Biochem Test 2
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How many phosphoester (phosphate ester) and phosphoanhydride bonds are in adenosine and adenosine triphosphate (ATP)? For reference, the structure of ATP is shown. phosphoester bonds in adenosine: phosphoanhydride bonds in adenosine: phosphoester bonds in ATP: phosphoanhydride bonds in ATP:
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Identify the components of the pictured single‑stranded DNA molecule.
cytidine deoxyguanosine
Identify the correct name or abbreviation for the given nucleoside or nucleotide. deoxyuridine uridine deoxycytidine CDP cytidine Identify the correct name or abbreviation for the given nucleoside or nucleotide. adenosine dGDP deoxyadenosine guanosine deoxyguanosine
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Identify the features of the DNA backbone.
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Name the monosaccharides in the images by placing the appropriate terms.
Hemoglobin: - the oxygen dissociation curve is sigmoidal in shape ("S" shaped) - as oxygen binds to this molecule the shape of the molecule changes, enhancing further oxygen binding - the binding pattern for this molecule is considered cooperative - this molecule delivers oxygen more efficiently to tissues Myoglobin: - the oxygen dissociation curve is hyperbolic in shape - this molecule has a greater affinity for oxygen Neither: - oxygen binds irreversibly to this molecule - carbon monoxide binds at an allosteric site, lowering oxygen binding affinity
Determine which statements apply to hemoglobin, myoglobin, or neither.
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Disaccharides are joined by glycosidic bonds formed between the anomeric carbon on one monosaccharide and a hydroxyl (−OH) group of another monosaccharide. Identify the types of linkages in each of the three disaccharides. Only place three of the linkages.
No, because myoglobin does not have a quaternary structure.
Does myoglobin exhibit a Bohr effect? Why or why not? No, because low pH increases myoglobin's affinity for oxygen. Yes, because ionic interactions between myoglobin's subunits stabilize the tense (T) state. No, because myoglobin does not have a quaternary structure. Yes, because carbon dioxide competes for myoglobin's oxygen‑binding site.
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Draw the hydrogen bonds between the bases. The letter R represents the rest of the nucleotide. The bases are positioned approximately as they would be in a DNA double helix. To indicate a hydrogen bond, simply draw a regular single bond between the atoms. Be careful not to modify the existing atoms or bonds in any way, nor add extraneous atoms or bonds. Draw the hydrogen bond(s) between thymine and adenine.
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Draw the nucleotide adenosine triphosphate. Include appropriate charges on the phosphate groups.
The product is a hemiketal.
Draw the product of the reaction between a ketone and an alcohol. Include all hydrogen atoms in the product. How would you classify the product of the reaction? Note that a hemiacetal formed from a ketone is also called a hemiketal; an acetal formed from a ketone is also called a ketal. The product is an alcohol. The product is a ketone. The product is a hemiketal. The product is a ketal.
β‑D‑galactopyranosyl 4 α‑D‑glucopyranose . lactose
Give the systematic name for the disaccharide shown in the image by selecting the appropriate terms. The systematic name is ____________-(1 ⟶ ______ )- _________________ The common name is __________
amino acid sequence: Leu-Ser-Val
Given the DNA template strand 3' AACAGACAG 5', write the amino acid sequence in the N‑terminal to C‑terminal direction. Note: Enter the amino acids using their three-letter designations. Put a hyphen between each amino acid. (for example, Glu‑Asp‑Val). Refer to a codon table.
Replication: - DNA polymerase - deoxynucleoside triphosphate - primer Transcription: - RNA polymerase - ribonucleoside triphosphate - promoter Translation: - ribosome - transfer RNA - ribosomal RNA
Identify the components of replication, transcription, and translation processes. You are currently in a sorting module. Turn off browse mode or quick nav, Tab to items, Space or Enter to pick up, Tab to move, Space or Enter to drop. Replication Transcription Translation
adenosine dADP
Identify the correct name or abbreviation for the given nucleoside or nucleotide. deoxyguanosine deoxyadenosine guanosine adenosine ADP Identify the correct name or abbreviation for the given nucleoside or nucleotide. GDP dGDP dADP ADP deoxyadenosine
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Identify the phosphate ester (phosphoester) bond and the N‑glycosidic bond in the nucleotide shown.
D C A Pi
Adenosine triphosphate (ATP) is considered the energy currency for the cell. This molecule is energy-rich, in part, due to its two phosphoanhydride bonds. Resonance structures for ATP are given, where X represents adenosine monophosphate (AMP). Which of the structures are appropriate resonance forms of ATP? D C A B Which molecule is more stable and has more resonance forms: ATP or orthophosphate (Pi)? They have equal stability and the same number of resonance forms. Pi ATP
the high partial pressure of oxygen in the lungs
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? the high partial pressure of oxygen in the lungs the low partial pressure of oxygen in the lungs the low partial pressure of carbon dioxide in the lungs the high partial pressure of carbon dioxide in the lungs
enantiomers
Analyze the following pair of compounds. Which of the terms explains the relationship between the two compounds enantiomers anomers diastereomers epimers
The two copper ions move together upon binding oxygen, and this motion could promote cooperativity.
