BioChem Quiz 9,10... (FOR T3)
Terms in this set (40)
1) The binding of oxygen to myoglobin and hemoglobin has what effect on the heme iron?
a. It has no effect because the iron is restricted to being within the plane of the porphyrin ring and cannot move on binding oxygen.
b. It causes the iron to move into the plane of the porphyrin ring.
c. It causes the iron to move out of the plane of the porphyrin ring.
It causes the iron to move into the plane of the porphyrin ring.
(The binding of oxygen partially balances the "pull" by the histidine ligand causing the iron to move into the plane of the ring.)
2) Which of the following indicates that the binding of oxygen to hemoglobin is cooperative?
a. The fact that hemoglobin consists of β dimers, designated 1β1 and 2β2, which form the tetrameric hemoglobin.
b. Hemoglobin has four subunits, each of which can bind oxygen.
c. A binding plot of Y (fraction of sites occupied) against pO2 is sigmoidal rather than hyperbolic.
A binding plot of Y (fraction of sites occupied) against pO2 is sigmoidal rather than hyperbolic.
(Sigmoidal binding plots are diagnostic for cooperative binding.)
3) Oxygen and 2,3-bisphosphoglycerate (2,3-BPG) cannot bind to hemoglobin at the same time because:
a. 2,3-BPG is a competitive inhibitor of oxygen binding to the hemoglobin molecule.
b. the structure of hemoglobin is changed when oxygen binds such that 2,3-BPG can no longer bind.
c. the binding of 2,3-BPG blocks a channel needed by oxygen to reach the heme.
the structure of hemoglobin is changed when oxygen binds such that 2,3-BPG can no longer bind.
(The binding of oxygen makes the site where 2,3-BPG would normally bind too small to accommodate it.)
4) Rapidly metabolizing tissues generate large amounts of protons and carbon dioxide. The result is that:
a. the oxygen-binding curve changes from sigmoidal to hyperbolic.
b. the oxygen-binding curve of hemoglobin (Y vs. pO2) is shifted to lower pO2 levels.
c. the oxygen-binding curve of hemoglobin (Y vs. pO2) is shifted to higher pO2 levels.
the oxygen-binding curve of hemoglobin (Y vs. pO2) is shifted to higher pO2 levels.
(This change results in more oxygen being provided to rapidly metabolizing tissues.)
5) In addition to transporting oxygen from the lungs to the tissues, hemoglobin is also involved in transporting carbon dioxide from the tissues to the lungs. How is this accomplished?
a. Carbon dioxide reacts with terminal amino groups on hemoglobin in a reversible manner.
b. Carbon dioxide competes for the oxygen-binding site on the heme.
c. Carbon dioxide competes for the 2,3-BPG binding site.
Carbon dioxide reacts with terminal amino groups on hemoglobin in a reversible manner.
(Carbon dioxide reacts with the N-terminal amino groups to form a carbamate in a reversible reaction.)
6) Each chain of hemoglobin can be viewed as existing in one of two states—the R state and the T state. What is the relationship of these states to oxygen binding?
a. Oxygen binds to the T state, converting it into the R state.
b. The conversion between the R and T states is not directly related to oxygen binding.
c. Oxygen binds to the R state, converting it into the T state.
Oxygen binds to the T state, converting it into the R state.
(Deoxyhemoglobin is predominately in the T form but converts to the R form upon binding oxygen.)
7) The difference between the "concerted" and "sequential" models of oxygen binding to hemoglobin is:
a. whether the transition between T and R states is "all-or-nothing" or has intermediate states (mixtures of R and T states in the same molecule).
b. whether one is considering the or β subunits.
c. whether one is considering a situation in which regulatory molecules such as 2,3-BPG are present or absent.
whether the transition between T and R states is "all-or-nothing" or has intermediate states (mixtures of R and T states in the same molecule).
(A concerted model implies an "all-or-nothing" phenomenon, while a sequential model allows for intermediate forms.)
8) In fetal hemoglobin the two β subunits are replaced with two γ subunits, resulting in fetal hemoglobin having a higher affinity for oxygen than the mother's normal adult hemoglobin. This is due to:
a. a different mode of oxygen binding to the heme in the γ subunits.
b. a decreased amount of cooperativity between the and β subunits.
c. decreased binding of 2,3-BPG.
decreased binding of 2,3-BPG.
