a hydrophobic pocket containing heme with an ferrous iron atom (Fe2) at its center This is the oxygen binding site 2+ o Fe is always bound to a histidine R-group of the alpha helix This binding stabilizes the plus 2 state of iron when it binds to oxygen, i.e., keeps it from being oxidized to Fe3+ o Heme is tightly bound to the globin so it is a prosthetic group. o consists of a single polypeptide chain that is homologous to the alpha- and beta-subunits of hemoglobin Hemoglobin (Hb):

o Hb is a heterotetramer (four "mers" and at least one monomer is different from the others). o Hb is composed of 2 alpha and 2 beta subunits. o Each subunit has its own heme and, therefore, its own O2 binding site. o When O2 binds to the Fe2+ at one of the Hb binding sites, it pulls on the histidine, which pulls on the alpha helix, changing the conformation of the globin slightly. This slight movement changes the conformation of the other three chains in the Hb. This changes the oxygen binding affinity of the other oxygen binding sites and results in cooperativity Cooperativity is positive, i.e., when oxygen binds to one heme, the other hemes are more likely to bind a second molecule of oxygen o When two oxygens are bound, the remaining two are more likely to bind oxygen than before. When three oxygens are bound, the remaining site is more likely to bind oxygen than before 15. Be able to draw an oxygen saturation curve for hemoglobin and myoglobin. See Figure 7.11, The curve shows the percent saturation with oxygen vs. oxygen pressure (concentration). The myoglobin curve is a rectangular hyperbola but the hemoglobin curve is an S-shaped (sigmoidal-shaped) curve because of cooperativity Hemoglobin does not bind oxygen as strongly as myoglobin. o Hemoglobin binds oxygen strong enough so that it is almost saturated with oxygen in the lungs where the partial pressure of oxygen is about 90 - 100 mm Hg o Hemoglobin releases oxygen in the tissues where the partial pressure of oxygen is about 20 - 40 mm Hg. When oxygen is released from hemoglobin, the loss of one molecule of oxygen facilitates the loss of additional oxygen molecules, This is the reverse of positive cooperativity If hemoglobin was bound to oxygen as strongly as myoglobin is bound to oxygen, hemoglobin would never release oxygen when it reached the tissues.

16. What is a ligand? What is an apoprotein? What is a holoprotein? What would an apoenzyme be? Ligand: Anything that binds to a protein o Oxygen, heme, cofactors, coenzymes, metal ions, substrates, hormones, or even other proteins are examples of ligands o Usually something that dissociates Apoprotein: A protein missing its ligand or ligands o Usually not functional without the ligand Holoprotein: A protein with its ligand so it is able to function Apoenzyme: An enzyme missing its cofactor or cofactors o When combined with the proper cofactors, usually a metal ion or a coenzyme, the apoprotein becomes an active enzyme (holoenzyme). 17. What is a prosthetic group? A prosthetic group is a tightly bound structure (usually a ligand) required for the activity of an enzyme or other protein o For example, the tightly held heme of hemoglobin o Another example is a cofactor that is tightly bound Cofactors are metal ions or coenzymes 18. Be able to explain why the oxygen saturation curve for hemoglobin is sigmoidal while the curve for myoglobin is a rectangular hyperbola. Include the terms subunits, O2, Fe2+, conformation, salt bridges, T-state, R-state. Reference Figure 7.11 and 7.14 Hyperbolic Curve: o Myoglobin is a single polypeptide chain so the conformational change that occurs when O2 binds reversibly to Fe2+ has no effect on the conformation or O2 binding on other myoglobin molecules Since there is not interaction between the myoglobin molecules, there is no T-state or R-state o As oxygen concentration is increased from zero, a rectangular hyperbola results At low concentrations (pressures) of O2 a large amount of the added oxygen binds to the myoglobin molecule because

there are plenty of binding sites available and the rate of dissociation of bound sites is low o For each increment of added oxygen pressure, a smaller increment of oxygen will be bound because there are fewer available binding sites and the rate of dissociation from bound sites will increase The closer one gets to 100% saturation, the more difficult it is to bind more oxygen Sigmoidal Curve: o The sigmoidal curve results from the cooperativity of O2 binding in hemoglobin (Hb) o Hemoglobin is a tetramer consisting of four polypeptide subunits (chains) that binds oxygen reversibly o Each of the monomers is a globin chain that looks and acts much like myoglobin with one big difference o The globin chains are associated with each other with salt bonds that decrease their affinity for oxygen o When the salt bonds are broken, the new conformation has a higher affinity for oxygen Hemoglobin can exist in two conformations: the T-state or the R-state. o In the T (tense) state, Hb has a low affinity for O2. o In the R (relaxed) state, Hb has a high affinity for O2. o The higher the oxygen pressure, the higher the proportion of hemoglobin molecules in the R-state. o At lower levels of pO2 (the T-state) Hb cannot bind O2 as well as myoglobin, therefore its % saturation is much lower. If no O2 was present, most of the Hb would be in the Tstate but some would be in the R-state and able to bind to O2. If O2 is added, it would bind to the Hb in the R-state and stabilize the R-state. The R-state will not go back to the T-state as long as one heme is bound to O2 so the proportion of Hb in the R-state would increase. Also, the other three oxygen binding sites on the Hb would have an increased affinity for O2.