Arthropods such as lobsters have oxygen carriers quite different from hemoglobin. The oxygen‑binding sites do not contain heme but, instead, are based on two copper (Cu) ions. The structural changes that accompany oxygen binding are shown in the reaction diagram. How might these changes be used to facilitate cooperative oxygen binding? The oxygen‑binding site does not structurally change but could promote cooperativity by rotating with respect to other oxygen‑binding sites. The imidazole groups move apart upon binding oxygen, and this motion could promote cooperativity. The two copper ions move together upon binding oxygen, and this motion could promote cooperativity. A hydrogen bond forms between the oxygen molecule and an imidazole group, which could promote cooperativity.
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Cytosine is a pyrimidine base component of DNA and RNA. Draw cytosine.
both DNA and RNA RNA only DNA only neither DNA nor RNA RNA only both DNA and RNA DNA only neither DNA nor RNA
DNA and RNA are structurally similar in some ways but different in others. Identify whether each of the statements applies to DNA, RNA, both RNA and DNA, or neither DNA nor RNA. It can contain the purine adenine. It can contain the pyrimidine uracil. This contains the sugar 2'-deoxyribose. In terms of base composition, %A=%G . This contains the sugar ribose. The bases are attached to sugars in a β‑N‑glycosidic linkage. It contains equal amounts of purines and pyrimidines. Sugars are connected with a 3'-5' phosphodiether link.
Initiation: - the small ribosomal subunit attaches to an mRNA molecule - an initial tRNA molecule pairs with the AUG codon of the mRNA molecule - the small and larger ribosomal subunits combine to form an initiation complex Elongation: - amino acids are transferred from tRNA molecules to the growing peptide chain - a peptide bond forms, linking the new amino acid to the growing chain - the ribosome moves down the mRNA strand to the next codon Termination: - the ribosome reaches a stop codon, preventing additional amino acids from being added - the tRNA, mRNA, and new protein are released
Determine whether each of the events involved in protein synthesis occurs in initiation, elongation, or termination. Initiation Elongation Termination
serine asparagine threonine
Select the amino acids that attach carbohydrates to proteins. valine serine asparagine threonine phenylalanine
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Classify the monosaccharides.
Aggrecan is heavily decorated with glycosaminoglycans, and the release of these glycosaminoglycans is a sign of aggrecan degradation. Another enzyme might be present that cleaves glycosaminoglycans from aggrecan without degrading aggrecan.
A contributing factor to the development of arthritis is the inappropriate proteolytic destruction of the aggrecan component of cartilage by the proteolytic enzyme aggrecanase. The immune system signal molecule interleukin 2 (IL‑2) activates aggrecanase; in fact, IL‑2 blockers are sometimes used to treat arthritis. Studies were undertaken to determine whether inhibitors of aggrecanase could counteract the effects of IL‑2. Pieces of cartilage were incubated in media with various additions and the amount of aggrecan destruction was measured as a function of time. Aggrecan degradation was measured by the release of glycosaminoglycan. What is the rationale for this assay? Aggrecan is a lectin that binds glycosaminoglycan ligands, and aggrecan releases these glycosaminoglycans as it is degraded. Aggrecan is heavily decorated with glycosaminoglycans, and the release of these glycosaminoglycans is a sign of aggrecan degradation. Aggrecan inhibits the removal of glycosaminoglycans from proteoglycans, and aggrecan degradation allows this removal to proceed. Aggrecan is a transferase enzyme that adds glycosaminoglycans to glycoproteins, and these glycosaminoglycans remain free if aggrecan is degraded. Why might glycosaminoglycan release not indicate aggrecan degradation? Glycosaminoglycans are hydrogen bonded to aggrecan and may dissociate from aggrecan at high temperatures. Aggrecanase may degrade the glycosaminoglycans of aggrecan without degrading the protein component. Glycosaminoglycan release also occurs with the degradation of glycogen in animal tissues. Another enzyme might be present that cleaves glycosaminoglycans from aggrecan without degrading aggrecan.
Aggrecan degradation is greatly enhanced. Aggrecan degradation is reduced to the background system level.