(The decreased affinity for 2,3-BPG makes the binding of oxygen more favorable.)
9) Hemoglobin S, the abnormal form of hemoglobin responsible for sickle cell anemia, is the result of a mutation in the gene for the β subunit. This mutation results in the change of:
a. a negatively charged amino acid R-group to a positively charged amino acid R-group.
b. a negatively charged amino acid R-group to a hydrophobic amino acid R-group.
c. a hydrophobic amino acid R-group to a positively charged amino acid R-group.
a negatively charged amino acid R-group to a hydrophobic amino acid R-group.
(The change is from a Glu to a Val.)
10) The molecular consequences of the hemoglobin S mutation are that:
a. the hemoglobin S forms aggregates and fibrous precipitates when oxygen is bound.
b. the hemoglobin S forms aggregates and fibrous precipitates when oxygen is released.
c. the hemoglobin S has a lower solubility and tends to precipitate in the lungs.
the hemoglobin S forms aggregates and fibrous precipitates when oxygen is released.
(When deoxygenated, hemoglobin S can form fibrous precipitates which result in clogged capillaries and impaired blood flow.)
1) D-glucose and L-glucose are:
(D- and L-glucose are mirror images of one another and thus enantiomers.)
2) Complete the statement: The furanose form of fructose is generated by formation of a hemiketal involving the attack of the hydroxyl group on carbon ____ with carbon ____.
a. 5, 2
b. 2, 6
c. 6, 1
(This generates the five-member furanose ring typical of fructose.)
3) Formation of pyranose and furanose forms of sugar result in the generation of a new asymmetric carbon giving rise to - and β -forms of the sugars. The carbon at which this newly created asymmetric center is generated is referred to as:
a. carbon number 1 in the numbering scheme for sugars.
b. the epimeric carbon.
c. the anomeric carbon.
the anomeric carbon.
(The anomeric position is defined as the newly generated asymmetric carbon.)
4) Which of the following is NOT true about cellulose?
a. Fibrils of cellulose have high tensile strength, due to hydrogen bonds between long straight chains.
b. Cellulose is a polymer of glucose joined via β -1,4 glycosidic linkages.
c. Branching in cellulose occurs about once in every 30 sugar residues, via -1,6-linkages.
c. Branching in cellulose occurs about once in every 30 sugar residues, via -1,6-linkages.
(This is NOT a true statement. Cellulose is a linear polymer (without branches). Branching occurs in starch, via -1,6-linkages.)
5) Once formed, the - and β -forms of D-glucose are:
a. not capable of interconversion.
b. interconvertible directly with no intermediate and are in equilibrium with one another.
c. interconvertible only through a linear, noncyclic intermediate with which they are both in equilibrium.
interconvertible only through a linear, noncyclic intermediate with which they are both in equilibrium.
(Although present in only a small amount, the linear form of glucose is in equilibrium with both anomers.)
6) Glycosidic bonds:
a. destroy the asymmetric configuration of the participating carbons.
b. only connect carbon-1 of one sugar to carbon-4 of another.
c. connect sugar molecules in both linear and branches of complex carbohydrates.
connect sugar molecules in both linear and branches of complex carbohydrates.
(Glycogen, for example, has both linear and branch glycosidic linkages.)
7) Human blood groups (ABO) are:
a. the result of the same oligosaccharides being attached to different proteins.
b. the result of differing glycotransferases.
c. the result of the presence or absence of certain sugars.
the result of differing glycotransferases.
(The A and B transferases differ in which monosaccharide is added to the O antigen.)
8) Glycoproteins are proteins to which carbohydrates have been covalently attached. The amino acid R-groups that serve as the sites for such attachment include:
a. hydroxyl-containing R-groups.
b. negatively charged amino acid R-groups.
(The O-linked glycoproteins involve ether linkages between sugars and hydroxyl groups of serine and threonine.)
9) Which two sugars are joined via an O-glycosidic bond to make the disaccharide known as lactose?
a. β -D-galactose and -D-glucose, joined via a 14 glycosidic bond
b. Two molecules of -D-glucose, joined via a 14 glycosidic bond
c. β -D-fructose and -D-glucose, joined via 12 glycosidic bond
β -D-galactose and -D-glucose, joined via a 14 glycosidic bond
(Lactose is composed of β -D-galactose and -D-glucose, joined via a 14 glycosidic bond.)
