Exam 3 Material
n‑Dodecanate
lauric acid WHY: n‑Dodecanate is a saturated fatty acid with 12 carbon atoms. The prefix dodeca‑ denotes 12. Lauric acid is the common name of n‑dodecanate, because it was first extracted from the seeds of the laurel tree.
Characteristics of triglycerols
3 fatty acid chains energy storage
Properties of unsaturated fatty acids
have bent fatty acid tails have at least one double bond in the fatty acid tails maintain some space between adjacent phospholipids allow the membrane to remain fluid and flexible at low temperatures
They are made up of a long‑chain amino alcohol joined to a fatty acid, either by a glycosidic linkage or a phosphodiester linkage. These do not contain glycerol. They are abundant in the nervous system
sphingolipids
Triglycerides that are liquid at room temperature
oils
Name 1 monounsaturated fatty acid
oleic acid
Name 2 saturated fatty acids
palmitic acid stearic acid
cis‑Δ9‑Hexadecenoate
palmitoleic acid WHY: cis‑Δ9‑Hexadecenoate is an unsaturated fatty acid with 16 carbon atoms. Number 16 can be derived from its name, because prefix hexa‑ denotes six and suffix deca‑ denotes ten. The single double bond is at position 9 from the carboxyl group. Palmitoleic acid is the common name for cis‑Δ9‑hexadecenoate. Palmitic acid is the saturated form present in palm oil. A desaturase enzyme converts saturated palmitic acid into unsaturated palmitoleic acid.
Explain how to determine which of the given ligands binds the tightest: a. ligand D, with a percent occupancy of 80% at 10 nM b. ligand B, with a dissociation constant (𝐾LKL) of 10^−3 M c. ligand A, with a dissociation constant (𝐾LKL) of 10^−9 M d. ligand C, with a percent occupancy of 30% at 1 μM
Compare KL values to determine which ligand binds tightest. -A lower value of KL corresponds to a higher affinity of a ligand for a protein. -The dissociation constant, KL , corresponds to the molar concentration of ligand at which half the binding sites are occupied. =That is, KL is equal to the ligand concentration [L] when the free protein concentration [P] is equal to the concentration of ligand‑bound protein [PL]. ==KL = [P][L]/[PL] ===Therefore, a lower value of 𝐾LKL corresponds to a higher affinity of a ligand for a protein. When given a percent occupancy: = Y = [L]/[L] + KL = Y represents percent occupancy/fractional saturation = where 𝑌 is the fraction of binding sites occupied. Note that some references use the term 𝑣 , instead of 𝑌 , to denote fractional saturation. Note that KL for ligands A and B is expressed in units of M. Therefore, solve for 𝐾LKL of ligands C and D in μM and then convert.
Classify the structural features of phosphoglycerides
glycerol backbone phosphorylated alcohol attached to backbone at C3 WHY: A phosphoglyceride has a glycerol backbone. Glycerol is a three-carbon molecule with one hydroxyl group on each carbon atom. In a phosphoglyceride, two fatty acids are esterified to the glycerol backbone at the C‑1 and C‑2 hydroxyl groups. The simplest phosphoglyceride, phosphatidate, has phosphoric acid esterified to the C‑3 hydroxyl group. Other phosphoglycerides form from the addition of an alcohol, such as choline, to the phosphate of phosphatidate.
Classify the structural features of sphingolipids
sphingosine backbone one fatty acid chain attached to backbone by an amide bond WHY: A sphingolipid has a sphingosine backbone. Sphingosine is an amino alcohol with a long, unsaturated hydrocarbon group. An amino alcohol contains an amino group and one or more hydroxyl groups. Other sphingolipids form when fatty acids are added to the amino group and the primary hydroxyl group of sphingosine. Sphingomyelin is a common sphingolipid in membranes. In sphingomyelin, a fatty acid links to the amino group through an amide bond, and phosphorylcholine links to the sphingosine hydroxyl group though an ester linkage.
Properties of saturated fatty acids
fatty acid tails are straight have no double bonds in the fatty acid carbon chains fatty acid tails pack tightly together make the membrane somewhat rigid at low temperatures
The building blocks for many lipids. They generally contain an even number of carbon atoms and an unbranched hydrocarbon chain.
fatty acids
Describe vitamin E
prevents oxidative damage to membrane lipids antioxidant WHY: Vitamin E describes the tocopherols, which protect membrane lipids from oxidation by free radicals. A deficiency in vitamin E can result in fragile erythrocytes, possibly leading to anemia.
Describe vitamin D
production takes place in the skin hormone precursor derived from a steroid regulates calcium uptake by intestine WHY: Vitamin D is converted to a hormone that regulates calcium uptake by the intestine and regulates calcium levels in the kidneys and in bone. Vitamin D is synthesized in the skin when 7‑dehydrocholesterol is exposed to UV light. Vitamin D deficiency (which can occur when the skin is blocked form receiving sunlight) can cause rickets, a weakening of the bones.
Accumulating in adipose tissue, they are the storage form of lipids. They can be used as metabolic fuel. These compounds have a polar part made of three ester groups and a nonpolar fatty acid tail.
triacylglycerols
Molecules that have three ester functional groups
triglycerides
This polysaccharide is the storage form of glucose in the human body.
Glucose
Explain why melting point of unsaturated fats are lower than melting points of saturated fats
Melting temperatures of unsaturated fatty acids are lower than melting temperatures of the saturated fatty acids of a similar length. This is due to a kink that double bonds introduce in the structure of fatty acid chains. The kink does not allow the chains to pack tightly, lowering their melting temperatures.
Structure of sphingosine
Sphingosine is an amino alcohol. The general structure of a sphingolipid is illustrated in the image, where X may be one of several different polar head groups, such as phosphocholine, and R is a fatty acid tail. Sphingosine is highlighted in green.
Name 1 anomeric pair
alpha-D-glucose and beta-D-glucose
Is the molecule shown hydrophilic, hydrophobic, or amphipathic (amphiphilic)?
amphipathic
Monosaccharides can undergo reactions to form other functional groups. Identify the monosaccharide derivative that fits the statement: Gluconic acid is ________________________
an aldonic acid WHY: Aldonic and uronic acids have carboxyl groups at a terminal carbon atom of the carbohydrate chain. Aldonic acids are formed from the oxidation of the terminal aldehyde, whereas uronic acids are formed from the oxidation of the terminal hydroxyl. Aldonic acids are typically formed from mild oxidizing reagents, such as Br2 and water. An example of an aldonic acid is gluconic acid. Aldaric acids result from the oxidation of both the terminal aldehyde and the terminal hydroxyl. Monosaccharides are oxidized by nitric acid to form aldaric acids.
Analyze the following pair of compounds. Which of the terms explains the relationship between the two compounds: -alpha-D-galactose -alpha-L-galactose a. diastereomers b. enantiomers c. anomers d. epimers
enantiomers WHY: Though it may be hard to tell from the Haworth projections, α‑D‑galactose and α‑L‑galactose are mirror images of each other. The uncyclized Fischer projections are shown to the right to help visualize the relationship better. The green line represents the mirror plane, showing that the compounds are enantiomers. The compounds are not anomers or epimers, since both require that the compounds differ in configuration at only one stereocenter.
Which of the terms explain the relationship between the two compounds? -alpha-D-galactose -beta-D-galactose a. epimers b. diastereomers c. anomers d. enantiomers
epimers disastereomers anomers Why: α‑D‑galactose and β‑D‑galactose differ only in configuration of the anomeric carbon atom (C1), which is the aldehyde carbon that forms a hemi‑acetal structure. Thus, they are anomers. Epimers are diastereomers that differ in configuration of only one stereogenic center. This does not have to be the C1 carbon, but can be any of the carbons in the sugar chain. Thus, the compounds are also epimers. Because the remaining stereocenters in each structure are identical, the compounds are also diastereomers.
Identify the anomeric carbons of sucrose: Is sucrose a reducing or non‑reducing sugar?
1 & 2' Non-reducing WHY: Sucrose is a disaccharide containing glucose and fructose monosaccharide moieties joined by a glycosidic linkage. When monosaccharides exist in their linear forms, the anomeric carbon carries the carbonyl group, C=O. Upon cyclization, the anomeric carbon creates a stereocenter. The position of the hydroxyl group below or above the stereocenter gives rise to α and β anomers. To find the anomeric carbon, locate the carbon directly attached to two oxygen atoms. In the glucose moiety, the anomeric carbon is at position 1. In the fructose moiety, the anomeric carbon is at position 2′. Anomeric carbons that have a free hydroxyl group can participate in oxidation‑reduction reactions. Therefore, sugars that have free anomeric carbons are reducing sugars. Anomeric carbons of sucrose form the glycosidic bond and cannot participate in oxidation‑reduction reactions. Therefore, sucrose is a non‑reducing sugar.