A contributing factor to the development of arthritis is the inappropriate proteolytic destruction of the aggrecan component of cartilage by the proteolytic enzyme aggrecanase. The immune system signal molecule interleukin 2 (IL‑2) activates aggrecanase; in fact, IL‑2 blockers are sometimes used to treat arthritis. Studies were undertaken to determine whether inhibitors of aggrecanase could counteract the effects of IL‑2. Pieces of cartilage were incubated in media with various additions and the amount of aggrecan destruction was measured as a function of time. What is the effect of adding IL‑2 to the system? Aggrecan degradation is greatly enhanced. Aggrecan degradation is greatly reduced. Aggrecanase activity is greatly reduced. Aggrecan degradation remains the same. What is the response when an aggrecanase inhibitor is added in addition to IL‑2? Aggrecan degradation remains the same as when IL‑2 was added. Aggrecan degradation completely stops. Aggrecan degradation is further enhanced. Aggrecan degradation is reduced to the background system level.
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Adenine is a purine base that is a component of several important molecules, including adenosine triphosphate (ATP), flavin adenine dinucleotide (FAD), DNA, and RNA. Draw adenine.
The control provides a baseline of "background" aggrecan degradation inherent in the assay. Some of the factors contributing to cartilage stabilization in vivo may not be present.
A contributing factor to the development of arthritis is the inappropriate proteolytic destruction of the aggrecan component of cartilage by the proteolytic enzyme aggrecanase. The immune system signal molecule interleukin 2 (IL‑2) activates aggrecanase; in fact, IL‑2 blockers are sometimes used to treat arthritis. Studies were undertaken to determine whether inhibitors of aggrecanase could counteract the effects of IL‑2. Pieces of cartilage were incubated in media with various additions and the amount of aggrecan destruction was measured as a function of time. What is the purpose of the control (cartilage incubated with no additions)? The control provides the maximum level of aggrecan degradation that the assay can detect. The control provides a baseline of "background" aggrecan degradation inherent in the assay. The control provides the level of aggrecan degradation that occurs when aggrecanase is inhibited. The control provides the level of aggrecan degradation that occurs in vivo. Why is there some aggrecan destruction in the control with the passage of time? Cartilage produces small amounts of IL‑2 on its own, which activates aggrecanase. Aggrecan is a glycoprotein, and glycoproteins are inherently unstable with a short half‑life. Some of the factors contributing to cartilage stabilization in vivo may not be present. Aggrecan destruction would only occur in the control if there was an experimental error.
a single‑stranded closed circle of 1000 nucleotides base‑paired to a linear strand of 500 nucleotides with a free 3′−OH terminus It has a primer with a free 3′−OH group. It has an available template strand.
A solution contains DNA polymerase and the Mg2+ salts of dATP, dGTP, dCTP, and TTP. When added to aliquots of the solution, which DNA molecule would lead to DNA synthesis? a single‑stranded closed circle containing 1000 nucleotide units a double‑stranded linear molecule of 1000 nucleotide pairs with a free 3′−OH group at each end a double‑stranded closed circle containing 1000 nucleotide pairs a single‑stranded closed circle of 1000 nucleotides base‑paired to a linear strand of 500 nucleotides with a free 3′−OH terminus Identify the reasons why the DNA molecule you selected would lead to DNA synthesis. ATP supplies the energy required. It has a primer with a free 3′−OH group. It is double‑stranded DNA. It has an available template strand.
chitin: - consists of N-acetylgcosamine residues - provides structural support for animals such as arthropods starch: - is a storage form of fuel in plant cells - is made up of two glucose polysaccharides, amylose and amylopectin glycogen: - functions in fuel storage in animal cells - is the storage form of glucose in animals cellulose - provides structural support for plants All four polysaccharides are homopolymers.
Assign each statement to the corresponding polysaccharide. Chitin Starch Glycogen Cellulose Polysaccharides are polymers of monosaccharides. Homopolysaccarides are made up of only one type of monomer, whereas heteropolysaccharides are composed of at least two different kinds of monomers. Are starch, glycogen, cellulose, and chitin classified as homopolymers or heteropolymers? All four polysaccharides are heteropolymers. Starch is a heteropolymer, whereas chitin, cellulose, and glycogen are homopolymers. Chitin is a heteropolymer, whereas starch, cellulose, and glycogen are homopolymers. All four polysaccharides are homopolymers.
