MCAT NS B/B
Which of the following do NOT have proteins with a nuclear localization signal? I. E. coli II. Homo sapiens III. Fungi IV. Archaea A. I only B. III only C. I and IV only D. I, III, and IV only
. E. coli (a species of bacteria) and archaea do not have nuclei and thus do not have a need for nuclear localization signal on their proteins. Homo sapiens and fungi are eukaryotes with nuclei. correct answer is C
how much does 1 amino acid weigh in Da
1 amino acid = 110 Da
What aspects separate single-crossover events from double-crossover events? A. Single-crossover events result in one-way displacement of chromosomal content from one chromosome to another, while double-crossover events always reverse this one-way displacement, resulting in chromosomes identical to the pre-crossover chromosomes. B. Single-crossover events occur during mitosis when a cell splits into two cells, while double-crossover events can only occur during meiosis when a cell splits into four cells. C. Single-crossover events affect only the ends of chromosome arms, while double-crossover events can affect segments in the middle of chromosome arms. D. Single-crossover events only affect one arm of each chromosome, while double-crossover events affect two arms of each chromosome.
A double-crossover event is one in which chromosomal arms of homologous chromosomes cross over in two different places along the arm. This results in a section in the middle of each chromosome being exchanged. choice A is wrong -> double cross over occur at 2 diff places and result in one big segment exchange, so no reversed choice B is wrong -> no cross over event in mitosis choice D is wrong -> single and double cross over will affect one arm of chromosome correct answer is C
granulocytes
A group of leukocytes containing granules in their cytoplasm; neutrophils, eosinophils, basophils. - involve in innate immune system
Integral membrane proteins (cell membrane)
A protein embedded in the lipid bilayer of a cell. These are typicallly cell surface receptors, channels, or pumps.
glucagon
A protein hormone secreted by pancreatic endocrine cells that raises blood glucose levels; an antagonistic hormone to insulin. - increase glycogenolysis - increase gluconeogenesis
Southern blotting
A technique that enables specific nucleotide sequences to be detected in a sample of DNA. It involves gel electrophoresis of DNA molecules and their transfer to a membrane (blotting), followed by nucleic acid hybridization with a labeled probe. - identify specific DNA sequences.
According to the data in Figure 1, what is the probability that a male Robertsonian translocation carrier who mates with a normal female will produce a viable offspring? ** #8 on word doc A. 1/6 B. 1/3 C. 1/2 D. 5/6
According to Figure 1, there are 6 possible outcomes for the gamete production in a 14/21 translocation carrier. Going from left to right, we have a normal gamete, a balanced translocation (all genetic info is present; this gamete would give rise to another ROB carrier), trisomy 21, monosomy 14 (because the gamete would only receive a copy of chromosome 21, not chromosome 14, from this parent), monosomy 21 (because the gamete would only receive a copy of chromosome 14, not chromosome 21, from this parent), and trisomy 14. There are no viable autosomal monosomies, and the only viable autosomal trisomies you should know are trisomy 21, 18, and 13 (though trisomy 8, 9, and 22 can also survive to term). Therefore, of the 6 gametes produced by the man, 3 of them contain genetic information that would produce a viable offspring (two normal, one with Down syndrome). 3/6 = 1/2. correct answer is C
Would the leg muscles of a patient paralyzed due to spinal cord transection be expected to exhibit MEPs (motor evoked response)? A. No, because the motor cortex is damaged B. No, because a muscle response requires an intact pathway C. Yes, if the stimulation intensity can be made sufficiently high D. Yes, because the motor cortex is intact
According to the passage, MEP is recording of the electrical muscular response elicited by artificially stimulating the motor cortex, which can cause muscle movement as well Spinal cord transection is a condition in which the nerves that run inside the vertebral column are damaged. This interrupts the conduction of efferent signals that cause the MEP. -> the path is destroyed so no motor response choice D is wrong -> motor cortex is not damaged, but if the path is destroyed, MEP cannot be generated Correct answer is B
Desmosomes
Anchoring junctions that prevents cells subjected to mechanical stress from being pulled apart; button like thickenings of adjacent plasma membranes connected by fine protein filaments - form internal tension reducing network of fibers
Which of the following would be expected as outcomes of ALDH knockdown gene therapy? I. Slower proliferation II. Reduced invasiveness III. Increased platinum sensitivity ** figure are #3 on word doc A. II only B. I and III only C. II and III only D. I, II, and III
II is wrong -> figure 2 shows ALDH knockdown gene does not reduce invasiveness lll is correct -> ALDH knockdown shows lower survival rate which means it's more sensitive to platinum correct answer is B
Which phase of mitosis is likely to be first interrupted if a cell has no microtubules? A. Prophase II B. Metaphase C. Telophase I D. Cytokinesis
In mitosis, the spindle fibers that move chromosomes are made up of microtubules. They are attached to chromosomes in metaphase and pull them apart in anaphase. choice A and C are wrong -> question ask for mitosis, and these are in meiosis correct answer is B
gene repression
Inhibition of a gene's expression caused by the change in activity of a regulatory protein - repressor bind to promotor to stop transcription inhibiton
If necessary to design a new experiment, which of the following best explains why researchers could use measurements of intracellular lactate levels (ILL) in cancer cells to assess efficacy of cancer drugs? passage info: Cancerous cells have much higher glycolytic flux (rate of glycolysis) than normal cells due in part to their rapid growth and cell division. Malignant cells rely more on fermentation than normal cells. A. High ILL would indicate that glycolysis is significantly inhibited. B. Low ILL would indicate that glycolysis is significantly inhibited. C. High ILL would indicate that the pentose phosphate pathway is significantly inhibited. D. Low ILL would indicate that the pentose phosphate pathway is significantly inhibited.