Characteristics of phosphoglycerides
2 fatty acid chains membrane component alcohol group phosphate group
This polysaccharide is the structural component of plant cell walls and cannot be digested by humans. AND The glucose units in this polysaccharide are joined by β‑1,4‑glycosidic bonds.
Cellulose Humans lack the necessary enzyme to catalyze the hydrolysis of these bonds, which means humans cannot digest cellulose.
Identify the pH for each oxygen dissociation curve in the graph. -options: pH = 7.2 pH = 7.4 pH = 7.6
Curve 1: pH = 7.6 Curve 2: pH = 7.4 Curve 3: pH = 7.2
Name 2 epimeric pair
D-glucose and D-mannose D-galactose and D-glucose
Name 3 aldose-ketose pairs
D-glyceraldehyde and dihydroxyacetone D-glucose and D-fructose D-ribose and D-ribulose
Triglycerides that are solid at room temperature
fats
Formed when a carbohydrate is glycosidically linked to a hydroxyl group of a lipid. Examples include gangliosides and cerebrosides. These are also found in biological membranes.
Glycolipids
BPG binds tighter to ________________(HbA/HbF), which _____________ (raises/lowers) its affinity for O2.
HbA lowers
How are waxes formed?
They are formed from dehydration reactions of a fatty acid and a long‑chain alcohol. The ester bond of the wax forms by the substitution of the −OH of the carboxylic acid with the −OR of the alcohol. The atoms that form the water product are highlighted in the image.
Compounds that contain a fused ring system. These have three 6‑membered rings and one 5‑membered ring. Some of these compounds are found in biological membranes.
Steroids
Monosaccharides can undergo reactions to form other functional groups. Identify the monosaccharide derivative that fits the statement: Fucose is ______________________
a deoxy sugar WHY: Deoxy sugars are lacking an alcohol on one of the carbon atoms. An example of a deoxy sugar is fucose.
Contain a carboxylic acid group and a hydrocarbon chain
fatty acids
Describe vitamin K
cofactor functions in blood clotting WHY: Vitamin K is an oxidation‑reduction cofactor active in the synthesis of active prothrombin. Prothrombin is an enzyme that converts fribinogen to fibrin, which is a necessary component of blood clots. A vitamin K deficiency can impair blood clotting.
Anomer
diastereoisomers in which the hemiacetal carbon of a cyclic monosaccharide is asymmetric.
cis, cis‑Δ9, Δ12‑Octadecadienoate
linoleic acid WHY: cis, cis‑Δ9, Δ12‑Octadecadienoate is a polyunsaturated fatty acid with 18 carbon atoms. Its prefix octa‑ means eight. The fatty acid contains two double bonds, which is indicated by the di in the ‑dienoate suffix. The double bonds are at the positions 9 and 12 from the carboxyl group. Linoleic acid is the common name for 𝑐𝑖𝑠,cis, 𝑐𝑖𝑠cis‑Δ9, Δ12‑octadecadienoate. Linoleic acid is present in linseed oil. The -oleic suffix also relates to oleic acid, which is an 18-carbon monounsaturated fatty acid abundant in olive oil. Low melting temperatures of linoleic acid chains control fluidity of the cell membrane
Name 2 polyunsaturated fatty acids
linoleic acid arachidonic acid
Fats, oils, and other fat-soluble molecules
lipids
Epimer
monosaccharide diastereoisomers that have a single asymmetric carbon.
When glycerol is esterified to two fatty acids and a phosphoric acid molecule, a _______________________________ is formed. These are found in biological membranes.
phosphoglyceride
Describe vitamin A
precursor of retinal hormone precursor and visual pigment regulates gene expression in epithelial tissue WHY: Vitamin A functions as a hormone and is a visual pigment. The vitamin A derivative retinoic acid affects gene expression; retinal functions in vision. Symptoms of vitamin A deficiency may include night blindness or dry skin.
Consider the Fischer projection (shown on the left in the picture with the answer)
A WHY: The six‑carbon monosaccharide represents the Fischer structure of D‑galactose. D‑galactose can cyclize to form either the furanose (five‑membered) ring or the pyranose (six‑membered) ring. The constituents on carbon atoms that are right in the Fischer projection are down in the Haworth projection. Constituents that are left in the Fischer projection are up in the Haworth projection. For a D‑sugar, the −CH2OH group on C5 will be up in the Haworth projection of a pyranose ring. For an L‑sugar, the −CH2OH group on C5 will be down in the Haworth projection of a pyranose ring. Next, the position of the hydroxyl group on the anomeric carbon is relative to the −CH2OH group at C5. If they are on the same side of the ring, the structure is β; if they are on opposite sides of the ring, the structure is α. Several observations can then be concluded for the Haworth projection of the β‑pyranose form of D‑galactose (β‑D‑galactopyranose). It is a six‑membered ring (pyranose). Constituents on carbon atom C2 will be down. Constituents on carbon atoms C3 and C4 will be up. The −CH2OH group at C5 will be up. The anomeric hydroxyl group will be up.
Explain how to name fatty acids
A fatty acid contains a long hydrocarbon chain and a terminal carboxyl group. In saturated fatty acids, all carbon bonds are single bonds. Unsaturated fatty acids contain one or more double bonds. A double bond is nearly always found in the cis configuration, where two of its adjacent carbon atoms are on the same side of the chain. The position of the double bond in the chain is given by the delta symbol, Δ, followed by the carbon atom number.
Monosaccharides can undergo reactions to form other functional groups. Identify the monosaccharide derivative that fits the statement: ___________________ forms when carbon 6 of a hexose is oxidized to a carboxylic acid.
A uronic acid WHY: Aldonic and uronic acids have carboxyl groups at a terminal carbon atom of the carbohydrate chain. Aldonic acids are formed from the oxidation of the terminal aldehyde, whereas uronic acids are formed from the oxidation of the terminal hydroxyl. Aldonic acids are typically formed from mild oxidizing reagents, such as Br2 and water. An example of an aldonic acid is gluconic acid. Aldaric acids result from the oxidation of both the terminal aldehyde and the terminal hydroxyl. Monosaccharides are oxidized by nitric acid to form aldaric acids.
3 Part Question: The mutated form of hemoglobin (hemoglobin S, or HbS) in sickle‑cell anemia results from the replacement of a glutamate residue by a valine residue at position 6 in the β chain of the protein. Normal hemoglobin is designated HbA. Under conditions of low [O2], HbS aggregates and distorts the red blood cell into a sickle shape. Sickled red blood cells are relatively inflexible and may clog capillary beds, causing pain and tissue damage. The sickled red blood cells also have a shorter life span, leading to anemia. Part 1: Which amino acids would be expected to produce a similar sickling effect if substituted for Val at position 6? a. lysine b. leucine c. phenylalanine d. arginine e. alanine Sickling occurs in deoxyhemoglobin S but not in oxyhemoglobin S. Oxyhemoglobin has a small, hydrophobic pocket in a β chain region located in the interior of the protein. In deoxyhemoglobin, however, this pocket is located on the surface of the protein. In deoxyhemoglobin S, Val 6 interacts with this surface pocket, leading to aggregation of HbS. Part 2: Choose two amino acids that would be reasonable candidates for the pocket-Val 6 interaction. a. leucine b. phenylalanine c. serine d. asparagine e. glutamate Part3: How does HbS aggregation occur in sickle‑cell anemia? Place the steps in the correct order. Note that deoxyhemoglobin is in the T state; oxyhemoglobin is in the R state. -explain the process from no aggregation to aggregation
ANSWER 1: b. leucine e. alanine ANSWER 2: a. leucine b. phenylalanine ANSWER 3: No aggregation- 1. [O2] decreases due to vigorous exercise or high altitudes 2. R state Hb shifts to T state Hb 3. Val interacts with the pocket of a beta chain on another HbS 4. Additional T state HbS interact with the growing aggregate to form an insoluble fiber Sickle red blood cell- WHY: HbS results from the insertion of the hydrophobic amino acid, valine, for the charged amino acid glutamate at position 6 in the beta chains. It would be reasonable to expect that other small, hydrophobic amino acids would produce the same sickling effect as seen in HbS. This was confirmed by studies that showed sickling in Hb variants with isoleucine or leucine at position 6. The lysine and arginine R groups are both charged while the phenylalanine R group is large and bulky. In the deoxygenated forms of hemoglobin HbA and HbS, there is a small, hydrophobic pocket on the surface of the protein. Aggregates of deoxy HbS form when Val 6 interacts with the hydrophobic pocket of another HbS molecule. The surface pocket-Val 6 interaction is driven by hydrophobic interactions; therefore, any hydrophobic amino acid is a reasonable candidate for a participating pocket residue. The formation of aggregates in HbS occurs when [O2] falls and the deoxygenated T state begins to accumulate. Deoxy HbS has a hydrophobic surface pocket that interacts with Val 6 from another HbS molecule. These Val 6-pocket interactions result in a string of HbS molecules that eventually form an insoluble fiber and distort the shape of the red blood cell into a sickle shape.
