TPR 5 Test Bio/Biochem
If a drug is applied to a rod cell which blocks sodium channels in the absence of light, this will result in: A. the hyperpolarization of the retinal rod plasma membrane. B. the repolarization of the retinal rod plasma membrane. C. an increase in the sodium concentration in the cell. D. the conversion of all trans-retinal to 11-cis-retinal.
A. the hyperpolarization of the retinal rod plasma membrane. Blocking the sodium channels in the absence of light would have the same effect on the cell as a burst of light. The membrane would hyperpolarize due to the fact that sodium can no longer enter the cell to depolarize it (choice A is correct; eliminate choice B). Sodium concentration would decrease, not increase, because the channel is blocked (eliminate choice C), and, finally, the conversion of all trans-retinal to 11-cis-retinal would occur only in the presence of light (eliminate choice D).
Which of the following conclusion(s) is/are supported by the results of the experiment shown in Figure 1? I. Energetic frustrations result in movement of the energy landscape to favor native interactions. II. The Go model in the absence of energetic frustrations has a higher Ea to move from an unfolded to a folded state for the H5A protein. III. Introduction of energetic frustrations results in an increase in the free energy at all stages of protein folding. A. I only. B. II only. C. I and III only. D. II and III only.
A. Item I is true: the results in Figure 1 show that the introduction of energetic frustrations results in a shift of the energy landscape to the right, toward increasing values of Q, which represents the fraction of native conformations present in the protein. Thus, a shift to the right represents an increase in native conformations (choices B and D are wrong). Item II is false: the Go model without frustrations for protein H5A shows an overall lower, not higher Ea, in moving from an unfolded to folded state (moving toward more native interactions). This is seen as a smaller "hump" at a Q value of 0.5. Item III is false: the passage states that the -ln(P(Q)) is proportional to free energy, and the Go model without frustrations is higher in free energy than the model with frustrations for certain values of Q and lower in free energy for other values (choice C is wrong and choice A is correct). Guessed, was lazy and did not read graph
The graph below shows the membrane potential of a rod-cell plasma membrane: Which one of the following is represented at time t = 1 sec? A. Hyperpolarization B. Depolarization C. Repolarization D. Nonpolarization
A. Makes sense if you look at graph ** Got this wrong because I was thinking of an action potential graph and not interpreting the graph that was given **USE INFORMATION GIVEN TO YOU
Acetylcholine binding to AChR directly leads to which of the following effects? A. Influx of Na+ B. Release of Ca2+ from the sarcoplasmic reticulum C. Influx of Cl- D. Efflux of K+
A. The nicotinic AChR in NMJ is a ligand-gated cation channel on the sarcolemma (choice B can be eliminated). Therefore, when acetylcholines binds to it, there is an influx of cations, especially Na+ (choice A is correct), which causes depolarization of the muscle fiber (choices C and D can be eliminated).
Which one of the following mutations would be most likely to convert a proto-oncogene into an oncogene? A. Silent mutation B. Missense mutation Correct Answer C. Nonsense mutation D. Deletion mutation
B. Missense mutation
Why is there a latent period in the production of antibodies during the primary response? A. The pathogen must first proliferate before the immune system can mount a response. B. Time is required for B lymphocytes to proliferate. C. Antibodies must be produced by T lymphocytes. D. The antigen must infect B lymphocytes prior to the production of antibodies.
B. Time is required for B lymphocytes to proliferate. POE & reasoning question choice A: wrong b/c pathogen needs to be present but not proliferate to cause disease choice C: wrong b/c AB's produced by B lymphocytes choice D. wrong b/c the antigen binds to B cells, doesn't infect them
The 11-cis-retinal chromophore associated with rhodopsin is derived from vitamin A. If a child is vitamin-A deficient, could this possibly lead to problems in night vision? A. Yes, because vitamins act as cofactors with proteins. B. Yes, because vitamin A is a precursor for 11-cis-retinal. C. No, because vitamins have no effect on visual excitation. D. No, because cones are responsible for night vision.
B. Yes, because vitamin A is a precursor for 11-cis-retinal. The question explains that 11-cis-retinal is derived from vitamin A, so vitamin A must be a precursor of 11-cis-retinal
An error is made during production of an RNA transcript which results in a change in a single amino acid in the final protein. This error is most likely a result of: A. incorrect translation of the mRNA into protein by the ribosomes. B. the poor editing ability of RNA polymerase. C. a frameshift mutation occurring during transcription. D. incorrect translation of the DNA into RNA by RNA polymerase.