10) Glycoproteins are targeted to different parts of the cell for secretion. This sorting, which targets the glycoproteins to their correct destination, is done in what organelle?
a. the Golgi apparatus
b. the lysozome
c. the endoplasmic reticulum
the Golgi apparatus
(In addition to continuing the glycosylation process, the Golgi apparatus is the major site where glycoproteins are sorted and targeted to their final destination.)
1) Lipid molecules are said to be amphipathic, meaning that:
a. they are capable of moving rapidly from one side of a lipid bilayer to the other.
b. they have a dual nature with part of the molecule being hydrophobic and the other part hydrophilic.
c. they have asymmetric carbons and can exist in left- and right-handed forms.
they have a dual nature with part of the molecule being hydrophobic and the other part hydrophilic.
(This dual nature is the key feature in understanding the role of lipids in membranes.)
2) Unsaturated fatty acids have double bonds that are in the cis, rather than the trans, configuration. One of the consequences of this is:
a. a bend in the molecule.
b. an alteration in the charge of the molecule.
c. an alteration in the number of carbons in the molecule.
a bend in the molecule.
(A cis double bond produces a bend in the molecule while a trans double bond does not.)
3) In phosphoglycerides, fatty acids are esterified at:
a. glycerol carbons 1 and 3.
b. glycerol carbons 1 and 2.
c. any two of the three glycerol carbons.
glycerol carbons 1 and 2.
(The third carbon is the site at which phosphate is esterified.)
4) Which of the following is NOT a difference of archaeal membrane lipids relative to those of other organisms?
a. Fatty acid esters are replaced with long chain alcohol ether links to the glycerol.
b. A backbone other than glycerol is used.
c. The long hydrophobic tails are branched rather than being linear.
A backbone other than glycerol is used.
(This is NOT a difference. Glycerol is used as the backbone although it has an inverted stereochemistry relative to that used in other organisms.)
5) Of the three major types of membrane lipids, which is not found in prokaryotes?
(Cholesterol is not present in prokaryotic membrane lipids. Even within eukaryotes, it is mainly limited to mammalian systems.)
6) Which of the following would be expected to lower the Tm for a phospholipid bilayer?
a. Adding ions to bind to charged groups in the polar head groups
b. Replacing a lipid containing 18-C fatty acids with one containing 16-C fatty acids
c. Replacing a lipid containing unsaturated fatty acids with one containing saturated fatty acids
Replacing a lipid containing 18-C fatty acids with one containing 16-C fatty acids
(Shorter chain fatty acids have weaker intermolecular interactions than longer fatty acids. The weakened interaction should lower the Tm.)
7) How many carbons do naturally occurring fatty acids typically have?
a. An even number of carbons, ranging from 14-24
b. An odd number of carbons, ranging from 13-25
c. Exactly 6, 10, or 14 carbons
An even number of carbons, ranging from 14-24
(Fatty acids from biological systems have an even number of carbons, ranging from 14-24. The most common fatty acids have 16 or 18 carbons.)
8) Which of the following is NOT true about lipids?
a. Lipids can be covalently attached to carbohydrates.
b. Lipids are used by organisms to store energy.
c. Lipids polymerize readily to form membranes.
Lipids polymerize readily to form membranes.
(This is not a true statement. Lipids do not polymerize. Rather, they self-associate to make noncovalent assemblies.)
9) Niemann-Pick disease can result from:
a. lack of docosahexaenoate (DHA).
b. a defect in farnesylation of certain membrane proteins.
c. lack of sphingomyelinase.
lack of sphingomyelinase.
(Sphingomyelinase is an enzyme involved in degradation of sphingomyelin.)
10) Which of the following is NOT categorized as a phospholipid?
(Eicosapentaenoate is the name for the ionized form of eicosapentaenoic acid (EPA), which has no phosphate group. EPA is an example of a ω-3 fatty acid, found in fish.)
1) In considering the movement of small molecules across a lipid bilayer, the permeability coefficient can be correlated with:
a. the size of the molecule.
b. the solubility of the molecule in a nonpolar solvent.
c. the interaction of the molecule with the polar head groups.
the solubility of the molecule in a nonpolar solvent.
(The greater the solubility in a nonpolar medium, the more rapidly a molecule crosses a lipid bilayer.)
2) The most common way in which integral membrane proteins span the membrane is in:
a. single strands with no secondary structure.
b. β -sheet structures.
c. a-helical segments.