Sucrose, a disaccharide, is often added as a sweetener, but it is not as sweet as its constituent monosaccharides D‑glucose and D‑fructose. Besides enhancing sweetness, fructose has hygroscopic properties that improve the texture of foods, reducing crystallization and increasing moisture. In the food industry, hydrolyzed sucrose is called invert sugar, and the yeast enzyme that hydrolyzes it is called invertase. The hydrolysis reaction is generally monitored by measuring the specific rotation of the solution, which is positive (+66.4∘) for sucrose, but becomes negative (inverts) with the formation of D‑glucose (specific rotation=+52.7∘) and D‑fructose (specific rotation=−92∘). Consider the chemistry of the glycosidic bond. Which methods could be used to hydrolyze sucrose to invert sugar nonenzymatically in your kitchen at home? a. Boil the sucrose solution in water. b. Add vinegar (acetic acid) to the solution in a warm water bath. c. Cool the solution and add water, stirring constantly. d. Add baking soda solution (pH≈9) to the solution in a water bath.
ANSWER: a. Boil the sucrose solution in water. b. Add vinegar (acetic acid) to the solution in a warm water bath. WHY: Boiling a solution of sucrose in water hydrolyzes some of the sucrose to invert sugar. Hydrolysis is accelerated and occurs at lower temperatures with the addition of a small amount of acid (vinegar or lemon juice, for example).
Cellulose could provide an abundant and cheap form of glucose for humans. Why is cellulose not a source of nutrients for humans? a. Humans, and most vertebrates, lack the enzyme cellulase. b. Humans cannot absorb the hydrolyzed form of cellulose. c. Animals use starch and glycogen as an energy source, whereas plants use cellulose for energy. d. Human gut bacteria break down cellulose in the large intestine, after nutrient absorption occurs. e. Vertebrate enzymes are unable to hydrolyze (β 1-4) linked glucose monomers.
ANSWER: a. Humans, and most vertebrates, lack the enzyme cellulase. e. Vertebrate enzymes are unable to hydrolyze (β 1-4) linked glucose monomers. WHY: Humans lack the enzyme cellulase in the gut and thus cannot break down cellulose, a polysaccharide made up of glucose monomers. Humans possess enzymes that break bonds between glucose units in the α configuration. However, human enzymes cannot hydrolyze the (β 1-4) glycosidic bonds in cellulose. Although humans cannot utilize cellulose, it is not correct to say that it cannot be used for energy. Some groups of animals are able to hydrolyze cellulose because gut symbionts produce cellulase. Humans can absorb and use the product of cellulose hydrolysis (glucose), but the amount produced by the flora of the digestive tract is small. The undigested cellulose is called fiber and serves to absorb water in the stool.
What effect will hemoglobin have on the bicarbonate reaction in blood in active muscle? a. It will increase HCO−3 production. b. It will increase CO2 production. c. It will increase the amount of O2 unloaded. d. It will decrease the pH level.
ANSWER: a. It will increase HCO−3 production. WHY: The Bohr effect aids in CO2 transport in the blood system. Carbon dioxide transport in the blood relies on three mechanisms: 1. CO2 dissolved in plasma, 2. CO2 bound to proteins as carbamino compounds, 3. as dissolved bicarbonate (HCO-3). The bulk of CO2 is transported as HCO−3, which is formed via a reaction catalyzed by carbonic anhydrase in red blood cells. When pH levels decrease, hemoglobin unloads more oxygen due to the Bohr effect. When hemoglobin is less saturated with oxygen, that is, more reduced, it more readily binds protons. This reaction directly affects the bicarbonate reaction. The removal of H+ in red blood cell cytosol helps drive more CO2 to react with water and form HCO−3. In active muscle, the increased H+ from increased metabolism causes hemoglobin to increase the amount of O2 unloaded. This in turn drives the conversion of CO2 to HCO−3, increasing the bulk transport of CO2 away from active tissues. The ability of deoxygenated blood to carry more CO2 is known as the Haldane effect. Conversely, when deoxygenated hemoglobin reaches the lungs, the high 𝑝O2 causes hemoglobin to load O2 and release H+. The increase in H+, or decrease in pH, drives the bicarbonate reaction in the opposite direction, converting HCO−3 to CO2 that is expelled via the lungs.
Which factors are associated with sickling of red blood cells in sickle cell disease? a. dehydration b. high O2 concentration c. high temperature d. low pH
ANSWER: a. dehydration c. high temperature d. low pH WHY: Sickling can occur as a result of low O2 levels, low blood pH, and increased blood viscosity (dehydration). The oxygen affinity of hemoglobin decreases when pH is low, causing hemoglobin to convert to the T state and release bound oxygen. As hemoglobin is deoxygenated, sickling may occur. Dehydration lowers blood volume and increases blood viscosity. Exertion and high temperature can result in dehydration, which increases blood viscosity and can cause sickling.
Select the molecules that can form the polar head group of phospholipids. a. inositol b. butanol c. serine d. leucine
ANSWER: a. inositol c. serine WHY: The polar head groups of phospholipids contain a hydroxyl group to attach to the phosphate and a hydrophilic portion to face the cytosol of a cell. If a molecule has only one of those two features, it cannot serve as the polar head group of phospholipids. Serine and inositol are common polar head groups of phospholipids. Serine has a terminal hydroxyl group at one end and a hydrophilic protonated amine at the other. Inositol is a sugar alcohol with a hydroxyl group attached to each carbon. One hydroxyl group attaches to the phosphate, whereas the other hydroxyl groups make the molecule hydrophilic. Leucine does not have a terminal hydroxyl group to attach to the phosphate, so it will not form the polar head of a phospholipid. Butanol does have a terminal hydroxyl group that can attach to the phosphate, but the rest of butanol is hydrophobic and only composed of hydrocarbons.
Are starch, glycogen, cellulose, and chitin classified as homopolymers or heteropolymers? a. All four polysaccharides are heteropolymers. b. All four polysaccharides are homopolymers. c. Chitin is a heteropolymer, whereas starch, cellulose, and glycogen are homopolymers. d. Starch is a heteropolymer, whereas chitin, cellulose, and glycogen are homopolymers.
ANSWER: b. All four polysaccharides are homopolymers. WHY: Polysaccharides are polymers of monosaccharides. Homopolysaccharides are made up of only one type of monomer, whereas heteropolysaccharides are composed of at least two different kinds of monomers. Chitin is a structural polysaccharide found in the exoskeleton, or hard outer covering, of arthropods (for example, insects and lobsters) and in the internal structures of some other invertebrates. It is a linear polysaccharide composed of N‑acetylglucosamine residues and is therefore a homopolysaccharide. Starch is made up of two glucose polymers, amylose and amylopectin. Amylose is a linear chain of D‑glucose, and amylopectin is a branched D‑glucose with a glycosidic linkage approximately every 30 glucose units. Because starch is made up of glucose monomers only, it is a homopolysaccharide. Starch is a glucose storage molecule for plants. Glycogen is the glucose storage molecule in animals. It is a branched polymer of glucose and is a homopolymer. Cellulose is a structural polymer present in the cell walls of plants and some algae. The beta configuration of the glucose residues in cellulose permit them to form long, straight chains which can interact through hydrogen bonds. It is a linear, unbranched polymer of D‑glucose and is a homopolysaccharide.
Which statements about reducing sugars are true? a. A reducing sugar will not react with the Cu2+ in Benedict's reagent. b. D‑Arabinose (an aldose) is a reducing sugar. c. Reducing sugars contain ketone groups instead of aldehyde groups. d. A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar. e. The oxidation of a reducing sugar forms a carboxylic acid sugar.