B. the poor editing ability of RNA polymerase. B. RNA polymerases have no editing function since RNA is a transient molecule and its information is not passed on to offspring (choice B is correct). The question states that the error is made during production of the RNA, so the change in amino acid sequence reflects the new nucleotide sequence and is not due to incorrect translation of the mRNA (choice A is wrong). Frameshift mutations are very severe and would result in all of the amino acids from the error on being changed (choice C is wrong), and DNA is not translated by RNA polymerase, it is transcribed (choice D is wrong).
In a follow-up experiment, normal cells were put under four different growth conditions. Which of the following best matches an experiment to its hypothesized outcome? A. Cells grown in low oxygen conditions down-regulate the function of all IDH isoforms. B. Cells stimulated with high ADP levels in their growth media have high IDH3 activity. C. Cells stimulated with high levels of ATP in their growth media have high levels of IHD3. D. Cell stimulated with high levels of NADH in their growth media have low IDH1 activity.
B. Cells grown in low oxygen conditions will resort to fermentation to survive, meaning IDH3 function would likely be low. However, IDH1 and IDH2 are not involved in the Krebs cycle, so there is no reason for the cell to also down-regulate these enzymes (eliminate choice A) . ADP accumulation stimulates cellular respiration, including IDH3 activity (choice B is correct). High ATP levels inhibit cellular respiration, so IDH3 activity would be negatively regulated (eliminate choice C). High NADH negatively regulates IDH3, but IDH1 and IDH2 use NADPH as a cofactor. NADH would therefore have little effect on IDH1 or IDH2 activity (eliminate choice D).
Diabetes insipidus is a condition in which an individual fails to produce antidiuretic hormone. If a person suffering from diabetes insipidus were to consume large amounts of a sugar-containing beverage, which of the following would most likely be observed after one hour? A. Production of concentrated urine B. Production of dilute urine C. Elevated plasma glucose levels D. Elevated urine glucose levels
B. An individual who cannot make ADH will be unable to reabsorb water from the collecting duct of the nephron, resulting in urine that is very dilute (choice B is correct, and choice A is wrong). Since diabetes insipidus is strictly a disorder involving ADH production, no effect would be seen on plasma glucose levels (very low after one hour due to the effects of insulin) or urine glucose levels, which should always be zero in a normal individual. Do not confuse this disorder with diabetes mellitus, a disorder involving insulin in which blood and urine glucose levels are abnormally high (choices C and D are wrong).
A person who suffers partial destruction of the lateral lemniscus on the left side would be expected to suffer which of the following? A. Ipsilateral hearing loss B. Bilateral hearing C. Total sensorineural deafness D. Defects in auditory processing and association
B. Bilateral hearing The passage states that the first decussation of the auditory pathway occurs at the connection between the cochlear nucleus and the superior olivary nucleus. Thus, any lesion that occurs past that point will result in binaural, rather than monaural, hearing loss (choice A is incorrect). By the same token, total sensorineural deafness would require bilateral destruction of the hearing pathway (choice C is incorrect). Auditory processing and association occur in the higher brain, after processing by the primary auditory cortex (choice D is incorrect). Bilateral hearing loss (choice B) is correct; a lesion on one side of the brain will result in partial but not complete hearing loss, since auditory information ascends bilaterally.
In an experiment, various concentations of streptomycin were added to a culture of E. coli. At low concentrations of streptomycin, increased amount of misreading of the mRNA was observed. At high streptomycin concentrations, the initiation of protein translation was inhibited. Which of the following statements is consistent with these results? A. Streptomycin inhibits the large subunit (50S) of the ribosome. B. Streptomycin inhibits the small subunit (30S) of the ribosome. C. Streptomycin binds to RNA polymerase and blocks its function. D. Streptomycin binds to the mRNA and targets it for degradation.
B. Streptomycin inhibits the small subunit (30S) of the ribosome. Since the errors are observed in translation and not in transcription, it is likely the ribosome that is being affected and not RNA polymerase (choice C is wrong). The role of small subunit (30S) of ribosome is to initiate translation by recognizing the first AUG start site and recruiting the tRNAfmet and the large subunit. It is also important for proofreading and maintaining fidelity during the translation process. Therefore, inhibition of small subunit would result in decreased initiation of protein translation and mRNA misreading (choice A is wrong and choice B is correct). If streptomycin bound to mRNA and caused its degradation, we would likely see less protein being made at low concentration (due to less mRNA available), not misreading of the mRNA (choice D is wrong).