Lactate is a product of fermentation, and the first paragraph states that cancerous cells rely heavily on fermentation for energy production. Inhibiting glycolysis or fermentation would reduce ILL and indicate that the metabolism of the cancer cell is being effectively inhibited by the drug. correct answer is B
The disruption of which membrane component is most likely to result in cellular traffic complications similar to those seen in gap junction disorders? A. Cholesterol B. Glycoproteins C. Glycolipids D. Phospholipids
Membrane transport is most likely to be affected if the disruption occurs in components that span the entire membrane. Transmembrane proteins (many of which are glycoproteins) are the only component listed that pass all the way through the cell membrane and facilitate membrane transport. choice C and D are wrong - Phospholipids and glycolipids are located on the surface of cell membranes and typically do not extend though the entire bilayer. Glycolipids act to provide energy and also serve as markers for cellular recognition. Phospholipids are a structural component of the membrane and are not involved in traffic/transport. correct answer is B
does Cl- ions involve in action potential
No
Do bacteria contain introns?
No, but eukaryotes have intron
Which nucleophile was likely used to prepare the oligonucleotides used in the purification procedure? passage info: First, the conjugation is performed in solution by mixing antibodies with an excess of azide-modified oligonucleotides. A. NaOH B. (CH3)3CCl C. HCN D. NaN3
Paragraph 2 states that the oligonucleotides are azide-modified. Azide (N3-) is a powerful nucleophile which can readily add to available electrophiles, such as nucleotide molecules. choice C is wrong -> HCN is hydrogen cyanide,, which is not mentioned in the passage correct answer is D ** know molecule structure
The concentration of intracellular signaling molecules fluctuates rapidly in dividing cells during the cell cycle. Which of the following experimental techniques would be best to elucidate the mechanism of regulation for these proteins? A. RT-PCR and Southern blot B. Southern blot and northern blot C. Western blot and RT-PCR D. Western blot and Southern blot
Rapidly dividing cells undergo mitosis under the influence of specific signaling molecules. These molecules are expressed when their genes are transcribed, then are translated into proteins. In order to gain the best understanding of how a signaling protein's levels are regulated, both the protein and mRNA levels would need to be studied. Western blotting gives us information about the amount of protein expressed in a cell, while RT-PCR gives us information about the amount of RNA expressed. choice A and D are wrong -> southern blot is measure DNA correct answer is C
single crossover
Sharing of DNA between homologs at one location
siRNAs (small interfering RNAs)
Short, double-stranded RNA molecules used in RNA interference. - silence genes by interrupting expression between transcription and translation.
Northern Blot
Similar technique [to Southern], except that Northern blotting involves radioactive DNA probe binding to sample RNA . - identify RNA sequence
MicroRNA (miRNA)
a small, single-stranded RNA molecule, generated from a double-stranded RNA precursor; the miRNA associates with one or more proteins in a complex that can degrade or prevent translation of an mRNA with a complementary sequence - silence genes by interrupting expression between transcription and translation.
Dalton (Da)
a unit of mass equal to 1 atomic mass unit - ex: carbon = 12 Da, oxygen = 16 Da
Which of the following is most likely to be a trait of cancer stem cells? paragraph 1: The cancer stem cell (CSC) theory proposes that resistant cells encompass only a minority of cells within a cancer, but are solely responsible for long-term recurrence. Therefore, chemotherapy must target these CSCs in order to show long-term effectiveness. A. Ability to differentiate into any cell type in the body B. Self-renewal C. Greater reliance on aerobic glucose metabolism D. Elevated levels of BAX expression
according to the info in paragraph 1, cancer stem cell theory explain that resistant stem cells can caused recurrence -> passage never discuss ability to differentiate into diff cell types choice A is wrong -> totipotent can differentiate to any cell type in the body, but passage never state this ** passage info correct answer is B
innate immune system
provide general protection, rather than protection against specific pathogens that have been previously encountered and "remembered." - non specific immune response - ex: anatomical barrier, granulocytes (neutrophils, eosinophils, basophils), white blood cells, lymphocytes, mast cells - ex: inflammation: general response to infection or injury
A researcher analyzes a nucleic acid sample. After looking only at its nucleotide composition, he decisively concludes that this sample represents single-stranded DNA and not another form of nucleic acid. Which of the following compositions most likely represents this sample? A. 17% A, 17% T, 33% G, 33% C B. 29% A, 14% U, 11% G, 46% C C. 4% A, 4% U, 46% G, 46% C D. 12% A, 12% T, 30% G, 46% C
question states that nucleic acid is single-stranded and is unlikely to be double stranded choice A is wrong - the composition could be double stranded DNA, so its hard to conclude if it's single or double stranded correct answer is D
Denaturation
separating DNA strand by heating or chemical agents - opposite of hybridization
smooth muscle contraction
- does not have troponin, so use calmodulin which initiate contraction by activating myosin cross-bridge - tropomyosin is present but have different roles - contraction is initiated by Ca regulatory phosphorylation myosin, instead of Ca activated troponin system
what hormone increase glucose level
- glucagon - cortisol from adrenal cortex for long term stress response - Epinephrine, from adrenal medulla for immediate stress response - growth hormone: antagonist response to insulin
Type III restriction enzymes
- cleave at sites a short distance from their recognition sites - require ATP (but do not hydrolyze it) - S-adenosyl-L-methionine stimulates this reaction, but is not required.
type l restriction enzyme
- cleave at sites remote from the recognition site - they require both ATP and S-adenosyl-L-methionine to function.