ANSWER: b. D‑Arabinose (an aldose) is a reducing sugar. d. A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar. e. The oxidation of a reducing sugar forms a carboxylic acid sugar. WHY: Some sugars are called reducing sugars because they can act as reducing agents. Benedict's reagent is used to test for reducing sugars. This reagent contains a mild oxidizing agent (Cu2+) at an alkaline pH. The oxidation of an aldose yields a carboxylic acid sugar. D‑Glucose is shown as an example. Reducing sugars are not limited to aldose monosaccharides. Ketose sugars can also be reducing sugars because they rearrange to form aldoses under the alkaline pH condition of the Benedict's reagent. Some disaccharides are reducing sugars, depending on whether both anomeric carbons are involved in the glycosidic linkage. If both anomeric carbons are part of the glycosidic bond, then it is a nonreducing sugar (e.g., sucrose). If one anomeric carbon is not part of the glycosidic bond, then the disaccharide is a reducing sugar (e.g., maltose, lactose) because the free anomeric carbon's aldehyde or ketone group can act as a reducing agent.
Which of the statements are true? a. Triacylglycerols (triglycerides) are carboxylic acids. b. Fats that contain more saturated fatty acid residues than unsaturated fatty acid residues are more likely to be solid at room temperature. c. Saturated fats are more likely than unsaturated fats to be solid at room temperature. d. Unsaturated fats have higher melting points than saturated fats. e. Triacylglycerols (triglycerides) contain ester bonds.
ANSWER: b. Fats that contain more saturated fatty acid residues than unsaturated fatty acid residues are more likely to be solid at room temperature. c. Saturated fats are more likely than unsaturated fats to be solid at room temperature. e. Triacylglycerols (triglycerides) contain ester bonds. WHY: Fats and oils are triacylglycerols (triglycerides). A triacylglycerol is composed of three fatty acid residues bonded to a glycerol molecule with ester bonds. The fatty acid residues can be the same or different. The molecule shown is a triacylglycerol with two saturated fatty acid residues and one unsaturated fatty acid residue. Saturated fats contain more saturated fatty acids than unsaturated fatty acids. Unsaturated triacylglycerols contain more unsaturated fatty acids than saturated fatty acids. The example molecule shown is a saturated fat because it has two saturated fatty acid residues and one unsaturated fatty acid residue. Saturated fats such as lard have higher melting points and are generally solid at room temperature; unsaturated fats such as vegetable oil are generally liquid at room temperature. Saturated fatty acids are relatively flexible and straight, and can stack together more closely, resulting in close intermolecular attractions. The double bonds in unsaturated fatty acids introduce inflexible bends in the carbon chain. As a result, the molecules cannot stack closely together, resulting in fewer intermolecular attractions and lower melting points.
Which statement that best explains the role of BPG in O2 transport from mother to fetus? a. HbF will extract O2 from oxygenated HbA because BPG enhances O2 binding to HbF. b. HbA‑bound O2 will tend to move to HbF because HbF has a lower affinity for BPG, an allosteric inhibitor of O2 binding. c. In the placental circulation, HbF will load up on O2 as BPG dissociates from HbA and binds to HbF. d. HbA‑bound O2 will tend to move to HbF because HbA binds an additional BPG per tetramer when it enters placental circulation.
ANSWER: b. HbA‑bound O2 will tend to move to HbF because HbF has a lower affinity for BPG, an allosteric inhibitor of O2 binding. WHY: The blue (+BPG) curves show that the HbF occupancy by O2 is higher than HbA over the entire range of pO2. The black (-BPG) curves are about the same for HbA and HbF. This implies that the magnitude of the rightward shift for the +BPG curves is dependent on the affinity of the hemoglobin form for BPG. Thus, HbF has a smaller rightward shift due to a lower affinity for BPG. BPG (2,3‑bisphosphoglycerate) has three negatively charged groups that bind to a positively charged pocket between the β‑subunits in deoxyhemoglobin. Since the BPG binding site spans the β‑subunits, there is only one binding site per deoxyhemoglobin. BPG does not bind to oxyhemoglobin because a shift of the β‑subunits in the R state obscures the BPG binding site. BPG is a negative allosteric regulator of O2 binding for both HbA and HbF, but the inhibitory effect is greater for HbA. This means that when HbA, HbF, and O2 are present together, O2 binding to HbF will be favored. This is illustrated in the placenta, where O2 flows from the mother's blood (containing HbA) to the fetus's blood (HbF).
A researcher has observed that the oxygen affinity of adult hemoglobin (HbA) is higher than expected when using a purified sample rather than red blood cell lysate. She plots the oxygen affinity curves for the purified HbA, cell lysate HbA, and purified HbA in the presence of 2.5 mM 2,3‑bisphosphoglycerate (2,3‑BPG). 2,3‑BPG is known to be an allosteric regulator of HbA in red blood cells. The researcher also compares the HbA data with data from a sample of purified recombinant fetal hemoglobin (HbF) and purified HbF in the presence of 2.5 mM 2,3‑BPG. All purified hemoglobin samples have the same protein concentration. The five dissociation curves are shown in the graph. Given the oxygen dissociation curves, which of the following statements are correct? a. The allosteric effects of 2,3‑BPG are homotropic. b. HbF loads oxygen at lower 𝑝O2pO2 than does HbA in the presence of 2,3‑BPG. c. 2,3‑BPG only weakly interacts with HbF. d. 2,3‑BPG is an allosteric activator of HbA. e. Purified HbA has a higher oxygen affinity than purified HbF.
ANSWER: b. HbF loads oxygen at lower 𝑝O2pO2 than does HbA in the presence of 2,3‑BPG. c. 2,3‑BPG only weakly interacts with HbF. e. Purified HbA has a higher oxygen affinity than purified HbF. WHY: A comparison of the HbA and HbF oxygen dissociation curves shows the difference in the allosteric effects of 2,3‑BPG on each molecule. 2,3‑BPG has a significant effect on the oxygen affinity of HbA, with the 𝑃50 changing from roughly 4 torr to 24 torr. Compare that to the difference between the 𝑃50 of purified HbF and HbF in the presence of 2,3‑BPG, which changes from 8 torr to 11 torr. The effect of 2,3‑BPG on the normal physiological function of HbA is notable. Purified HbA has a greater oxygen affinity than purified HbF, as shown by the relative positions of their dissociation curves. The purified HbA dissociation curve lies to the left of the purified HbF curve. If this were the case in the body, HbF would be unable to deliver oxygen to the fetus's tissues. Note that the cell lysate HbA dissociation curve lies to the right of the all the HbF dissociation curves. The experiment shows that HbA unloads roughly two‑thirds of its oxygen between 47 torr and 18 torr when 2,3‑BPG is present. At 18 torr, HbF is still roughly 80% saturated. It can load the oxygen released by HbA and deliver it to fetal tissues. In the absence of 2,3‑BPG, purified HbF unloads oxygen before purified HbA, denying the fetus oxygen. However, as the dissociation curves show, the oxygen affinity of HbF is higher than HbA in the presence of 2,3‑BPG.
How does hemoglobin function as a pH buffer? a. Hemoglobin binds hydrogen ions when carbon dioxide exits the red blood cell. b. Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell. c. Hemoglobin releases hydrogen ions after carbon dioxide enters the red blood cell. d. Hemoglobin releases hydrogen ions when oxygen exits the red blood cell.
ANSWER: b. Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell. WHY: Hemoglobin is part of the protein buffer system, which maintains the pH of body fluids around 7.4. It functions as a buffer by binding to hydrogen ions after carbon dioxide diffuses into the red blood cell (RBC), thus preventing a drop in pH. In the systemic capillaries, carbon dioxide diffuses into the blood. Most of this carbon dioxide enters the RBCs and combines with water to form carbonic acid. Some of the carbonic acid dissociates into hydrogen ions and bicarbonate ions. Hemoglobin prevents a drop in pH by binding to and buffering these hydrogen ions. Hemoglobin does not bind to hydrogen ions when carbon dioxide exits the RBC. In the pulmonary capillaries, carbon dioxide diffuses out of the blood. Hemoglobin then releases hydrogen ions, which combine with bicarbonate ions to form carbonic acid. Carbonic acid is converted to carbon dioxide and water. The newly formed carbon dioxide then diffuses out of the blood. Hemoglobin does not release hydrogen ions when carbon dioxide enters the RBC. The entry of carbon dioxide increases the concentration of hydrogen ions, and hemoglobin binds to and buffers these hydrogen ions. Hemoglobin does not release hydrogen ions when oxygen exits the blood. In the systemic capillaries, oxygen exits the blood and carbon dioxide enters. The hydrogen ions generated by the entry of carbon dioxide bind to hemoglobin, change its shape, and allow oxygen to be released.