Using the data provided in the passage, what is the best explanation for the high viability rate of the DU145 prostate cancer cells when treated with Orlistat, compared to the other two prostate cancer cell lines? A. The DU145 cell line will not take up [1-C14] acetate. B. The DU145 cell line expresses less FAS than other cell lines. C. The DU145 cell line is more malignant than the other two prostate cancer cell lines. D. The LNCaP and PC3 cell lines contain a form of FAS more susceptible to Orlistat.
B. The DU145 cell line expresses less FAS than other cell lines. data interpretation & reasoning question use POE A. this is a false statement. Graph does show that DUI45 cell line did uptake 1-c14 acetate (even if less) B. This is true & is indicated by the graph C. False, the passage does not even mention this as a possibility D. This could be true, but the data does not support this statement so it's not the answer
Each of the following experiments preceded the work described in the passage EXCEPT: A. mRNA was harvested from bovine tissue and treated with reverse transcriptase. B. restriction fragment length polymorphism analysis was done, to determine which introns and exons are retained in each cDNA. C. affinity chromatography using an IDH3 antibody was used to isolate the two proteins that cooperate with IDH3. D. mass spectrometry was used to identify the proteins that function with IDH3 in a trimer.
B. restriction fragment length polymorphism analysis was done, to determine which introns and exons are retained in each cDNA. The passage says that bovine cDNA was studied to determine how IDH1 was spliced. cDNA (or complementary DNA) is made by reverse transcribing mRNA (choice A is true and eliminate choice A). RFLP uses restriction endonucleases to cut 10-100 base-pair stretches of polymorphic DNA, called minisatellites, into small fragments. It can be used for DNA fingerprinting, but would not be useful here (choice B is false and is the correct answer). Affinity chromatography can be used to specifically isolate a biomolecule and mass spectroscopy can be used to identify peptide chains (choices C and D are true and eliminate choices C and D).
If a primary immune response confers immunity, why do individuals get the common cold more than once? A. The common cold virus does not trigger the proliferation of T lymphocytes. B. The common cold virus does not have receptors. C. Individuals are not exposed to the same virus. D. There are too few common cold viruses that circulate in the blood.
C. Individuals are not exposed to the same virus. reasoning and POE question choice A is true. Common cold viruses DO trigger proliferation of T lymphocytes choice B is also true, but receptors are needed by B and T cells to recognize antigens (choice B is true but does not answer the question, so it is eliminated). choice D doesn't answer the question & is not even mentioned or inferred in the passage Choice C makes the most sense, because There are many different viruses that can cause the symptoms we associate with the common cold: runny nose, sore throat, etc. Exposure to one of these viruses will cause the disease and confer immunity to that particular virus, but not to the hundreds of other viruses capable of causing the common cold
Which of the following represents the major products of the saponification of linolein, a triacylglycerol in which glycerol is esterified with linoleic acid? A. 1 Glycerol and 1 linoleic acid B. 3 Glycerol and 1 potassium linoleate C. 1 Glycerol and 3 potassium linoleate D. 1 Glycerol and 3 linoleic acid
C. 1 Glycerol and 3 potassium linoleate A triacylglycerol is composed of a glycerol backbone and three fatty acids, in this case three linoleic acid molecules. Saponification of linolein yields one glycerol molecule (eliminate choice B) and three linoleic acid salts, so three potassium linoleate equivalents are expected (eliminate choice A). Linoleic acid is not seen, as fatty acids do not remain protonated under basic conditions (eliminate choice D).
Fatty acid synthase is most likely suppressed by all of the following hormones EXCEPT: A. glucagon. B. thyroid hormone. C. insulin. D. cortisol.
C. Fatty acid synthesis will only be active if the cells have enough energy and are ready to build storage molecules, especially long term storage molecules. Insulin is released in response to elevated blood glucose and the glucose can be used for cellular respiration and storage (choice C is correct).