inositol pentaphosphate, because of its negative charge
Blood cells from some birds do not contain 2,3‑bisphosphoglycerate but, instead, contain one of the compounds shown, which plays an analogous functional role. Which compound do you think is most likely to play this role? Why? choline, because of its positive charge spermine, because it is electrically neutral inositol pentaphosphate, because of its negative charge indole, because of its hydrophobic ring system
D‑glucose and D‑mannose
Compare the structures of D‑glucose, L‑glucose, D‑mannose, and D‑galactose. Which two carbohydrates are epimers? D‑mannose and D‑galactose L‑glucose and D‑mannose D‑glucose and D‑mannose D‑glucose and L‑glucose
DNA: - a single molecule that can be over 10,000 nucleotides long - makes up the genome for eukaryotic organisms - usually double-stranded - includes deoxyribose sugar RNA: - usually single-stranded - includes the base uracil - can be translated into a protein - a single molecule that can form a complex secondary structure
Classify each description as corresponding to DNA or RNA. You are currently in a sorting module. Turn off browse mode or quick nav, Tab to items, Space or Enter to pick up, Tab to move, Space or Enter to drop. DNA RNA
DNA polymerase: - uses deoxyribonuleoside triphosphates - exhibits semiconservative replication - needs primer RNA polymerase: - uses ribonucleoside triphosphates - does not need a primer
Classify each feature as characteristic of DNA polymerase or RNA polymerase. DNA polymerase RNA polymerase
aldotetrose aldohexose ketohexose
Classify each monosaccharide according to the position of the carbonyl group and the number of carbon atoms in the molecule. ketopentose ketotriose aldotetrose aldotriose ketotetrose aldopentose aldopentose aldohexose ketopentose ketotetrose aldotetrose ketohexose fructose aldohexose ketohexose ketotetrose ketopentose aldopentose aldotetrose
purine, deoxyribose pyrimidine, deoxyribose purine, ribose
Classify each nucleotide or nucleoside by the type of base and sugar present in each image. Classify the molecule. pyrimidine, deoxyribose purine, ribose purine, deoxyribose pyrimidine, ribose Classify the molecule. pyrimidine, deoxyribose purine, ribose pyrimidine, ribose purine, deoxyribose Classify the molecule. purine, ribose pyrimidine, ribose pyrimidine, deoxyribose purine, deoxyribose
tRNA: - anticodon - transports amino acids mRNA - product of transcription - codon rRNA: - combines with protein to form ribosomes
Classify each term and description based on whether it is associated with transfer RNA (tRNA), messenger RNA (mRNA), or ribosomal RNA (rRNA). tRNA mRNA rRNA
Nucleosides: - do not contain a phosphate group - are the product when a base bonds a C-1' of ribose or deoxyribose Both: - are found in DNA and RNA - contain a base and a monosaccharide - may contain either a purine or pyrimidine Nucleotides: - contain a base, a monosaccharide, and a phosphate group - are the monomers of nucleic acids - exemplified by adenosine 5'-monophosphate
Classify the descriptions as pertaining to nucleosides, nucleotides, or both nucleosides and nucleotides. Nucleosides Both Nucleotides
T state: - stabilized by increased ion pairings at the a1B2 and a2B1 interfaces - iron ion protrudes from heme towards His F8 - tense state of hemoglobin R state: - narrowed pocket between B subunits - iron ion assumes a planar conformation - relaxed state of hemoglobin
Classify the overall structure of hemoglobin in its two conformational states, based on images depicting the conformational changes in heme.
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Classify the sugars as either aldoses or ketoses.
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Complete the Haworth projection of α-D-galactose by placing labels on the pyranose ring. The anomeric carbon is shown.
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Complete the Haworth projection of 𝛽‑D‑fructose. The anomeric carbon is shown
heme porphyrin Fe 2+ deoxyhemoglobin six four histidine
Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is _______ The organic ring component of heme is _______ Under normal conditions, the central atom of heme is ________ In __________, the central iron atom is displaced 0.4 A out of the plane of the porphyrin ring system. The central atom has _______ bonds: _________ to nitrogen atoms in the porphyrin, one to a _______ residue, and one to oxygen
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Complete the structure of the monosaccharide present in DNA. The sugar should be in its β-furanose form.
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Consider the Fischer projection. Which Haworth projection corresponds to the β‑pyranose form of the Fischer projection?
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Consider the Haworth projection. Which of the chair conformations is represented by the Haworth projection?
epimers diastereomers
Consider the Haworth projections of β‑L‑galactose and β‑L‑glucose shown here. Which terms describe the relationship between these two sugars? enantiomers anomers epimers diastereomers
Nucleosides do not contain a phosphate group.
Consider the difference between nucleosides and nucleotides. Select the true statement. Nucleosides are monomers of nucleic acids. Nucleotides can be named deoxyguanosine. Nucleosides do not contain a phosphate group. Nucleosides are only found in DNA.
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Consider the image of protein synthesis (translation). Identify the two different RNA molecules involved in the process, as well as the amino acids being produced.
2' 1 non-reducing
Consider the structure of sucrose with labeled carbon atoms. Identify the anomeric carbon atoms of sucrose. 4′ 2′ 1 4 Is sucrose a reducing or non‑reducing sugar? reducing non‑reducing
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Convert the Haworth projection for a monosaccharide to its corresponding Fischer projection. Select the Fischer projection that is the open chain version of the Haworth projection.