The researcher obtains the sequence used to generate the recombinant HbF and also sequences her sample of purified HbA. Knowing that the HbA β subunit C‑terminus is important for the interaction between HbA and 2,3‑BPG, the researcher compares this portion of HbA with the C‑terminal portion of the HbF γ subunit. Which residue in HbA β do you think contributes the most to the increased interaction between HbA aand 2,3‑BPG? a. Ala-142 b. His-143 c. Asn-139 d. Gly-136 e. Lys-144
ANSWER: b. His-143 WHY: Adult hemoglobin (HbA) is an α2β2 tetramer. Fetal hemoglobin (HbF) is an α2γ2 tetramer. The gamma subunits of HbF are homologous to the beta subunits of HbA. Although HbA β and HbF γ have 73% sequence identity and 86% structural similarity, the primary sequence differences have significant physiological consequences. 2,3‑Bisphosphoglycerate (2,3‑BPG) is an allosteric regulator of HbA. A single molecule of 2,3‑BPG binds in a pocket at the center of the α2β2 tetramer. Binding involves electrostatic interactions between the anionic 2,3‑BPG molecule and positively charged amino acid R groups on each HbA β subunit. These R groups are present on amino acid residues His‑2, Lys‑82, and His‑143 of the β subunit. 2,3‑BPG has a weak affinity for HbF. His‑143 of the HbA β subunit has a +1 imidazole group. Ser‑143 in the HbF γ subunit is uncharged. This specific amino acid change in HbF lessens the positive charge in each γ subunit, weakening the interaction with 2,3‑BPG. Several other residues in the C‑terminus of HbF γ are different from the equivalent residues in HbA β. However, these changes involve R groups of similar types. For example, both glycine and alanine have small, nonpolar R groups, so the G136A substitution does not affect functionality. Only the H143S substitution significantly affects the electrostatic properties of the hemoglobin binding pocket. 2,3‑BPG is an allosteric inhibitor of hemoglobin. 2,3‑BPG shifts the hemoglobin dissociation curve to the right, increasing the 𝑝O2 at which half the binding sites are occupied by O2 (the 𝑃50). The function of hemoglobin is to bind O2, so O2 is its substrate. 2,3‑BPG is an allosteric regulator, not a substrate, so its allosteric effects on hemoglobin are termed heterotropic. In contrast, homotropic effects are the result of substrate binding. The effects of O2 on hemoglobin binding to more O2 are homotropic effects.
Which of the statements are true? a. Cell membranes in reindeer legs (near the hooves) are kept flexible because they have a large number of saturated fatty acids. b. A plant that produces more monounsaturated fatty acids than polyunsaturated fatty acids in its membranes in winter has an increased resistance to freeze damage. c. Cell membranes in cold tolerant winter wheat plants have a higher ratio of unsaturated fatty acids to saturated fatty acids than do cold intolerant wheat varieties. d. Yeast cells that produce more unsaturated fatty acids than saturated fatty acids in response to cold have greater cold tolerance.
ANSWER: c. Cell membranes in cold tolerant winter wheat plants have a higher ratio of unsaturated fatty acids to saturated fatty acids than do cold intolerant wheat varieties. d. Yeast cells that produce more unsaturated fatty acids than saturated fatty acids in response to cold have greater cold tolerance. WHY: Unsaturated fatty acids allow flexibility in cell membranes at lower temperatures. Unsaturated fatty acids have a bent, inflexible shape around their double bonds and cannot fit as closely together as saturated fatty acids. Therefore, unsaturated fatty acids have fewer intermolecular attractions and generally have lower melting points. Among unsaturated fatty acids, melting point generally decreases with an increasing number of double bonds. Cell membranes that contain a higher proportion of unsaturated fatty acids to saturated fatty acids have lower melting points and therefore require lower temperatures to freeze (become solid). These cell membranes remain flexible at lower temperatures. Winter wheat plants have a higher ratio of unsaturated fatty acids to saturated fatty acids and are, as a result, more cold tolerant than other wheat varieties. Some yeast cells produce more unsaturated fatty acids in response to cold, and thereby gain greater cold tolerance. The cell membranes in the legs of reindeer (near the hooves) have a higher proportion of unsaturated fatty acids than do cell membranes in other areas of the body. These cells need to be more cold tolerant because the lower legs and hooves of reindeer are often in contact with snow. Plants that produce more polyunsaturated fatty acids (more than one double bond) than monounsaturated fatty acids (one double bond) in response to cold have increased cold tolerance.
These two branched polysaccharides contain both α‑1,4 and α‑1,6‑glycosidic bonds.
Amylopectin and glycogen
The glucose units in this polysaccharide are joined only by α‑1,4‑glycosidic bonds.
Amylose
These two polysaccharides make up starch, the storage form of glucose in plants.
Amylose and amylopectin
Monosaccharides can undergo reactions to form other functional groups. Identify the monosaccharide derivative that fits the statement: _________________________ forms when carbon 1 of a hexose is oxidized to a carboxylic acid.
An aldonic acid WHY: Aldonic and uronic acids have carboxyl groups at a terminal carbon atom of the carbohydrate chain. Aldonic acids are formed from the oxidation of the terminal aldehyde, whereas uronic acids are formed from the oxidation of the terminal hydroxyl. Aldonic acids are typically formed from mild oxidizing reagents, such as Br2 and water. An example of an aldonic acid is gluconic acid. Aldaric acids result from the oxidation of both the terminal aldehyde and the terminal hydroxyl. Monosaccharides are oxidized by nitric acid to form aldaric acids.
Draw the Fischer projections of the four aldotetroses. Draw the D‑sugar on the left and its L‑isomer directly to the right of it. Be sure you select the appropriate hydroxy group so that the bond carbon is connected to the oxygen bond.
An aldotetrose is a four-carbon monosaccharide that contains an aldehyde group. There are four aldotetrose sugars, D‑threose, L‑threose, D‑erythrose, and L‑erythrose. The "bottom" chiral carbon atom (the chiral carbon atom farthest from the C=OC=O) determines whether the sugar is in the D‑ or L‑ configuration: In the D‑configuration, the OHOH group on this carbon atom is on the the right, whereas in the L‑configuration the OHOH group is on the left.
Explain an oxygen-binding curve
An oxygen‑binding curve illustrates the relationship between saturation of oxygen binding sites and the partial pressure of oxygen. In other words, it is a graph of saturation versus the partial pressure of oxygen. P50 is the partial pressure of oxygen at 50% saturation. The hyperbolic myoglobin curve is to the left of the sigmoidal (S‑shaped) hemoglobin curve. Therefore, at all oxygen concentrations, myoglobin has a higher saturation than hemoglobin does, which means that myoglobin has the higher oxygen affinity.
What is happening at each curve?
Curve 1 represents the outcome of a loss of quaternary structure. Curve 2 can represent both a decrease in CO2CO2 in solution and an increase in pH. Curve 3 is the normal state of hemoglobin with no perturbation. Curve 4 shows the effect of an increase in 2,3-BPG concentration. Curve 1 is not sigmoidal and represents a loss of cooperativity in the oxygen‑binding behavior of hemoglobin. Hemoglobin derives its cooperativity from the interactions driven by conformational changes among the subunits of the tetramer. A loss of quaternary structure means that the individual subunits of hemoglobin no longer contact one another, making cooperativity impossible. The combination of CO2 and water creates carbonic acid (H2CO3).(H2CO3). The dissociation of newly produced carbonic acid lowers the pH. Conversely, decreasing the amount of CO2 in solution will raise the pH. Hemoglobin is sensitive to changes in pH due to the presence of several key chemical groups in the tetramer that have pKa values of around 7.7. When these groups are protonated, salt bridges form within hemoglobin that stabilize the tense (T) state. At higher pH values, these groups deprotonate. Deprotonation leads to destabilization of the T state and increases the affinity of hemoglobin for oxygen. This behavior is described by the Bohr effect. 2,3-bisphosphoglycerate (2,3-BPG), sometimes called 2,3-diphosphoglycerate (2,3-DPG), lowers the oxygen affinity of hemoglobin by stabilizing the T state of hemoglobin. Higher oxygen saturation is required to induce the transition from the T state to the relaxed (R) state of hemoglobin and the release of 2,3-BPG.