A graduate student in a biochemistry lab studies D-amino acid oxidase, an FAD-containing peroxisomal enzyme. She commonly uses a series of D-amino acid oxidase inhibitors in her experiments, but a new volunteer in the lab mixed up the tubes of inhibitors and she is not clear which inhibitor is in which tube. She runs a series of experiments to sort out the inhibitors. The data look like this: SEE GRAPH IN QUESTION A. The solid line on the data graph is uninhibited D-amino acid oxidase, the dotted line is a competitive inhibitor, and the dashed line is a mixed inhibitor. B. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a noncompetitive inhibitor, and the dashed line is an uncompetitive inhibitor. C. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a competitive inhibitor, and the dashed line is an uncompetitive inhibitor. D. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a competitive inhibitor, and the dashed line is a noncompetitive inhibitor.
C. The dotted line on the data graph is uninhibited D-amino acid oxidase, the solid line is a competitive inhibitor, and the dashed line is an uncompetitive inhibitor. misread answer choices & picked the wrong one
Rad21 functions in double-strand break repair and is also a subunit of the chromatid cohesin complex, necessary to keep sister chromatids connected. Cells with non-functional Rad21 due to mutation will display all the following phenotypes EXCEPT: A. increased sensitivity to UV and reactive oxygen species. B. decreased ability to maintain chromosomes in the 2x state. C. faster than usual mitotic phase due to increased chromosome motility. D. increased probability of cell cycle arrest at the G2/M transition.
C. UV light and reactive oxygen species can both induce double-strand breaks in DNA. If cells have defective Rad21, they will have a diminished ability to repair these breaks (choice A is true and can be eliminated). Since Rad21 is important in chromatid cohesion, without functional Rad21, the sister chromatids may not be able to remain joined after DNA replication. This means cells would revert to a 4n (tetraploid) 1x (one chromatid per chromosome) state instead of the 2n 2x state that is normal in G2 and the first part of mitosis (choice B is true and can be eliminated). If double-strand break repair cannot occur to full capacity, the cell would likely arrest at the G2/M transition (choice D is true and is can be eliminated). However, there is no reason why decreased double-strand break repair or poor chromatid cohesion would cause a faster mitotic phase or increased chromosome motility. If anything, mitosis would be arrested as the cell tried to deal with disorganized chromosomes (choice C is a false statement and is therefore the correct answer).
All of the following observations would support the energy landscape theory of protein folding EXCEPT: A. Lys and Glu residues that interact in the protein's native structure are the first amino acids to interact during protein folding. B. Two non-native amino acids, Phe and His, temporarily interact near the end of protein folding. C. Val and Ser, two non-native residues, interact early in the folding process, and drive further folding of native and non-native residues D. The significant majority of Ile residues are found on the protein interior once protein folding is complete.
C. Val and Ser, two non-native residues, interact early in the folding process, and drive further folding of native and non-native residues energy landscape theory: native interactions in the beginning stages of folding lead to a further energetically favorable conformational change. Find the statement that DOES NOT support the energy landscape theory A. True: Mentions native interactions that occur during protein folding B. True: If native AAs interact at the beginning of protein folding then nonnative AAs interacting at the end of protein folding is true C. False: Non native residues DO NOT interact in early PF process (does not agree with tenants of theory) D. True: Ile is hydrophobic therefore it would be found in the interior after pf. This is energetically favorable & supports theory
A gene has 900 base pairs, and the protein translated from this gene has 210 amino acids. This is mostly likely due to the fact that: A. exons have been removed from the RNA transcript. B. the stop codon in mRNA is found at the 210th base pair. C. the DNA molecule was not synthesized with the appropriate enzymes. D. introns have been removed in the formation of the mRNA.
D. introns have been removed in the formation of the mRNA. Exons are not removed from the transcript; they are ligated together to form the mature mRNA (choice A is wrong). A stop codon at the 210th base pair would result in a protein only 70 amino acids long (3 nucleotides = 1 amino acid), so choice B is wrong. There is only one enzyme that can synthesize DNA (DNA polymerase) and in any case, if the DNA molecule was synthesized incorrectly, it would be highly unlikely that any protein could be made from that gene at all (choice C is wrong). Choice D is correct: In the formation of mature mRNA, introns are removed and exons are ligated together. To make a protein of 210 amino acids, exons amounting to 210 × 3 = 630 base pairs must have been ligated together, along with 3 base pairs for a stop codon (633bp total); introns amounting to 900 - 633 = 267 base pairs were spliced out.