(Alpha helices offer an energetically favorable way for a protein to span the membrane in terms of having hydrophobic R-groups on the outside and the polar peptide bond atoms hydrogen-bonded together on the inside.)
3) Which membrane component exhibits the fastest movement within the bimolecular sheets?
a. Phospholipids undergoing lateral diffusion
b. Proteins flip-flopping
c. Phospholipids undergoing transverse diffusion
Phospholipids undergoing lateral diffusion
(Within membranes, phospholipids can travel along one side of the bimolecular sheet very rapidly, in a process called lateral diffusion.)
4) Which of the following factors makes the membrane environment more ordered and less fluid?
a. The presence of at least one cis double bond in the hydrocarbon chain of phospholipids
b. Elevated temperatures, close to or above Tm
c. The presence of cholesterol
The presence of cholesterol
(Cholesterol makes eukaryotic membranes less fluid. The bulky ring structure of cholesterol interferes with the typical motions of the fatty acid tails of phospholipids. In addition, cholesterol appears to have some specific interactions with certain components (such as sphingolipids), forming less mobile raft-like complexes.)
5) In passive transport (also known as facilitated diffusion), the energy driving the movement of materials across a membrane is provided by:
a. nothing because there is no energy requirement for passive transport.
c. the concentration gradient across the membrane.
the concentration gradient across the membrane.
(Materials move from regions of high concentration to low concentration with the driving force being the concentration gradient itself.)
6) The Na+-K+ pump (Na+-K+ ATPase) is an active transport system that:
a. pumps sodium ions out of the cell and potassium ions into the cell.
b. provides energy such that sodium and potassium ions can move freely in either direction.
c. pumps sodium ions into the cell and potassium ions out of the cell.
pumps sodium ions out of the cell and potassium ions into the cell.
(Because the movement of each ion is against its concentration gradient, energy is needed to move them in the direction indicated.)
7) Which of the following is NOT true regarding P-type ATPases?
a. All use a similar mechanism involving a phosphorylated intermediate in their mechanism of action.
b. The drug called digitalis improves contraction of cardiac muscle by inhibiting its P-type ATPase.
c. Because of their common mechanism, they lack specificity and a given P- type ATPase can often transport different ions or molecules.
Because of their common mechanism, they lack specificity and a given P- type ATPase can often transport different ions or molecules.
(This statement is not true. Although they share a common mechanism, each different P-type ATPase has a high specificity for the particular molecule transported.)
8) Secondary transporters (or cotransporters) can be divided into symporters and antiporters. What is the basis for this distinction?
a. Symporters allow the same molecule to be moved in both directions, while antiporters allow it to move in only one direction.
b. Symporters use the energy from one molecule moving down its concentration gradient to drive the movement of a second molecule in the opposite direction against its concentration gradient, while antiporters drive the movement in the same direction.
c. Symporters use the energy from one molecule moving down its concentration gradient to drive the movement of a second molecule in the same direction against its concentration gradient, while antiporters drive the movement in the opposite direction.
Symporters use the energy from one molecule moving down its concentration gradient to drive the movement of a second molecule in the same direction against its concentration gradient, while antiporters drive the movement in the opposite direction.
(This correctly defines the difference between symport and antiport systems.)
9) Which of the following is NOT true of ion channels?
a. Ion channels are gated, only being opened in response to specific signals.
b. Movement of ions through ion channels is more rapid than through pumps or secondary transporters.
c. Ion channels are continually open while pumps or secondary transporters depend on an energy source to open.
Ion channels are continually open while pumps or secondary transporters depend on an energy source to open.
(This statement is not true. Channels can exist in both open and closed states.)
10) The three-dimensional structure of the potassium channel provided a rationale for the selectivity of potassium ions and the rejection of sodium ions. Which of the following is the basis for the ability of the potassium channel to discriminate between these two ions?
a. The sodium ion is too small to pass through the channel.
b. The opening of the channel is structurally arranged such that potassium ions can bind but it is too small for sodium ions.
c. A restriction in the channel allows for potassium ions to be resolvated by protein ligands but is too large to effectively resolvate sodium ions.
A restriction in the channel allows for potassium ions to be resolvated by protein ligands but is too large to effectively resolvate sodium ions.
(The size of the restriction is such that potassium ions can be bound effectively but the smaller sodium ions cannot.)
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