Type II restriction enzymes
- cleave within or at short specific distances from their recognition sites - often require magnesium
How many chromosomes are found in a liver cell of a Robertsonian translocation carrier? A. 23 B. 45 C. 46 D. 47
As stated by the passage, in ROB, two acrocentric chromosomes have broken at the beginning of the short arm near the point where it meets the long arm. The long arms then fuse together. The resulting chromosome consists of two long arms but no short arms. The short arms are lost, but as all the genes on a short arm are available on the short arms of other acrocentric chromosomes, a Robertsonian translocation carrier will have no health problems due to his or her chromosome rearrangement. However, while other people usually have 46 chromosomes, Robertsonian translocation carriers have 45. We can also see that in Figure 1, in order to create ROB two chromosomes are combined and create two more, one of which is lost. Thus we have a net loss of 1 chromosome. The normal human cell has 46 chromosomes, so 46 - 1 = 45. correct answer is B
Two known competitive inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then, lysozyme was also added into the reaction mixture. Based on Figure 1, where would the resulting enzyme kinetics curve for this experiment fall? ** #9 on word doc A. Above line 1 B. Below line 1 but above line 3 C. Below line 3 D. The same as line 1
Be sure to read the question stem carefully! Inhibitor A is a strong acid (H3O+) and inhibitor B is a strong base (OH-). Since these inhibitors are mixed together before being added to the reaction mixture, they would simply neutralize each other and become water before having any effect at all. Thus, the enzyme kinetics would not be affected, and the curve would fall along line 1. correct answer is D
Methanol poisoning occurs when the body converts a large amount of methanol to harmful chemicals that attack the optic nerves, leading to acute blindness. Ethanol is the antidote for methanol poisoning because ethanol is preferentially processed by the body. Ethanol is most likely to be a(n): passage info: Competitive inhibition occurs when the inhibitor binds on the enzyme active site, preventing substrates from binding. Noncompetitive inhibition is observed when the inhibitor binds equally well to the enzyme whether or not a substrate is already bound. Allosteric inhibition refers to cases in which the inhibitor binds to a site on the enzyme that is not the active site, and irreversible inhibition occurs when the inhibitor reacts with or interferes in some way with the enzyme to change it chemically. Irreversible inhibition often occurs via a covalent bond. A. allosteric inhibitor. B. competitive inhibitor .C. noncompetitive inhibitor. D. irreversible inhibitor.
Competitive inhibitors bind to the active site of the enzyme in question, blocking the substrate from attaching. The question stem states that ethanol is "preferentially processed," or processed by one or more enzymes instead of methanol. Thus, it is highly likely that ethanol binds to the active site where methanol otherwise would, making it a competitive inhibitor. choice D is wrong - Irreversible inhibition occurs when the inhibitor interacts with the enzyme in a way that chemically alters it. correct answer is B
What type of control does siRNA exert on G6PD expression? passage info: the introduction of small interfering RNA (siRNA) that suppresses the expression of the G6PD into adipocytes reduces insulin resistance and adipogenesis. Though the biochemical mechanisms of obesity are not fully understood, continued research in this area may help ease the burden of obesity-associated diseases such as cardiovascular disease and diabetes. A. Transcriptional control B. Promotion C. Repression D. Post-transcriptional control
Due to its structure, siRNA is only able to bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred, but before translation. Specifically, it prevents the translation of mRNA corresponding to the target protein. correct answer is D choice B and C are wrong - both happens before transcription
Which of the following is an element of humoral immunity? A. Phagocytes B. Immunoglobulins C. T cells D. MHC I
Humoral immunity is part of the body's adaptive immune response. It is provided by B cell activity, which promotes an antibody, or immunoglobulin, response. Antibodies can recognize polysaccharide, phospholipid, and nucleic acid antigens to help the body fight extracellular bacteria, viruses, and toxins. choice A, C, and D is wrong -> they are cell mediated immunity correct answer is B
According to the passage, which feature of CTZ presents the most significant obstacle to its use as a cancer drug? ** #7 in word doc A. Low solubility in hydrophilic media B. Low solubility in hydrophobic media C. Aromatic structure D. Electron delocalization
Figure 1 shows that CTZ is a nonpolar compound with multiple hydrocarbon groups (shown below), lots of electron delocalization, and no charge, meaning that it is poorly soluble in polar (hydrophilic) solvents like water. Because blood and the cytosol are largely made of water, the drug likely has trouble traveling in the bloodstream. correct answer is A
High levels of LDL and OX-LDL increase the proportion of cholesterol in cell membranes. If the trisomy 21 data in Figure 2 can be attributed to the effect of cholesterol on these membranes, which of the following statements is most likely true? ** #8 on word doc A. hTERT cells treated with OX-LDL display more rigid membranes than hTERT cells treated with LDL B. Untreated hTERT cells display more rigid membranes than hTERT cells treated with LDL. C. At moderate to high temperatures, ethanol increases membrane fluidity. D. hTERT cells simultaneously exposed to HDL and LDL display increased membrane fluidity relative to untreated cells.