Which of the following is/are the disadvantages of passive induction of EAMG? Requires regular infusion of antibody Breaks host tolerance for AChR by stimulating T cells Needs large doses of IgG A. I only B. II only C. I and II only D. I and III only
D. Item I is correct: The antibodies have a certain half-life, so in order to maintain EAMG in the rats through the passive method, they need to be injected with the antibodies on a regular basis (choice B can be eliminated). Item II is incorrect: Administration of AChR would break the tolerance but anti-AChR would not produce reactive T cells against AChR (choice C can be eliminated). Item III is correct: Higher dosage of the antibodies is required to have a significant effect, as inhibition of few AChR would not produce pathology (choice A can be eliminated and choice D is correct).
PCl5(g) PCl3(g) + Cl2(g) ΔHo = 92.5 kJ/mol Two identical reaction vessels, X and Y, are charged with equal amounts of gaseous PCl5, and both trials are allowed to reach equilibrium. If flask X is held at 300 K while flask Y is maintained at 500 K, which of the following will be true about the reactions? A. Equilibrium is reached faster in flask Y because higher temperatures shift the reaction toward products. B. Equilibrium is reached faster in flask Y because ΔGY is more negative than ΔGX. C. Equilibrium is established at the same rate in both flasks because the K value of a reaction is a constant. D. Equilibrium is reached faster in flask Y due to the increased collision frequency of reactant molecules.
D. Equilibrium is reached faster in flask Y due to the increased collision frequency of reactant molecules. TRICKY QUESTION The wording in this question is extremely tricky & this is a terrible test question IMO We are asked to find out which of the following is true so the correct answer is dependent on understanding the answer choices given Right away we can eliminate C. K stands for Kalvin & is not a constant Now all the other answer choices talk about equilibrium is reached FASTER..... speed/ rate of a reaction= kinetic stability of reaction=thermodynamics since we now know that all the remaining choices describe a kinetic reaction, we need to look at the rest of each answer choice to select the TRUE statement shifting of a reaction (A) and delta G (B) are thermodynamic concepts, these statements are false so eliminate Choice D mentions increased collision frequency, this is a kinetic factor & is true of a kinetic reaction. D is correct.
If the Control 4 western blot tested for α-tubulin expression, which of the following is/are true? I. COX4, or cytochrome c oxidase subunit IV would be a good protein for Control 3. II. IDH1 is expressed in the cytosol and peroxisomes. III. Control 1 could probe for histone H3, and control 2 could probe for catalase expression. A. I and II B. I and III C. II and III D. I, II and III
D. Item I is true: The passage says that IDH3 catalyzes the fourth step of the tricarboxylic acid cycle, which occurs in the mitochondria, so fraction/control 3 must be the mitochondrial compartment. Cytochrome c oxidase subunit IV isn't a protein you've necessarily heard of before, but cytochrome c is part of the electron transport chain and is thus localized to the mitochondria. It makes sense this enzyme is localized there too (eliminate choice C). Item II is true: The passage says that fraction/control 2 are peroxisomal and α-tubulin is a component of microtubules in the cytoskeleton. IDH1 is expressed in these two compartments (eliminate choice B). Item III is true: The passage says that IDH is not expressed in the nucleus and none of the isoforms have expression in fraction 1. This fraction is thus nuclear, and histone H3 would be a good marker. Histone proteins function in eukaryotic genome packaging. Catalase is a peroxisomal protein (eliminate choice A and choice D is correct).
Which of the following would most likely become a major concern of mortality in MG patients? A. Cardiac failure B. Muscle weakness in the upper limbs C. Paralysis of the facial muscles D. Respiratory distress
D. Respiratory distress MG is a disease affecting skeletal muscle; therefore, all skeletal muscles in the body are susceptible (eliminate choice A). A very important skeletal muscle in humans is the diaphragm; in severe MG patients, death can occur due to respiratory distress or arrest. Paralysis of facial muscles and weakness of upper limbs can occur; however, they are generally not the primary cause of mortality (eliminate choices B and C; choice D is correct).
What type of cell is the origin of an acoustic neuroma? A. Microglia B. Astrocyte C. Oligodendrocyte D. Schwann cell
D. The passage states that an acoustic neuroma is a cancer of the myelin-producing cells of the vestibulocochlear nerve, a cranial nerve. Cranial nerves are part of the peripheral nervous system, so they are myelinated by Schwann cells (choice D is correct) rather than oligodendrocytes (eliminate choice C). Microglia have an immune role in the CNS (eliminate choice A), while astrocytes carry out various maintenance and regulatory activities in the CNS (eliminate choice B).