3′-CGTAAGTTC-5′
Enter the complementary sequence to the DNA strand shown. 5′- GCATTCAAG -3′
No, because both anomeric carbons are involved in the glycosidic linkage A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar. D‑Arabinose (an aldose) is a reducing sugar. The oxidation of a reducing sugar forms a carboxylic acid sugar.
Glucose and fructose are reducing sugars. Sucrose, or table sugar, is a disaccharide consisting of both fructose and glucose. Is sucrose a reducing sugar? Why or why not? No, because both anomeric carbons are involved in the glycosidic linkage. Yes, because the fructose unit can convert to the open‑chain form. No, because only one anomeric carbon is involved in the glycosidic linkage. Yes, because the glucose unit can convert to the open‑chain form. Which statements about reducing sugars are true? A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar. D‑Arabinose (an aldose) is a reducing sugar. The oxidation of a reducing sugar forms a carboxylic acid sugar. A reducing sugar will not react with the Cu2+ in Fehlings's reagent. Reducing sugars contain keto groups instead of aldehyde groups.
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Glucose generally exists in ring (cyclic) form. A Haworth projection shows the orientations of the hydroxyl groups and the hydrogen atoms on the ring. Draw the α and the β forms of glucose by placing the groups (H,OH,or CH2OH) in the appropriate positions.
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Guanine is a component of several important molecules, including guanosine triphosphate (GTP), DNA, and RNA. Draw guanine.
D
Hemoglobin Kansas (described by Bonaventura and Riggs in 1968) is a hemoglobin variant in which a Thr residue is substituted for Asn at residue 102 of the β chain. This mutation destabilizes the R state. Which graph could represent hemoglobin Kansas's dissociation curve?
accumulation of carbon dioxide decrease in pH
Hemoglobin is a protein in red blood cells that binds to oxygen. Which physiological changes that naturally occur in the body reduce hemoglobin's affinity for oxygen? accumulation of carbon dioxide decrease in temperature accumulation of nitrogen decrease in pH
The addition of galactose to the O antigen specifies type B. The addition of N‑acetylgalactosamine to the O antigen specifies type A. The absence of a glycosyl unit on the O antigen specifies type O.
How does glycosylation influence the specification of blood type? The removal of N‑acetylglucosamine from the O antigen specifies type A. The addition of galactose to the O antigen specifies type B. The addition of N‑acetylgalactosamine to the O antigen specifies type A. The absence of a glycosyl unit on the O antigen specifies type O. The removal of fructose from the O antigen specifies type B. The removal of sialic acid from the O antigen specifies type O.
1 2
How many phosphoester (phosphate ester) and phosphoanhydride bonds are in adenosine triphosphate (ATP)? For reference, the structure of ATP is shown here. phosphoester bonds: phosphoanhydride bonds:
1 greater than nonreducing
How many reducing ends are in one glycogen molecule? number of reducing ends: Complete the passage comparing the nonreducing and reducing ends in a molecule of glycogen. The number of nonreducing ends in a glycogen molecule is _________ the number of reducing ends in the molecule. As we will see in Chapter 21, glycogen is an important fuel‑storage form that is rapidly mobilized. You would expect most metabolism to take place at the ________ end(s) of glycogen.
one of four iron‑containing parts of hemoglobin that bind to oxygen molecules
Identify the best description of a heme group. protein that contains four molecules which carry oxygen inside red blood cells single metal atom bound to each of the four oxygen binding parts of hemoglobin family of hormones that stimulate hemoglobin production under hypoxic conditions one of four iron‑containing parts of hemoglobin that bind to oxygen molecules
The iron in the heme group can form six bonds. The fifth is with an imidazole ring of a histidine, and the sixth is with oxygen. Hemoglobin contains four heme groups. The R state of hemoglobin has greater affinity for O2 than the T state because the iron atom is in the plane of the porphyrin ring. A heme group contains four pyrrole rings, which are linked via methene (methine) bridges. The heme group is conjugated.
Identify the true statements regarding the heme group of myoglobin and hemoglobin. The globin chains of myoglobin and hemoglobin prevent the oxidation of Fe2+ to Fe3+ , which irreversibly binds oxygen. The cooperativity of oxygen binding in hemoglobin arises from a shorter Fe−O2 bond length in the T state compared to the R state. The iron in the heme group can form six bonds. The fifth is with an imidazole ring of a histidine, and the sixth is with oxygen. Hemoglobin contains four heme groups. The R state of hemoglobin has greater affinity for O2 than the T state because the iron atom is in the plane of the porphyrin ring. A heme group contains four pyrrole rings, which are linked via methene (methine) bridges. The heme group is conjugated.
contain four ring nitrogen atoms. contain two heterocyclic rings. contain only two ring nitrogen atoms. contain one heterocyclic ring.