Figure 2 shows that LDL- and OX-LDL-treated cells have a markedly higher incidence of trisomy 21 than untreated hTERT cells. The question stem also indicates that cells treated with these lipoproteins have more cholesterol in their membranes. At moderate to high temperatures (including normal physiological temperature), cholesterol increases the rigidity of cell membranes by attracting adjacent phospholipid tails. Thus, a more rigid membrane appears to correlate with a higher incidence of trisomy 21. From Figure 3, however, we see that ethanol appears to counteract the trisomy-inducing effects of LDL and OX-LDL. It is therefore reasonable to conclude that ethanol decreases the rigidity (or increases the fluidity) of cell membranes under the conditions in this study. choice A is wrong -> based on figure 2, no significance is found between the two choice B is wrong -> opposite is true choice D is wrong -> figure did not show expose to HDL and LDL at the same time correct answer is C
Which of the following changes occur immediately following the consumption of a carbohydrate-rich meal? passage info: Glucose 6-phosphate dehydrogenase (G6PD) is the rate-limiting enzyme in the pentose phosphate pathway (PPP), an alternative path to glycolysis for cellular glucose A. Insulin secretion decreases; G6PD activity increases .B. Insulin secretion decreases; G6PD activity decreases. C. Insulin secretion increases; G6PD activity increases. D. Insulin secretion increases; G6PD activity decreases.
From outside knowledge, we know that insulin secretion increases after a meal to help the cells take up glucose from the bloodstream. The passage states that the PPP is a parallel (alternative) path to glycolysis, so it makes sense that it would also be activated when blood sugar levels rise after a meal. or think about LeChateliers principle, after eating high carb meal, there are a lot of glucose which converted to glucose 6-P thru glycolysis, so there are too much glucose 6-P, and we need G6PD to convert them choice D is wrong -> though last paragraph states G6PD expression is inhibited causing reduce insulin resistance of adipocyte cells. insulin resistance is more related to receptors, and it has nothing to do with production from pancreas correct answer is C
AEA is most accurately categorized as a: passage info: fatty acid binding proteins (FABPs) are intracellular lipid carriers that, in addition to binding free fatty acids, interact with other endogenous lipids including the endocannabinoid N-arachidonoylethanolamine (AEA). AEA is synthesized in response to neuronal depolarization from a phospholipid precursor situated within the lipid bilayer of the neuronal membrane. Following reuptake into the postsynaptic neuron, it is acted upon by the catabolic enzyme fatty acid hydrolase (FAAH), yielding ethanolamine and a polyunsaturated fat, arachidonic acid (AA). synaptic cannabinoid receptor 1 (CB1), producing pain-diminishing analgesic and anti-inflammatory effects. FABP5 and FABP7 bind to endocannabinoids with high affinity, mediating the intracellular trafficking of endocannabinoids to FAAH. A. steroid hormone. B. neurotransmitter. C. fatty acid. D. amine hormone.
From the first paragraph of the passage: "Following depolarization, AEA is synthesized from a phospholipid precursor situated within the lipid bilayer of the neuronal membrane ... Following reuptake into the postsynaptic neuron, it is [degraded] by ... FAAH." This, combined with information in the second paragraph that endocannabinoids, such as anandamide, bind CB1 receptors located on neuronal membranes, makes clear that, despite its somewhat unique structure, anandamide is a neurotransmitter. choice A and D are wrong -> hormones are secreted from gland, but AEA is secreted from neuron choice C is wrong - by looking at the structure it's not fatty acid b/c the amide present in the structure correct answer is B
In a population of Amish people, the frequency of the recessive autosomal allele for polydactyly is 1.2%. What percent of the population are heterozygotes for the polydactyly allele? A. 0.0144% B. 1.19% C. 2.37% D. 97.6%
Hardy-Weinberg equation: - total number of allele: A+a = 1 - total number of genotypes: AA + 2Aa + aa = 1 if a=0.012 and A = 0.988 2Aa = 2*0.012*0.988 = 0.024*0.988 hard calculation, so use approximation since 0.988 is very close to 1 correct answer is C
vitamin E
Helps protect the skin from the harmful effects of the sun's UV light. - antioxidant
The easiest method to separate the two subunits of the botulinum protein for subsequent analytical purposes would be: A. gas chromatography. B. mass spectrometry. C. thin-layer chromatography .D. size-exclusion chromatography.
Size-exclusion chromatography could therefore be used to separate the two proteins by size. choice A is wrong -> gas chromatography separate molecules based on affinity, and molecules will be vaporized. but protein is too big to be vaporized choice c is wrong -> Thin-layer chromatography separates molecules based on affinity and would not be useful to separate molecules by size. choice B is wrong -> Mass spectrometry is used to measure the size of one molecule correct answer is D
cholesterol in cell membrane
Structural unit of cell membrane that keeps it from collapsing. - adds stiffness and flexibility - stabilize membrane fluidity - not found in plant cell, only in animal cell
Based on the information in the passage, compared to the T11 residue, the T3 residue is: A. closer to the C-terminus, and that terminus is likely to be positively charged in vivo. B. closer to the C-terminus, and that terminus is likely to be negatively charged in vivo. C. closer to the N-terminus, and that terminus is likely to be positively charged in vivo. D. closer to the N-terminus, and that terminus is likely to be negatively charged in vivo.
The amino acid structure of a protein is conventionally written from the N-terminus to the C-terminus, meaning that the T3 residue (which is the third residue in the protein) must be closer to the N-terminus than T11. correct answer is C
While the blood is buffered primarily through the equilibrium between carbon dioxide and carbonic acid, coupled with hemoglobin, the blood may also be buffered through other plasma proteins. Which of the following is true? A. A shift in the pH can alter the tertiary or quaternary structure of the protein, allowing it to buffer the pH by precipitating out of plasma in response to pH shifts. B. The amino acid residues that make up the protein may act as Brønsted acids or bases, reducing shifts in pH. C. Plasma soluble proteins have enzymatic function allowing them to sequester hydronium ions from the blood inside membrane-bound organelles in the podocytes lining the capillaries. D. In the presence of altered pH, any plasma-soluble proteins will undergo either acid- or base-catalyzed cleavage, thus depleting the acid or base causing the disruption to blood pH.