Solutions of D-glucose consist of mixtures of α- and β-anomers. The α- and β-anomers individually display +112° and +18.7° optical rotation values, respectively. If the optical rotation of a solution of D-glucose at 40°C is +52.3° in water and +61.6° in DMSO, we can conclude that: A. the optical rotation of DMSO is +9.3°. B. the ratio of α- and β-anomers is greater in water than in DMSO. C. the ratio of α- and β-anomers is the same in water as in DMSO. D. the ratio of α- and β-anomers is lower in water than in DMSO.
D. the ratio of α- and β-anomers is lower in water than in DMSO. DMSO is an achiral molecule and therefore has no optical rotation (eliminate choice A). The optical rotation of mixtures is given by the concentration-weighted average of the optical rotations of individual components. (This is why racemic mixtures of enantiomers have no net optical rotation). Since the optical rotations of the two solutions are different, the ratios of α- and β-anomers must also be different (eliminate choice C). The observed (concentration-weighted average) optical rotation is lower in water than in DMSO indicating that the component with the lower optical rotation has greater concentration in water than in DMSO. The β-anomer has the lower optical rotation and therefore is present at higher concentration in water making choice D correct. **guessed correctly, did not know topic at all, review
Which of the following is true based on the chart below? PHOSPHOLIPIDHEAD GROUPPhosphatidic acidPhosphatePhosphatidylethanolaminePhosphate and ethanolaminePhosphatidylcholinePhosphate and cholinePhosphatidylserinePhosphate and serine A. Lipid and protein macromolecules can both contain serine, while lipids and nucleic acids have no biological building blocks in common. B. Components of the plasma membrane can contain nitrogen atoms and amino acids. Correct Answer C. Phospholipids are amphoteric, in that they have hydrophilic and hydrophobic properties or regions. Your Answer D. At least some phospholipids contain three fatty acids connected to glyercol via ester linkages.
Proteins and phosphatidylserine both contain serine, and phospholipids and nucleic acids both contain phosphate groups (eliminate choice A). Based on the chart in the question stem, choice B is correct. Phospholipids can contain nitrogen atoms (as in ethanolamine and serine) and also amino acids (as in serine). Lipids are amphipathic because they contain both hydrophilic and hydrophobic regions. Amphoteric means that a molecule can act as either an acid or a base (eliminate choice C). Phospholipids contain a backbone (usually glycerol) with two fatty acids and a phosphate group. The phosphate group can be linked to other polar groups. Lipids that contain a glycerol backbone and three fatty acids are called triacylglycerides (eliminate choice D).
Individuals with immune deficiencies are found to be more susceptible to cancer than are healthy individuals. Which of the following reasons is the most likely explanation? A. The lymph vessels contain tumor cells. B. These individuals produce an abundant amount of lymphocytes in their bone marrow. C. The immune system is compromised and cannot eliminate cancerous cells. D. The lymphocytes are no longer able to make antigens.
The job of the immune system is to remove any foreign substances (pathogens, toxins, etc.) from the body to prevent possible harm to the body. It also patrols the body for normal cells displaying abnormal functions (i.e., cancer cells) and eliminates them. If the immune system is deficient, it cannot do this very effectively (choice C is correct). Immune-deficient individuals would produce fewer lymphocytes than normal individuals, not more (choice B is wrong), and lymphocytes do not make antigens, they fight antigens (choice D is wrong). While choice A may be true, it is not the reason these individuals are more susceptible to cancer, but rather is the outcome of being more susceptible to cancer (choice A is wrong).
All of the following are true concerning the side chain of cysteine EXCEPT: A. the sulfur is sp3 hybridized. B. it can help stabilize tertiary protein structure. C. the sulfur has two lone pairs of electrons. D. it undergoes hydrogen bonding.
forgot structure of this AA REVIEW & MEMORIZE Because the S has two lone electron pairs, it is sp3 hybridized (eliminate choices A and C). When two cysteine residues are close in space in a protein, the formation of a disulfide bond between them helps to stabilize the globular structure of the protein (eliminate choice B). The thiol group of cysteine does not undergo hydrogen bonding because it contains no O—H, N—H, or F—H bonds. **This was an easy question if I paid attention. Obviously Cysteine can't undergo H bonding b/c it does not contain an O-H, F-H, or N-H