Identify two structural features of purines and pyrimidines. Purines contain four ring nitrogen atoms. contain only two ring nitrogen atoms. contain only three ring nitrogen atoms. contain one heterocyclic ring. contain two heterocyclic rings. Pyrimidinescontain only two ring nitrogen atoms. contain four ring nitrogen atoms. contain one heterocyclic ring. contain two heterocyclic rings. contain only three ring nitrogen atoms. Classify each molecule as a purine or a pyrimidine. Purines Pyrimidines
cytosine
In Watson-Crick base pairing in DNA, guanine pairs with cytosine uracil adenine thymine
partial pressure of oxygen
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? partial pressure of oxygen concentration of nitric oxide body temperature blood pH partial pressure of carbon dioxide
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Label blood types O and A with the correct monosaccharides. The defining monosaccharide is noted by the gray and orange bond. Each monosaccharide may be used more than once.
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Label blood types O and B with the correct monosaccharides. The defining monosaccharide is noted by the gray and orange bond. Each monosaccharide may be used more than once.
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Label the diagram of a growing polynucleotide chain.
base stacking hydrogen bonding hydrophobic interactions
Select the stabilizing factors that hold the two strands of DNA together in a double helix. covalent bonding base stacking hydrogen bonding hydrophobic interactions ionic bonding
glycogen cellulose amylose cellulose amylopectin, glycogen amylose, amylopectin
Match each description with the correct polysaccharide. This polysaccharide is the storage form of glucose in the human body This polysaccharide is the structural component of plant cell walls and cannot be digested by humans The glucose units in this polysaccharide are joined only by a-1,4-glycosidic bonds the glucose units in this polysaccharide are joined by B-1,4-glycosidic bonds These two polysaccharides make up starch, the storage form of glucose in plants
carbohydrates epimers glycogen starch proteoglycan cellulose mucoprotein lectins glycosyltransferases enantiomers
Match each term with its description. has the molecular formula of (CH2O)n _______ monosaccharides that differ at a single asymmetric carbon atom ________ the storage form of glucose in animals _______ the storage form of glucose in plants __________ glycoprotein containing glycosaminoglycans __________ the most abundant organic molecule in the biosphere _________ N-acetylgalactosamine is a key component of this glycoprotein ___________ carbohydrate-binding protein ______ enzymes that synthesize oligosaccharides ________ stereoisomers that are mirror images of each other ______
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Nucleotides are molecules that make up the structures of RNA and DNA. Modify the compounds to create adenosine monophosphate (AMP).
GTC
Place an asterisks (*) next to the 3' carbon atoms in the polynucleotide shown. What is the name of this polynucleotide? CGT TCG CTG GTC
The hemoglobin-oxygen binding curve is sigmoidal. Hemoglobin consists of four heme‑bound subunits.
Select all the statements that correctly describe the cooperative binding of oxygen to hemoglobin. The hemoglobin-oxygen binding curve is sigmoidal. Hemoglobin consists of four heme‑bound subunits. Salt bridges between hemoglogbin subunits allow oxygen to bind tightly. Hemoglobin acts as an oxygen store under anaerobic conditions. Hemoglobin always binds oxygen tightly.
D‑galactose, D‑glucose, and D‑fructose. D‑galactose and sucrose not a reducing
Raffinose is a trisaccharide and a minor constituent in sugar beets. Consider the structure of raffinose, and complete the passage about the composition and properties of raffinose. Raffinose is composed of ____________ Treatment of raffinose with β‑galactosidase produces ____________. Raffinose is _________ sugar.
HbF HbA, lowers HbA‑bound O2 will tend to move to HbF because HbF has a lower affinity for BPG, an allosteric inhibitor of O2 binding.
Studies of oxygen transport in pregnant mammals show that the O2‑saturation curves of fetal (HbF) and maternal (HbA) hemoglobin are markedly different when measured under the same conditions. Which hemoglobin has a higher affinity for O2 at the tissue pO2 of around 4 kPa? cannot be determined from this data HbF HbA The allosteric inhibitory effector BPG binds to both the A form and the F form. Both forms show a lower affinity for O2 in the presence of BPG. However, the inhibitory effect of BPG is larger for HbA. The different O2 affinity indicated by the curves has been attributed to different BPG affinity for the two Hb forms. Complete the statement. BPG binds tighter to ________, which _______ its affinity for O2 . Which statement that best explains the role of BPG in O2 transport from mother to fetus? HbA‑bound O2 will tend to move to HbF because HbF has a lower affinity for BPG, an allosteric inhibitor of O2 binding. HbA‑bound O2 will tend to move to HbF because HbA binds an additional BPG per tetramer when it enters placental circulation. HbF will extract O2 from oxygenated HbA because BPG enhances O2 binding to HbF. In the placental circulation, HbF will load up on O2 as BPG dissociates from HbA and binds to HbF.