It follows that as the oxygen pressure is increased, the affinity of Hb for O2 would increase. That is, as O2 concentration is increased in increments, starting from zero, each increment of O2 increases the incremental amount of bound O2 until the hemoglobin solution is half saturated. After the hemoglobin is half saturated with O2, a smaller increment of O2 will be bound because there are fewer available binding sites and the rate of dissociation from bound sites will increase. The closer one gets to 100% saturation, the more difficult it is to bind more O2. 19. Be able to describe the structure of IgG in terms of the number of chains, the number and types of domains, and the forces that hold the chains together. Which part binds to an antigen and why is it specific? What fold family do the domains belong to. Reference Figure 7.15 The IgG immunoglobulin (antibody) consists of 2 smaller polypeptide chains (light chains) and 2 larger polypeptide chains (heavy chains). o Each light chain is bound to a heavy chain by disulfide bonds and the heavy chains are bound to each other by disulfide bonds. Each heavy chain has 4 domains and each light chain has 2 domains so the IgG has a total of 12 domains There are two types of domains: o Variable Domains: variable regions for antigen binding The variable domains of a light and heavy chain works specifically in binding to antigens liking o Constant Domains: the same for all immunoglobins Each light and heavy chain has on variable domain and the rest are constant domains. o There are millions of different variable regions found in a person, each one specific for one or a few antigens. The variability is due to many processes, including DNA recombination.

o Each variable domain is a beta-barrels with loops that stick out and bind to antigens They bind to specific antigens because they have different conformations with different charge distributions o All the domains, both variable and constant, are collapsed betabarrel made from multiple beta-sheets They all belong to the immunoglobulin fold family (all evolved from a common gene) o The forces that hold the chains together in immunoglobulins are hydrophobic interactions, hydrogen bonding, ionic bonds (salt bridges), and disulfide bonds.If one hydrolyzes the disulfide bonds, the chains drift apart and normal function is lost 20. What is the general name for the class of proteins that help other proteins fold into their native tertiary conformation? Heat shock proteins o Keep hydrophobic areas from bonding and forming incorrect secondary structures Native conformation conformation proteins usually have in a cell 21. What is denaturation? Name a few ways that proteins are denatured in humans. Denaturation a process in which the folding structure of a protein is altered and become biologically inactive due to exposure to certain chemical or physical factors o Physical forces and chemical reactions disrupt ionic, hydrogen, hydrophobic bonds, and disulfide bonds maintaining the native conformation Examples: o Nonenzymatic glycosylation: glucose binds to an exposed amino group on a protein to form a denatured glycosylated protein. This mechanism of denaturation is thought to be important in the pathology of diabetes Sugar forms a covalent bond with the terminal valine Glucose determines the rate of this reaction o Acid denaturation (pH denaturation): HCl in gastric juice interferes with the protein's hydrogen and ionic bonds

Disrupts the native conformation to make the protein a better substrate for digestive enzymes o Heat denaturation: Increasing the temperature will increase vibrational and rotational energies in the bonds and break them apart (ex. cooking an egg: albumin converts from its native translucent state to a denatured white precipitate) Cold temperatures can also denature some proteins o Enzymes Denaturation: Enzymes unfold and lose their active state when exposed to denaturing agents (e.g. strong acids or bases, heat, solvents, and salts) When enzymes lose their structure, they lose their function as catalysts, i.e., substrates can no longer bind to the active site If enough denaturation takes place, the enzyme reaction and the pathway it is part of fails Other: o Denaturation usually results in the permanent loss of protein function. However, under certain conditions, some denatured protein can refold into their native conformations and regain their original functions. This is called renaturation. o For example, ribonuclease is denatured with organic solvents such as urea that disrupt hydrogen-bonding patterns and convert the protein to a soluble random coil that has no activity. If the denaturing agent (urea) is removed,, ribonuclease, which is a simple single-subunit protein, can refold spontaneously into its native conformation. Even some complex multisubunit proteins containing bound cofactors can sometimes renature spontaneously under the right conditions. 22. What is a prion? A prion (proteinaceous infectious agent) is a protein that causes slow but progressive brain degeneration in a host organism resulting in death o Example mad cow disease

23. Explain how a prion can cause dementia and death. Include the terms PrPc, PrPsc, template, activation energy, cascade, amyloid protein, and proteolytic degradation. PrPc normal conformation of the protein PrPSc the disease-causing conformation o The Sc stands for scrapies, the name given to the disease when it occurs in sheep o It is autocatalytic and resistant to proteolysis Although PrPSc and PrPc have the same amino acid composition, the PrPsc conformer is substantially enriched in B-sheet structure compared with the PrPc conformer, which has little or no B-sheet structure and is about 40% alpha-helix Template (molecular template) a molecule (as of DNA) that serves as a pattern for the generation of another macromolecule Activation energy energy that must be put into a chemical reaction in order for the chemical reaction to occur o Higher the activation energy, the less likely the reaction is to occur Cascade series of chemical reactions that occur a result of a single trigger reaction or compound Amyloid proteins insoluble fibrous protein aggregates found in amyloid plaques (gnarled, stringy proteins) o In this case, a large number of molecules of PrPsc interact and precipitate together to form amyloid plaques o The name amyloid comes from the early mistaken identification of the substance as starch Proteolytic degradation the conversion of proteins into amino acids o PrPSc precipitates out Explanation: o Most cases of Prion disease are sporadic, which means they occur in people without any known risk factors or gene mutations so that will be the type explained here. Somehow a molecule of PrPSc is made from PrPc to begin the disease. o PrPSc contains no nucleic acids; therefore, it cannot self-replicate Rather, PrPSc acts as a template to misfold normal cellular proteins PrPc into PrPSc, a form that cannot undergo proteolytic degradation.

The template lowers the activation energy barrier for the conformational change of PrPc toPrPSc, a reaction that would otherwise never take place o The refolding of PrPc to PrPSc initiates a cascade as each new PrPSc formed acts as a template for the refolding of additional PrPc molecules to disease-causing ones Amplification of the reaction increases with time, so a single PrPSc ends up making millions and then billions of itself o As the concentrati