The amino acids that make up a protein may include many acidic or basic side chain groups. Those side chains can either release or absorb protons, allowing them to help buffer the blood through action as a Brønsted-Lowry acid or base. choice D -> no such thing as acid or base catalyzed protein cleavage correct answer is B
humoral immune response
The branch of acquired immunity that involves the activation of B cells and that leads to the production of antibodies, which defend against bacteria and viruses in body fluids.
cell-mediated immune response
The branch of acquired immunity that involves the activation of cytotoxic T cells, which defend against infected cells.
Which of the statements below is supported by the experimental results, as shown in Figures 1 and 2? ** #4 on word doc A. The duration of Eos co-culture with NK cells directly and linearly correlates to the amount of ECP found in the supernatant after centrifugation. B. Cells cultured with a 1:1 NK-to-Eos ratio displayed statistically similar levels of activation to cells cultured with a 5:1 NK-to-Eos ratio, as measured by CD69 expression. C. NK co-culture stimulates Eos activation while inhibiting degranulation. D. Co-culture with NK cells significantly increased Eos degranulation in all groups, as compared to Eos cells cultured alone.
The figure shows no statistically significant difference between samples with a 1:1 NK-to-Eos ratio and those with a 5:1 ratio, as evidenced by the fact that no horizontal line stretches from directly above the 1:1 group to directly above the 5:1 group (shown below). As such, the 1:1 and 5:1 groups are statistically similar, and choice B is supported by the experimental results. choice A is wrong -> no linear relationship is observed correct answer is B choice C is wrong -> NK co-culture stimulate degranulation choice D is wrong -> not significantly increased in all group, such as from 0:1 ratio to 1:1 ratio
Some proteins and antibodies are susceptible to pH damage. In order to prevent this damage, what would be the most efficient and effective additional step to perform at the end of the elution process? A. 0.1 M glycine•HCl, pH 3.0 B. 0.1 M NaCl, pH 7.0 C. 1 M Tris•HCl, pH 8.5 D. 0.7 M KOH, pH 13.85
The passage describes the final step of the process as lowering the pH to elute the proteins off of the beads. To reduce the risk of damage, we need to neutralize any excess acid. 1 M Tris•HCl, pH 8.5 is the buffer that will most effectively dissociate most protein:protein and antibody:antigen binding interactions without permanently affecting protein structure. Eluted protein fractions are best neutralized immediately by addition of alkaline buffers. choice A is wrong -> we tend to avoid low pH damages the molecules choice B is wrong -> neutral pH will not help neutralize acidity of final elute choice D is wrong -> though its pH is basic, it's too high correct answer is C
Given the information presented in the passage and the results of the serum electrophoresis, what is the most likely subunit composition of the CK 2 isoenzyme? ** #5 on word A. MM B. MB C. BB D. The subunit composition cannot be determined by the given information.
The passage indicates three possible isoform compositions: BB, MB, and MM, each of which corresponds to either CK1, CK2, or CK3. The passage states that the cytosolic CK B subunit contains a significantly smaller proportion of hydrophobic residues and a greater proportion of acidic residues with low pI values than does the cytosolic CK M subunit, which is enriched in asparagine and lysine. The BB dimer, then, must have the lowest pI of the three possibilities and will migrate farthest toward the positive electrode in gel electrophoresis. (Remember, acidic residues tend to be negatively-charged, so the BB dimer will remain negative at a lower pH than the other options.) Based on Figure 2, we can deduce that the BB dimer is CK1. By a similar rationale, the MB dimer should be intermediate in pI, and should migrate closer to the positive electrode than MM, but not as close as BB. Thus, the MB dimer must be CK2 (choice B). choice C is wrong -> CK 1 is most likely composed by BB correct answer is B
Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD? I. Competitive II. Allosteric III. Irreversible passage info: Most importantly, the PPP regenerates the universal reducing agent for anabolic pathways, NADPH, from NADP+ and H+. NADPH is critical for preventing cellular damage from free radicals derived from oxygen. G6PD is inhibited by high levels of NADPH. A. I only B. II only C. II and III only D. I, II, and III
The passages states that high levels of NADPH inhibit G6PD, and Equation 1 shows that the substrate of G6PD is G6P. Because the structures of G6P and NADPH are very different, it is unlikely that that NADPH competes with G6P at the active site; thus, this is not competitive inhibition. The passage also never suggests that the inhibition of G6PD is irreversible; rather, it is likely to be dynamic based on the amount of NADPH available. Therefore, NADPH most probably binds to a site that is not the active site, which is characteristic of allosteric inhibition. A generic model of allosteric inhibition is included below. correct answer is B
What is the expected ploidy of a cyst that has completed an encystation? A. 2n B. 4n C. 8n D. 16n
The protozoa will typically be diploid, and as stated in the second paragraph, cyst formation involves a round of DNA replication, followed by nuclear replication, followed by a second round of DNA replication. This process is illustrated below, and the presence of 4 4n nuclei yields a final total ploidy of 16n. correct answer is D
Robertsonian translocation
Translocation in which the long arms of two acrocentric chromosomes become joined to a common centromere, resulting in a chromosome with two long arms and usually another chromosome with two short arms.
Phospholipid (cell membrane)
Two layers Gives the cell it's permeable characteristic. The lipid bilayer is semi-permeable, allowing only certain molecules to diffuse across the membrane.
Given the locations of the MlyI cleavage site and recognition sequence, MlyI is NOT a: ** #1 on word doc A. Type I restriction enzyme. B. Type II restriction enzyme. C. Type III restriction enzyme. D. Type IV restriction enzyme.