How does partial pressure of oxygen drive the diffusion of oxygen?
Gas exchange occurs throughout the body for the delivery of oxygen to the cells for metabolism and for the disposal of carbon dioxide produced in cells during metabolism. Oxygen and carbon dioxide are gases and are made up of molecules having mass; therefore, they exert pressure. When oxygen is one component in a mixture of gases, such as air, the amount of pressure exerted by the oxygen molecules alone is called the partial pressure of oxygen. The same principle is true for carbon dioxide. Thus, each component of a gas mixture contributes to the total pressure of the gas. In the respiratory system, the partial pressure of oxygen drives oxygen diffusion, or net movement. Like other gases, oxygen diffuses down its partial pressure gradient, moving from a higher partial pressure to a lower partial pressure. For example, in the lungs, the partial pressure of oxygen is low, with deoxygenated blood returning from the tissues. The partial pressure of oxygen in the inhaled air is higher than that of the blood in the lungs, so oxygen diffuses from the air into the blood in the alveolar capillaries. Although some oxygen is dissolved and carried in the plasma, most of the oxygen is loaded onto hemoglobin in red blood cells for transport. In the tissues, the partial pressure of oxygen is lower than that of the oxygenated blood circulating through the tissues. Oxygen diffuses down the pressure gradient; therefore, oxygen is unloaded from the hemoglobin and moves from the oxygen-rich blood into the oxygen-poor tissues, where it is used in cell metabolism. Although blood pH, body temperature, and carbon dioxide affect oxygen loading and unloading by changing hemoglobin's affinity for oxygen, none is a driving factor for the diffusion of oxygen. The concentration of ozone in the blood has no effect on hemoglobin function or oxygen diffusion in the lungs or tissues.
Select the scenarios which, according to the Bohr effect, result in hemoglobin's release of bound oxygen. a. Oxygen levels in the blood increase. b. The concentration of hydrogen ions in the blood increases. c. The pH of blood increases. d. Carbon dioxide levels in the blood increase.
HINT: ANSWER: b. The concentration of hydrogen ions in the blood increases. d. Carbon dioxide levels in the blood increase. WHY: The Bohr effect describes hemoglobin's reduced affinity for oxygen (O2) when pH decreases. Such a decrease is often due to increased levels of carbon dioxide (CO2). As shown in the graph, changes in blood pH affect hemoglobin's O2 saturation curve. -Low blood pH, meaning blood is more acidic, causes hemoglobin to take up hydrogen ions and have a lower affinity for O2. Thus, more O2 is released at low pH. -Conversely, an increase in pH results in a higher O2 affinity, so more O2 is taken up by hemoglobin. =It should be noted that at high concentrations of O2, hemoglobin is nearly saturated, so a change in pH has less of an effect on O2 release. Blood pH is highly dependent on the amount of CO2 in the blood. -Active tissues produce CO2 as a metabolic byproduct. CO2 reacts with water to form carbonic acid, which dissociates and releases hydrogen ions, lowering pH in the blood and surrounding tissues. -The lower pH decreases hemoglobin's affinity for O2. Thus, in conditions of high carbon dioxide and low pH, such as in actively respiring cells, the Bohr effect accounts for the fact that the lower pH makes it easier for O2 to unbind from hemoglobin before diffusing into the tissues where it is needed.
Which three statements accurately describe the blood buffering system in humans? a. The blood buffering system maintains the pH of blood near 7.4. b. The blood buffering system is facilitated by the enzyme carbonic anhydrase, which interconverts carbon dioxide and water to carbonic acid, ionizing into bicarbonate and H+. c. The blood buffering system utilizes the H2CO3/HCO-3 conjugate acid/base pair. d. The blood buffering system depends on the ionization of H2PO-4. e. The blood buffering system utilizes the acetic acid/acetate conjugate acid/base pair.
HINT: A buffer is a solution of a weak acid and its conjugate base. Buffers resist changes to pH when an acid or a base is added to the solution. Start by considering which types of buffers may be present as biological buffers. Then, determine which buffer is specific to the blood buffering system. ANSWER: a. The blood buffering system maintains the pH of blood near 7.4. b. The blood buffering system is facilitated by the enzyme carbonic anhydrase, which interconverts carbon dioxide and water to carbonic acid, ionizing into bicarbonate and H+. c. The blood buffering system utilizes the H2CO3/HCO-3 conjugate acid/base pair. WHY: In blood, carbon dioxide, CO2 , combines with water, H2O, to form carbonic acid, H2CO3. This reaction is reversible and is aided by the enzyme carbonic anhydrase. The carbonic acid formed may then ionize into H+ and HCO−3. The blood buffering system is based on the ionization of carbonic acid, H2CO3. Although most human body cells have an internal pH ranging from 6.9 to 7.4, the blood buffering system maintains the pH of blood near 7.4. Although phosphate groups serve as biological buffers, they are not part of the blood buffering system. Acetic acid is not part of the blood buffering system. The liver metabolizes ingested ethanol into acetic acid which is further metabolized via the citric acid cycle.
Which of the following statements about Ngb-H64Q is true? a. Ngb-H64Q has a higher affinity than hemoglobin for carbon monoxide, whereas hemoglobin has a higher affinity than Ngb-H64Q for oxygen. b. Ngb-H64Q binds to oxygen and carbon monoxide with equal affinity. c. Ngb-H64Q has a higher affinity for oxygen than it does for carbon monoxide. d. Ngb-H64Q has a higher affinity than hemoglobin for both carbon monoxide and oxygen. However, Ngb-H64Q binds to carbon monoxide with a higher affinity than to oxygen.
HINT: An antidote for carbon monoxide poisoning should efficiently scavenge carbon monoxide from hemoglobin. ANSWER: d. Ngb-H64Q has a higher affinity than hemoglobin for both carbon monoxide and oxygen. However, Ngb-H64Q binds to carbon monoxide with a higher affinity than to oxygen. WHY: The mutation of His 64 to glutamine increased the binding affinity of human neuroglobin for carbon monoxide and oxygen. The mutant neuroglobin, Ngb-H64Q, binds carbon monoxide more tightly than hemoglobin does, allowing it to scavenge carbon monoxide from carboxyhemoglobin. Additionally, Ngb-H64Q binds to carbon monoxide with higher affinity than to oxygen, thus preventing oxygen in the blood from outcompeting carbon monoxide. Both wild-type and mutant neuroglobin bind oxygen with higher affinity than hemoglobin does.
All of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxygen is loaded onto hemoglobin molecules in the lungs and unloaded from the hemoglobin molecules in the tissues. What drives the loading of oxygen onto hemoglobin molecules in the lungs? a. the low partial pressure of oxygen in the lungs b. the high partial pressure of oxygen in the lungs c. the high partial pressure of carbon dioxide in the lungs d. the low partial pressure of carbon dioxide in the lungs
HINT: Gas diffuses down a concentration gradient. (remember - hemoglobin is in the blood) ANSWER: b. the high partial pressure of oxygen in the lungs WHY: Gas diffuses down a concentration gradient, moving from an area of higher partial pressure to an area of lower partial pressure. This diffusion drives gas exchange within the respiratory system. Cells in the body use oxygen during metabolic processes and produce carbon dioxide as a waste product. The blood that arrives at the lungs has a lower concentration of oxygen than the air inhaled into the lungs. Oxygen is loaded onto hemoglobin molecules in the lungs because the air in the lungs has a higher partial pressure of oxygen than the blood entering the lungs. Similarly, the unloading of oxygen into the body's tissues is driven by the lower partial pressure of oxygen in the tissues compared to the oxygenated blood arriving at the tissues. As hemoglobin transports carbon dioxide as well as oxygen, the partial pressure of carbon dioxide affects hemoglobin's affinity for oxygen. However, oxygen and carbon dioxide diffuse down their own concentration gradients; thus, the partial pressure of carbon dioxide does not drive the movement of oxygen. Carbon dioxide diffuses from the tissues into the bloodstream because the partial pressure of carbon dioxide is higher in the tissues than in the blood. Carbon dioxide is then unloaded in the lungs and exhaled.