No, because Zn2+ cannot partially transfer an electron to oxygen to form a complex.
Suppose myoglobin is prepared with Zn2+ in the place of Fe2+ in the porphyrin. Would you expect this modified myoglobin to bind oxygen? No, because the larger size of Zn2+ will prevent the ion from fitting within the porphyrin ring. No, because Zn2+ cannot partially transfer an electron to oxygen to form a complex. Yes, because oxygen binding to myoglobin depends solely on the presence of a distal histidine. Yes, because oxygen can form a bond with either Zn2+ or Fe2+.
tritiated thymidine
Suppose that you want to radioactively label DNA, but not RNA, in dividing and growing bacterial cells. Which radioactive molecule would you add to the culture medium? tritiated thymidine radioactive phosphorus tritiated uridine radioactive sulfur
Shifts Left: - the adult hemoglobin (HbA) is replaced by an infant's fetal hemoglobin (HbF) - Hemoglobin is isolated from red blood cells and stripped of 2,3-bisphosphoglyceration - tetrameric hemoglobin is dissociated into its subunits Shifts Right: - the blood pH drops for 7.4 to 7.2 - the CO2 concentration in the blood increases - the concentration of 2,3-bisphosphoglycerate increase during acclimation to high altitude
The graph represents the adult hemoglobin binding curve (in green) at pH 7.4 in the presence of 2,3‑bisphosphoglycerate. The hemoglobin binding curve has a sigmoidal shape, due to four interacting oxygen‑bound sites. For comparison, the myglobin binding curve has only one oxygen‑bound site and has a hyperbolic curve. For each of the six scenarios, determine whether the hemoglobin binding curve would shift left or shift right.
2 30 myoglobin
The graph shows the oxygen‑binding curves for myoglobin and hemoglobin. Label the graph and answer the questions. Use the curves to determine the partial pressure of oxygen at 50% saturation for hemoglobin and myoglobin. myoglobin 𝑃50= hemoglobin 𝑃50= Which protein has a higher affinity for oxygen?
D‑Mannose (an aldose) is a reducing sugar. The oxidation of a reducing sugar forms a carboxylic acid sugar. A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar.
Which statements about reducing sugars are true? A reducing sugar will not react with the Cu2+ in Benedict's reagent. D‑Mannose (an aldose) is a reducing sugar. Reducing sugars contain ketone groups instead of aldehyde groups. The oxidation of a reducing sugar forms a carboxylic acid sugar. A disaccharide with its anomeric carbons joined by the glycosidic linkage cannot be a reducing sugar.
leucine alanine leucine phenylalanine No aggregation - O2 decreases due to vigorous exercise or high altitude - R state Hb shifts to T state Hb - Val interacts with the pocket of B chain on another HbS - Additional T state HbS interact with the growing aggerate to form an insoluble fiber sickled red blood cell
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. See image of eight aggregated HbS molecules. 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. Which amino acids would be expected to produce a similar sickling effect if substituted for Val at position 6? leucine phenylalanine lysine alanine arginine 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. Choose two amino acids that would be reasonable candidates for the pocket-Val 6 interaction. leucine glutamate lysine phenylalanine asparagine 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.
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The process of DNA transcription uses one nucleic acid (DNA) as the template for creating another nucleic acid (RNA). Since DNA and RNA are both nucleic acids, each is made up of a combination of common and unique components. Match each term to the appropriate structures on the diagram of DNA and RNA. Some terms will be used more than once.
B -> 4, galactose B2 <-> al, fructose glucose O-glycosidic bond
The structures of lactose and sucrose are shown. Compare and contrast these carbohydrates. Lactose Sucrose Consider a Venn diagram for both sugars, where the left circle represents lactose and the right circle represents sucrose. Identify which terms describes only lactose, which terms describes only sucrose, and which terms describe both molecules. The linkage in lactose is __________ . The component in lactose is ________. The linkage in sucrose is ___________. The component in sucrose is ___________. The component in both lactose and sucrose is ____________ . The type of bond in both lactose and sucrose is ___________
3′-ACGTAGATC-5′
The three-dimensional structure of DNA is shown in the interactive. You are currently in a graphing module. Turn off browse mode or quick nav. Enter the complementary sequence to the DNA strand shown. 5′- TGCATCTAG -3′
Hydrophobic residues on the surface of hemoglobin subunits interact with similar regions on the other subunits through van der Waals interactions.