Type I enzymes cleave at sites remote from the recognition site; they require both ATP and S-adenosyl-L-methionine to function. Since the passage shows a cleavage site very near a recognition site, MlyI must not be a Type I enzyme correct answer is A
The catalyst for TGFβ production most likely plays a direct role in: A. gene silencing. B. ribosomal inactivation. C. transfer of an acetyl group from acetyl-CoA. D. gene activation.
according to the passage, catalyst HDAC1 remove acetyl group from N-acetyl lysine group on histone. -> this affect gene expression deacetylation of histones cause DNA to bound histones more tightly and inhibit transcription that silence gene expression -> in contrast, histone acetylation promote gene expression Choice B is wrong -> passage never states HDAC1 affect ribosome, plus ribosome locates in cytoplasm, not nucleus choice C is wrong -> passage states acetyl group is removed from lysine group, and it will be transfered to acetyl-Coa choice D is wrong -> acetylation promote gene activation correct answer is A
What is the most likely transmission mechanism for the passage of Leigh syndrome from parent to child? A. Cytoplasmically inherited RNA B. Cytoplasmically inherited transcription errors C. Mitochondrial DNA D. Nuclear DNA
according to the passage, female and all her offspring have Leigh syndrome This means Leigh syndrome is a maternally-transmitted disease. Because it is also a cytoplasmically inherited disorder, it will be transmitted through cytoplasmic components, of which mitochondria are examples. choice A is wrong -> should be DNA, not RNA choice B is wrong -> If there is an error in transcribing DNA into RNA, it will only affect that particular RNA molecule. Meaningful effects on cytoplasmic inheritance will only be seen if there is an error or a mutation in the DNA itself. choice D is wrong -> all offspring carry disease, which mean this is not DNA inheritence correct answer is C
In which steps in Figure 3 is the fatty acid oxidized? I. Step 1 II. Step 2 III. Step 3 ** #7 on word doc passage info: Step 1 produces an FADH2 and step 3 produces an NADH + H+. Table 3 shows how many ATP are produced per NADH, FADH2 and per acetyl-CoA. A. I only B. III only C. I and III only D. I, II, and III
according to the passage, step 1 and step 3 produce FADH2 and NADH, which means there are oxidation and reduction rxn occur in step 1 and step 3 - step 1: FAD 2+ -> FADH2 - step 3: NAD+ -> NADH so, fatty acid is oxidized in step 1 and 3 Step 2 is simply the addition of water across a double bond, which adds two protons and one oxygen to the fatty acid, meaning there is no net oxidation or reduction of the fatty acid (the addition of an oxygen to a carbon does oxidize the carbon, but that is balanced out by the conversion of an alkene to an alkane, which is a reduction). correct answer is C
restriction enzyme
an enzyme that cuts DNA at or near specific recognition nucleotide sequences, known as restriction sites - occur in prokaryotes and archaea - sticky end is desirable because they ensure that the DNA fragments are ligated, or connected together, in the proper orientation
What is a likely neurological symptom caused by the toxin's effect on acetylcholine release? A. Tetanus B. Muscle spasms C. Flaccid paralysis D. Nausea
by inhibiting release of acetylcholin, the toxin interferes with nerve impulses and causes flaccid (sagging) paralysis of muscles that inhibit muscle contraction choice A is wrong -> tetanus is bacteria infection that causes muscle spasm choice B is wrong -> Inhibiting acetylcholine release would not cause repeated depolarization/activation of the muscle. correct answer is C
vitamin D
calcium and phosphate absorption from the gastrointestinal tract - sunshine vitamin
Glycoproteins
proteins that have carbohydrates covalently bonded to them - They are embedded in the cell membrane and help in cell to cell communications and molecule transport across the membrane.
double crossover
chromosomal arms of homologous chromosomes cross over in two different places along the arm. This results in a section in the middle of each chromosome being exchanged.
decrease tidal volume will cause blood pH decrease or increase
decrease blood pH because less CO2 will be exhaled
insluin
decreases plasma glucose level - increase glycogen synthesis - increase esterification of fatty acids to store in fat - decrease gluconeogenesis
adaptive immune system
directed against particular pathogens to which the body has been previously exposed. These pathogens are "remembered" by specialized cells, leading to a quicker response in the case of a second exposure. - ex: B and T cells
Which of the following conclusions is most strongly supported by the data presented in the passage? ** figure are #3 on word doc A. ALDH expression is associated with additional factors that promote invasiveness distinct from platinum resistance. B. Platinum resistance is the mechanism through which ALDH promotes invasiveness. C. ALDH expression is exclusively responsible for the platinum-resistant phenotype. D. BAX expression is promoted by ALDH.
figure 1 -> lower level ALDH (ALDH-) shows low invasiveness and less platinum resistance figure 2 -> no diff in invasiveness between normal and ALDH knock-out gene, but low platinum resistance we can conclude that platinum resistance is associated with ALDH expression, but it cannot be the whole story of how ALDH is associated with greater invasiveness. That is, native ALDH overexpression is a marker of an invasive phenotype, meaning that there must be other factors associated with ALDH expression that promote invasiveness. choice B is wrong -> data does not show any mechanism that platinum resistance promote invasiveness -> figure 2 evoke this statement b/c ALDH knock-out show same invasivenss choice C is wrong -> figure 1 and 2 shows ALDH-non-expressing and ALDH-knockdown cells, respectively, do show some platinum resistance -> so ALDH is not the sole factor responsible for platinum resistance; there are other factors involved correct answer is A
Which of the following is most likely NOT a symptom of acylcarnitine translocase deficiency? A. Hyperglycemia B. Muscle weakness C. Liver damage D. High ammonia levels in blood
from passage we learn that acylcarnitine translocase involves in fatty acid catabolism, and w/o it there would be too much fatty acid in the body and less energy to be used choice B is wrong -> less energy will lead to muscle weakness choice C is wrong -> Excess fat can be stored in the liver and cause fatty liver disease. choice D is wrong -> An abundance of acyl-carnitine that is unable to enter the mitochondria will result in high ammonia levels. The reason for this is that the body will have to rely more heavily on protein metabolism if it cannot carry out fatty acid metabolism, and ammonia is a waste product of protein metabolism. correct answer is A
Structural proteins (cell membrane)
give the cell support and shape.