Which molecules are bound to hemoglobin when hemoglobin is in the R state? a. CO2 b. oxygen c. 2,3‑bisphosphoglycerate d. Fe3+ e. Fe2+
HINT: Hemoglobin has two conformations: the R state and the T state. Consider which state has a higher affinity for oxygen. ANSWER: b. oxygen e. Fe2+ WHY: Hemoglobin has two conformations: the R state and the T state. -The T state is the main conformation of deoxyhemoglobin, and... -the R state is the main conformation of oxyhemoglobin. When oxygen binds to the T state, it causes a conformation change to the R state. Oxygen binding stabilizes the R state. As CO2 binds to hemoglobin, the affinity of hemoglobin for O2 decreases. The heme group contains iron(II), or Fe2+, which binds oxygen reversibly. Fe3+ does not bind oxygen. 2,3‑Bisphosphoglycerate binds to the cavity between the β subunits of hemoglobin in the T state and stabilizes the T state. Conversion to the R state changes the shape and size of the cavity and prevents the binding of 2,3‑bisphosphoglycerate.
Select the phrases that describe the structure of hemoglobin. a. has a single polypeptide containing multiple oxygen‑binding heme groups b. is formed from four globular protein subunits c. contains ferrous ions that bind to both oxygen and carbon dioxide d. contains four ferrous ions, each with a binding site for oxygen e. undergoes conformational changes when oxygen is bound or released
HINT: Hemoglobin is an example of the highest level of protein structure. Hemoglobin is made of multiple folded polypeptides, each with one oxygen‑binding component that is receptive only to oxygen molecules. Carbon dioxide binds to the amino acid chains of the polypeptides. ANSWER: b. is formed from four globular protein subunits d. contains four ferrous ions, each with a binding site for oxygen e. undergoes conformational changes when oxygen is bound or released WHY: Hemoglobin is a complex quaternary protein structure made of four folded polypeptide chains, or globular proteins. These protein subunits consist of two α‑globin and two β‑globin chains. These globins are held together noncovalently by hydrogen bonds, hydrostatic interactions, and ionic bonds to form the final quaternary structure of hemoglobin. -In the image, the α‑globin chains are colored tan and the β‑globin chains are colored purple. Each globin molecule also contains a heme group, which is a nonprotein pigment molecule. The four heme groups are colored red in the image. Heme is responsible for the red color of blood, and each heme group has one iron atom in its center. Each iron atom exists as a ferrous ion (Fe2+) and constitutes an oxygen‑binding site. In this way, one hemoglobin molecule is able to bind a total of four oxygen molecules (O2). Although hemoglobin can also carry carbon dioxide, the heme binding sites are oxygen‑specific. Instead, carbon dioxide binds to the amino acids of the globin subunits. As each O2 binds to Fe2+ in a hemoglobin molecule, it disrupts some of the electrostatic bonds between the globin subunits. This slightly changes the conformation of hemoglobin such that the other oxygen‑binding sites are further exposed, thus increasing hemoglobin's affinity for binding more O2 until all four binding sites are full. As O2 molecules are released to tissues, the conformational changes reverse. Thus, the hemoglobin molecule's affinity for binding O2 diminishes, which promotes oxygen unloading to the tissues.
The oxygen-hemoglobin dissociation curve describes the percent oxygen saturation of hemoglobin as a function of the partial pressure of oxygen (PO2) in the surrounding fluid. As the PO2 increases on the horizontal axis, the oxygen saturation percentage also increases on the vertical axis. In comparison to that of adult hemoglobin, how is the fetal oxygen-hemoglobin dissociation curve shifted? a. to the right of the adult curve, indicating a lower oxygen affinity of fetal hemoglobin b. to the right of the adult curve, indicating a higher oxygen affinity of fetal hemoglobin c. to the left of the adult curve, indicating a higher oxygen affinity of fetal hemoglobin d. to the left of the adult curve, indicating a lower oxygen affinity of fetal hemoglobin
HINT: Maternal blood in the placenta has a lower partial pressure of oxygen than most arterial blood in the rest of the body. ANSWER: c. to the left of the adult curve, indicating a higher oxygen affinity of fetal hemoglobin WHY: The fetal oxygen-hemoglobin dissociation curve is shifted to the left in comparison to the adult curve. The leftward shift indicates that fetal hemoglobin has a higher affinity for oxygen than adult hemoglobin. Maternal blood in the placenta has a lower partial pressure of oxygen (PO2) than normal arterial blood, which decreases the driving force for oxygen diffusion between maternal and fetal blood. However, the higher oxygen affinity of fetal hemoglobin facilitates the extraction of oxygen from maternal blood and helps achieve adequate oxygen saturation of fetal blood. The steepness of the fetal oxygen-hemoglobin dissociation curve at PO2 values below around 40 mmHg indicates that fetal hemoglobin readily binds oxygen, even in relatively low-oxygen environments. This explains how fetal hemoglobin can achieve relatively high oxygen saturation when the PO2 of maternal blood in the placenta is relatively low. The curve steepness also indicates that fetal hemoglobin can readily unload oxygen to fetal tissues at lower PO2 levels. The comparison of hemoglobin oxygen affinities can be understood by choosing a PO2 value and comparing the percent oxygen saturation on each curve for that particular PO2 value. The curve shifted to the left will have a higher percent oxygen saturation than the curve on the right at that PO2 value. Therefore, fetal hemoglobin (the left-shifted curve) has higher oxygen affinity because it binds more oxygen than adult hemoglobin (the curve on the right) at the same PO2.
Suppose Gina climbs a high mountain where the oxygen partial pressure in the air decreases to 70 torr. Assume that the pH of her tissues and lungs is 7.4 and the oxygen concentration in her tissues is 20 torr. The P50 of hemoglobin is 26 torr. The degree of cooperativity of hemoglobin, 𝑛, is 2.8. Estimate the percentage of the oxygen‑carrying capacity that she utilizes. Calculate your answer to one decimal place. Capacity = _________ % After Gina spends a day at the mountaintop, where the oxygen partial pressure is 70 torr, the concentration of 2,3‑bisphosphoglycerate (2,3‑BPG) in her red blood cells increases. Why does increasing the concentration of 2,3‑BPG in Gina's blood cells help her function well at high altitudes? a. Excess 2,3‑BPG binds oxygen when hemoglobin becomes saturated, acting as an oxygen transporter. b. Her hemoglobin P50 decreases, causing more blood‑to‑tissue oxygen offloading. c. Her oxygen‑binding curve shifts to the right, promoting oxygen delivery to tissues. d. The extra 2,3‑BPG stabilizes the relaxed, or R, state of hemoglobin, increasing oxygen binding.
HINT: To determine the percentage of oxygen‑carrying capacity, calculate the difference between the fractional saturation of the lungs and the fractional saturation of the tissues. To determine the percentage of oxygen‑carrying capacity that she utilizes, calculate the difference between the fractional saturation of the lungs and the fractional saturation of the tissues. Fractional saturation, 𝑌, is represented by: Y = [PO2]^n / [PO2]^n + P50^n The term 𝑝O2 is the partial pressure of oxygen. Assume the 𝑝O2 in her lungs is the same as in the air. -The 𝑃50 is the partial pressure of oxygen when hemoglobin is half‑saturated. The 𝑃50 is 26 torr. First, calculate both 𝑌lungs and 𝑌tissues. Then, find the difference between these values. Enter a percentage calculated to one decimal place. ANSWER P1: Capacity = 62% ANSWER P2: c. Her oxygen‑binding curve shifts to the right, promoting oxygen delivery to tissues.
Select all statements that correctly describe hemoglobin and myoglobin structure. a. Each iron atom can form six coordination bonds. Two of these bonds are formed between iron and oxygen. b. The heme prosthetic group is entirely buried within myoglobin. c. Hemoglobin and myoglobin are heterotetramers. d. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron. e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron (Fe)(Fe) atom. f. Each hemoglobin molecule can bind four oxygen molecules; each myoglobin can bind one oxygen molecule. g. Molecular oxygen binds reversibly to the Fe(II) atom in heme.
HINT: Both myoglobin and hemoglobin are globular proteins. Consider whether all globular proteins have quaternary structure. Consider the structure of heme. In addition, there is a histidine residue outside the plane of the heme group that is bonded to Fe . Consider whether the heme group is polar (or ionized), nonpolar, or contains both polar and nonpolar groups. Answer: d. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron. e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron (Fe)(Fe) atom. f. Each hemoglobin molecule can bind four oxygen molecules; each myoglobin can bind one oxygen molecule. g. Molecular oxygen binds reversibly to the Fe(II) atom in heme. WHY: -a. is incorrect because: Each iron atom can form six coordination bonds. One of these bonds are formed between iron and oxygen. Heme is a planar group with both polar and nonpolar functional groups. The polar propionate groups are at the surface of the protein, and the nonpolar groups are buried. Heme contains a central iron atom (Fe2+,Iron(II))(Fe2+,Iron(II)) . Both myoglobin and hemoglobin require heme to bind and carry oxygen. Oxygen binds reversibly to the iron of heme. The iron atom is liganded by the four N atoms of heme and one His residue from the protein, leaving a sixth coordination site for O2 . If heme were not part of a protein, the O2 bonded to heme would oxidize Fe2+ to Fe3+, which cannot bind O2, through a heme‑O2‑heme intermediate. However, the formation of this intermediate is sterically hindered by the interactions between the protein and the oxygen-carrying heme group. Hemoglobin is a heterotetramer, with two α chains and two β chains. Each of the four subunits contains a heme group. Therefore, hemoglobin can bind four O2 molecules. Myoglobin consists of one polypeptide chain (therefore it is a monomer) and can carry one O2 molecule.