The α and β subunits of hemoglobin bear a remarkable structural similarity to myoglobin. However, certain residues that are hydrophilic in myoglobin are hydrophobic in the subunits of hemoglobin. Why might this be the case? Hemoglobin forms long, extended structures that feature repeated sequences, whereas myoglobin forms globular structures. Hydrophilic residues on the surface of myoglobin form ionic interactions with similar regions on other myoglobin molecules. Hydrophobic residues on the surface of hemoglobin subunits interact with similar regions on the other subunits through van der Waals interactions. Myoglobin is a water‑soluble protein, whereas hemoglobin is found in the hydrophobic environment of membranes.
DNA has a double‑stranded structure that ensures an accurate mechanism of duplication. DNA is more resistant against enzymes that degrade nucleic acids.
There are two types of nucleic acids, DNA and RNA. Nearly all organisms use DNA, not RNA, as the central repository for genetic information. Choose the statements that explain this phenomenon. DNA has a double‑stranded structure that ensures an accurate mechanism of duplication. DNA contains a hydroxyl group at the 2′ carbon. DNA is flexible and forms complex catalytic structures. DNA contains adenine as one of its nitrogenous bases. DNA is more resistant against enzymes that degrade nucleic acids.
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Threose is an aldose monosaccharide. The Fischer projection of D‑threose is shown. Draw D‑threose using wedge and dash bonds around the chiral carbon atom(s).
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Thymine is a pyrimidine base that is a component of DNA. Draw thymine.
Amino acid sequence: Met-Ala-Arg-Ser-Ala
Write the amino acid sequence that would result from the RNA sequence: (shown 5' to 3') AUG GCA CGA UCA GCU Refer to a codon table. Note: Enter the amino acids using their three‑letter designations. Put a hyphen between each amino acid. Example: Met‑Ala‑Leu‑Cys‑Ala
5′-GAGACT-3′
Write the complementary sequence for AGTCTC. Write the new sequence in the standard 5′‑to‑3′ notation.
5′−UAACGGUACGAU−3′ N‑terminus Leu-Pro-Ser-Asp-Trp-Met C‑terminus poly(Leu‑Leu‑Thr‑Tyr)
What is the sequence of the mRNA molecule synthesized from the DNA template strand 5′−ATCGTACCGTTA−3′? Enter the mRNA sequence using the one‑letter abbreviations for the nucleotides. What amino acid sequence is encoded by the mRNA molecule 5′−UUGCCUAGUGAUUGGAUG−3′? Assume that the reading frame starts at the 5′ end and that the peptide is synthesized from the amino teminus to the carboxyl terminus. Use a codon table to determine the identity of the translated peptide. Enter the amino acid sequence using three‑letter abbreviations for the amino acids separated by a dash. What is the sequence of the polypeptide formed if poly(UUAC) is added to a cell‑free, protein‑synthesizing system? poly(Leu) poly(Tyr) Asn‑Glu‑stop poly(His‑Ser‑Phe‑Ile) poly(Leu‑Leu‑Thr‑Tyr) Val‑Ser‑Lys‑stop
RNA, protein DNA, RNA DNA, DNA
What transformation occurs during each of these processes? Translation: _________ ⟶ _________ Transcription: ____________ ⟶ _______________ Replication: _________ ⟶ ____________
Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron atom. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron atom. Each hemoglobin molecule can bind four oxygen molecules; each myoglobin can bind only one oxygen molecule. Heme is composed of an organic protoporphyrin component and a metal atom. Each iron atom can form six coordination bonds. One of these bonds is formed between iron and oxygen.
Which five statements about hemoglobin and myoglobin structure are true? Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron atom. Molecular oxygen binds irreversibly to Fe2+ in heme. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron atom. Each hemoglobin molecule can bind four oxygen molecules; each myoglobin can bind only one oxygen molecule. Hemoglobin and myoglobin are heterotetramers. Heme is composed of an organic protoporphyrin component and a metal atom. Each iron atom can form six coordination bonds. One of these bonds is formed between iron and oxygen.
epimers anomers diastereomers
Which of the terms explain the relationship between the two compounds? enantiomers epimers anomers diastereomers
Transcription is the process of synthesizing RNA from DNA. Translation is the process of synthesizing an amino acid sequence from RNA.
Which statement describes the central dogma of biology? Transcription is the process of synthesizing DNA from RNA. Translation is the process of synthesizing an amino acid sequence from DNA. Transcription is the process of synthesizing amino acids from DNA. Translation is the process of synthesizing amino acids from RNA. Transcription is the process of synthesizing RNA from DNA. Translation is the process of synthesizing an amino acid sequence from RNA. Translation is the process of synthesizing RNA from DNA. Transcription is the process of synthesizing an amino acid sequence from RNA.