vitamin K
help blood clotting
receptor proteins (cell membrane)
help cells communicate with their external environment through the use of hormones, neurotransmitters, and other signaling molecules.
what process will silence gene expression (inhibit transcription)
histone deacetylation and DNA methylation
In miRNA-directed gene silencing, a small RNA binds to an mRNA and directs degradation of the mRNA or prevents translation of the mRNA. Which of the following terms describes the process through which binding occurs? A. RNA polymerization B. Hybridization C. Elongation D. Transcription
hybridization mean binding 2 nucleotide strands choice A is wrong -> polymerization: the process by which nucleotides are strung together to form a single-stranded RNA strand correct answer is B
The researchers chose to co-culture all samples in the presence of interleukin-5, a cytokine. What is the most likely reason for this decision? A. Interleukin-5 facilitates degranulation in NK cells. B. Interleukin-5 inhibits the cytotoxic effects that NK cells have against eosinophils. C. Eosinophils die rapidly when not exposed to interleukin-5. D. The researchers were directly testing the effect of interleukin-5 on eosinophil activity.
if eosinophils die very quickly when not exposed to IL-5, then most or all cells may die before the 3- or 12-hour interval is even complete, making it difficult to assess the impact of NK co-culture. In reality, IL-5 does facilitate eosinophil survival. choice A is wrong -> research is to study eosinophil degranulation, not NK degranulation choice B is wrong -> according to the passage, NK cells kill "stressed, transformed, or infected cells", and if NK can get along well with other leukocytes, such as T cells, B cells, monocytes, and neutrophils, why would it kill eosinophil choice D is wrong -> interleukin is never the research interest correct answer is C
Considering the results of the laboratory testing given in the passage, which of the following diagnoses concerning the patient is most likely correct? passage info: CK1 is found exclusively in the brain, CK2 is found exclusively in cardiac muscle, and CK3 is present in both skeletal and cardiac muscle. ** #6 on word A. The patient is unlikely to be suffering a heart attack, because the elevated CK isoenzymes are more closely associated with damage to the brain than with damage to cardiac muscle. B. Additional testing is necessary, because the CK isoenzymes that are elevated are not specific for cardiac muscle damage, and may also indicate damage to skeletal muscle. C. Additional testing is necessary, because CK isoenzymes that are markers of cardiac damage appeared at normal levels, but may have been elevated prior to the time at which the samples were drawn. D. The patient is likely suffering a heart attack, because his CK isoenzyme levels are indicative of recent cardiac muscle damage in a patient with no signs of kidney disease.
lab result shows extremely high level of CK3 and high level of CK2, that indicated pt's cardiac muscle is damaged. -> CK3 result shows damage can be either in skeletal or cardiac muscle, but CK2 confirm damage in cardiac muscle choice B is wrong -> CK2 result confirm damage in cardiac muscle correct answer is D
PCR (polymerase chain reaction)
lab technique used to make a large number of copies of a DNA sequence of interest first step: denaturation of the template DNA by heating the reaction mixture to around 95°C - destroy hydrogen bonding - denaturation temperature second step: annealing that takes place at lower temp
Patients with excess fat are more likely to require larger therapeutic doses of which vitamin? A. Vitamin B1 B. Vitamin C C. Vitamin D D. Vitamin B3
lipid soluble vitamin: A, D, E, K water soluble vitamin: B and C Greater amounts of subcutaneous fat sequester more of the lipid-soluble vitamins and lower their release into the circulation. Thus, excess fat increases the initial dose of vitamin required to achieve a particular effect. correct answer is C
Glycolipids in the cell membrane
located on cell membrane surfaces and have a carbohydrate sugar chain attached to them. They help the cell to recognize other cells of the body.