For the given changes in environment, choose whether the oxygen binding affinity of hemoglobin will increase, decrease, or not change. -Increase in body temperature. -Decrease in blood pH. -Decrease in partial pressure of CO2.
HINTS: The partial pressure of CO2 has a direct influence on the pH of the blood. If CO2 concentration increases, due to an increased partial pressure, then the pH will also increase. Increasing body temperature causes oxygen to be less soluble in the blood. How would this affect the oxygen concentration? A decrease in blood pH does have an effect on the oxygen binding affinity of hemoglobin. Consider what happens to H+H+ concentration as blood pH decreases. ANSWERS: -Increase in body temperature: decrease -Decrease in blood pH: decrease -Decrease in partial pressure of CO2: increase WHY: Increasing the body temperature affects the solubility of oxygen in the blood system. -This is attributed to Henry's law; by increasing the temperature of the blood system, O2 becomes less soluble. A decrease in the pH causes an increase in H+ concentration, which leads to a decrease in oxygen binding affinity. This is known as the Bohr effect. This is displayed in the given chemical equation, where Hb−4O2 is hemoglobin bound to four oxygen atoms and Hb−2H+ is hemoglobin unbound to oxygen. Increasing the H+ concentration or decreasing the pH will push the reaction to the release of oxygen. When the partial pressure or concentration of CO2 decreases, the oxygen binding affinity of hemoglobin increases. The levels of CO2 influence both the intercellular pH (Bohr effect) and can cause the formation of carbamino compounds (for which hemoglobin has a higher affinity). Decreasing the levels of CO2 will increase the pH and decrease the carbamino compound, and therefore increases the oxygen affinity.
Oligosaccharide residues on cell surfaces provide recognition sites for a variety of functions, including cell-cell adhesion, movement of immune cells through capillary walls, viral attachment and infection of cells, and targeting of red blood cells and serum proteins for destruction. ________________________ are the proteins involved in recognition of specific oligosaccharide structures.
Lectins WHY: Lectins, a word derived from the Latin word "legere," means "to select." Lectins bind carbohydrates and have a high specificity for specific oligosaccharides. These proteins are found in both plants and animals and do not have enzymatic activity. Leptins are hormones that signal satiety. Antibodies are proteins of the immune system that can recognize oligosaccharide structures but can have much broader specificity. Globulins and albumins are broad classes of proteins based on structure and solubility properties, not on function. Although some hormones are peptides, hormones have a general signaling function rather than a recognition function.
Classify the sugars based on whether they show mutarotation or not. galactose glucose lactose sucrose trehalose
Shows mutarotation: galactose glucose lactose Does not show mutarotation: sucrose trehalose WHY: Mutarotation is the term given to the change in the specific rotation of a cyclic sugar as it reaches an equilibrium between its α and β anomeric forms. The open chain aldehyde or keto form is an intermediate in the isomerization. Therefore, an anomeric carbon atom must have the hydroxyl group free to isomerize to the open chain aldehyde or keto form. The monosaccharides galactose and glucose both have free anomeric hydroxyl groups. The disaccharide lactose is 𝛽-D-galactopyranosyl-(1⟶4)-D-glucopyranoseβ-D-galactopyranosyl-(1⟶4)-D-glucopyranose, with the anomeric carbon on the second glucose residue free. The disaccharide sucrose is 𝛼-D-glucopyranosyl-(1⟶2)-𝛽-D-fructofuranosideα-D-glucopyranosyl-(1⟶2)-β-D-fructofuranoside, with the anomeric hydroxyls of both glucose and fructose involved in the disaccharide linkage, so they cannot equilibrate to the open chain form. The disaccharide trehalose is 𝛼-D-glucopyranosyl-(1⟶1)-𝛼-D-glucopyranosideα-D-glucopyranosyl-(1⟶1)-α-D-glucopyranoside, also with the anomeric hydroxyls of both glucose residue involved in the disaccharide linkage, so they also cannot equilibrate to the open chain form.
How are Haworth projection for a monosaccharide converted to its corresponding Fischer projection?
The constituents on carbon atoms that are "down" in the Haworth projection are "right" in the Fischer projection. Constituents on carbons that are "up" in the Haworth projection are "left" in the Fischer projection. When carbon 6 is "up" in the Haworth projection, it is a "D" sugar so the hydroxyl on carbon 5 is "right".
Consider the fatty acid (picture is with the answers) Which of the designations are accurate for the fatty acid? a. ω‑6 fatty acid b. 18:2(Δ9,12) c. 8,11-octadecadienoic acid d. 17:2(Δ8,11) e. 18:2(Δ6,9)
a. ω‑6 fatty acid b. 18:2(Δ9,12) WHY: Several notations are given: a systematic name; a notation that describes the length, number of double bonds, and locations of double bonds; and a notation that uses the term ω. This fatty acid is linoleic acid (cis-,cis-9,12-octadecadienoic acid). Because it has 18 carbon atoms and 2 double bonds (at positions 9 and 12), this fatty acid can be designated by the notation 18:2(Δ9,12). The carbon atom most distant from the carboxyl group is the ω (omega) carbon. If the first double bond is between carbon 3 and carbon 4 from this end (counting the ω carbon as number 1), then it is an omega‑3 (ω‑3) fatty acid. If the first double bond is between carbon 6 and carbon 7, starting at the ω end, then the fatty acid is an omega‑6 (ω‑6) fatty acid. This fatty acid is an ω‑6 fatty acid.
Monosaccharides can undergo reactions to form other functional groups. Identify the monosaccharide derivative that fits the statement: Muramic acid is _____________________
an amino sugar WHY: Amino sugars usually have an amino group replacing the hydroxyl on carbon 2 of the chain. One example is muramic acid.
In the lungs, oxygen diffuses into the blood and is loaded onto hemoglobin for transport. In the tissues, oxygen is unloaded from hemoglobin and diffuses from the blood into nearby cells. What drives the diffusion of oxygen? a. concentration of carbon dioxide b. partial pressure of oxygen c. body temperature d. blood pH e. concentration of ozone
b. partial pressure of oxygen
A glycolipid is: a. a lipid molecule that contains a phosphate group b. a molecule produced in the reaction between a glycerol molecule and a lipid molecule c. a lipid molecule that contains at least one carbohydrate unit d. a lipid molecule produced during glycolysis
c. a lipid molecule that contains at least one carbohydrate unit
Consider the lipid shown. Identify the fatty acids incorporated in the lipid. a. myristic acid b. palmitic acid c. stearic acid d. lauric acid To which class of lipid does this belong? Which properties are true of this lipid? a.The hydrophobic chains are attracted to water due to hydrogen bonding. b. The lipid can form part of a cell membrane. c. The lipid has two nonpolar tails. d. The phosphate ester is hydrophobic. e. The lipid has a polar head group.
c. stearic acid d. lauric acid phospholipid b. The lipid can form part of a cell membrane. c. The lipid has two nonpolar tails. e. The lipid has a polar head group. WHY: The two esters in the lipid are derived from fatty acids. One of the esters comes from C17H35COOH(stearic acid) whereas the other comes from C11H23COOH (lauric acid). Because the lipid contains a phosphate group, it is a phospholipid. A triacylglycerol would contain a triester and a steroid would contain a four fused ring system. The lipid contains a polar hydrophilic (water loving) head group consisting of the protonated amino group and the phosphate group. The head group would be attracted to water due to hydrogen bonding. The nonpolar component of the lipid is the two esters that contain long hydrophobic tails. Because they are hydrophobic, they will not be attracted to water. Phospholipids are often found as part of the phospholipid bilayer in cell membranes. They arrange themselves so that the hydrophobic tails avoid water and the hydrophilic heads are exposed to the aqueous cell exterior or the aqueous cytosol.