Based on mass, which of the following molecules will most easily pass through a gap junction? passage info: Gap junctions are composed of two hemichannels (connexons), one embedded in each cell's plasma membrane and anchored to the cytoskeleton of the cells. These hemichannels interact to form a tight channel. My experiments show that molecules smaller than 500 Da pass through the channel. A. An RNA sequence of 3 uracil nucleotides B. Rotigaptide C. An Ala-Leu dipeptide D. A triglyceride with 3 hexadecanoic acid molecules
look for the smallest molecules the smallest molecule listed is the two-amino-acid chain of alanine and leucine. The heaviest amino acid, tryptophan, weights 204 Da. Thus, two bound residues must weigh 408 Da or less. This is much smaller than the other answer choices. If you did the math, you would find that each amino acid has 2 carbon, 2 oxygen, 1 nitrogen, and 4 hydrogen atoms. Alanine has a methyl side chain (1 C, 3 H), while leucine has an isobutyl side chain (4 C, 9 H), for an additional 5 carbon and 12 hydrogen atoms and a total of 9 C, 2 O, 2 N, and 20 H also 1 amino acid = 110 Da, so 2 amino acid is about 220 da choice A is wrong - One nucleotide consists of a ribose sugar, a phosphate group, and a nitrogenous base. The phosphate group contains 1 P and 4 O atoms. The ribose consists of 3 O and 5 C atoms, while the nitrogenous base includes at least 1 N and 6 C atoms. Adding this up, we get 35 + 7(16) + 11(12) + 14 = 293. So, three attached nucleotides of any kind would put us above 800 Da, well over choice C. choice B is wrong - the are more amino acids in rotigaptide, and see #10 on word doc for structure choice D is wrong - Hexadecanoic (aka palmitic) acid has the formula C16H32O2. The carbon alone weighs 192 Da, so three hexadecanoic acid molecules would weigh over 576 Da. correct answer is C
Which of the following is most likely to use a protein channel to cross the eukaryotic cell membrane? A. Aldosterone B. Ca2+ C. O2 D. CO2
membrane is hydrophobic as it is composed by phospholipid bilayer. Thus, it will prevent ion and polar solutes diffuse across the membrane - but it will allow hydrophobic molecules pass through - transmembrane channel is necessary because the molecule may be too large or polar for simple diffusion => ex:sodium-calcium exchanger Calcium ions would be unlikely to pass directly through the cell membrane and thus require an ion channel. choice A is wrong -> aldosterone is steroid hormone built from nonpolar cholesterol, so it will pass the membrane choice C and D is wrong -> gas are small molecules that can easily pass (simple diffusion) correct answer is B
Western blotting
procedure that uses labeled antibodies to detect specific antigens in a mixture of proteins separated according to their molecular weight - identify protein
Peripheral membrane proteins (cell membrane)
proteins associated with but not embedded within the plasma membrane - connected to the membrane by interactions with other proteins
Which step of the procedure corresponds to the release of DNA-coupled conjugates? ** graph is #2 on word doc passage info: First, the conjugation is performed in solution by mixing antibodies with an excess of azide-modified oligonucleotides. The resulting mix containing antibody-DNA conjugates and remaining free antibodies and oligonucleotides is incubated, and the products are immobilized on agarose beads. After washes to remove unconjugated antibodies, the captured antibody-DNA conjugates along with unconjugated oligonucleotides are both released by enzymatic cleavage using the restriction enzyme MlyI that cleaves at the border of the complementary region (Figure 2) between the capture and conjugate oligonucleotides to release single-stranded DNA-coupled conjugates. The cleavage process required both ATP and S-adenosyl-L-met in order to run to completion. The eluate was then captured on protein G beads and washes were conducted to remove any free oligonucleotides having no antibodies attached. Finally, purified conjugates were eluted from the protein G beads by lowering the pH. A. B B. C C. E D. D
step A: DNA oligonucleotides are hybridized to the oligonucleotides in the mixture. step B: agarose beads are used to capture the biotinylated capture oligonucleotides, both in the form of conjugates and free oligonucleotides. step C: the unconjugated antibodies are removed by washes. step D: the MlyI enzyme is used to cleave the captured oligonucleotide hybrids, allowing both conjugates and oligonucleotides to be eluted from the solid support. step E: the eluate is then incubated with protein G beads, while free oligonucleotides are removed by washes. step F: purified conjugates are eluted from the solid support. correct answer is D
Which of the following is NOT a function of the sympathetic nervous system? A. Increased heart rate B. Pupillary constriction C. Vasodilation D. Vasoconstriction
sympathetic NS: - pupil dilation: need the light to access the circumstance - increased rate and force of heart contraction - blood vessel dilation in skeletal muscle - blood constriction in gastrointestinal organs, - inhibition of peristalsis by the digestive tract choice B is correct
Type IV restriction enzymes
target modified (e.g. methylated, hydroxymethylated) DNA.
elongation
the adding of subunits to make a longer strand of a macromolecule; similar to polymerization
polymerization
the process by which nucleotides are strung together to form a single-stranded RNA strand
ketone bodies
the product of the incomplete breakdown of fat when glucose is not available in the cells - ketone bodies synthesis occur b/c during high rate of fatty acid metabolism, large amt of acetyl-CoA is produced, and TCA cycle can't take care of this (think about freeway during rush hr in LA) - occur in liver mitochondria
Assuming that ECP (positive charge) was named for its electrical charge at physiological pH, which of the following must be true? A. The primary structure of ECP contains more acidic residues than uncharged residues. B. The primary structure of ECP contains more basic residues than uncharged residues. C. The primary structure of ECP contains more acidic residues than basic residues. D. The primary structure of ECP contains more basic residues than acidic residues.
the question stem is asking which statement MUST BE TRUE, so be careful with wording choice B is wrong -> basic residues can be less than uncharged residues to make ECP cationic correct answer is D
passive immunity
the short-term immunity that results from the introduction of antibodies from another person or animal.
Transport proteins (cell membrane)
transport molecules across cell membranes through facilitated diffusion. - ex: globular protein
Troponin isoenzymes are used as an alternative biomarker in the diagnosis of heart attacks. In which of the following muscle types does the troponin complex function in contraction? I. Skeletal muscle II. Smooth muscle III. Cardiac muscle A. I only B. I and II only C. I and III only D. I, II, and III
troponin only present in skeletal and cardiac muscle, and not in smooth muscle correct answer is C
hybridization
two nucleic acid strands bind to form a double-stranded structure - opposite of denaturation - ex: siRNA(short single strand RNA) can hybridize with mRNA in a process that induces gene silencing and prevent translation
vitamin A
vision
hypertonic
when comparing two solutions, the solution with the greater concentration of solutes - if cell is placed in hypertonic solution, water will move from cell and cause it to shrink
hypotonic
when comparing two solutions, the solution with the lower concentration of solutes - if cell is placed in hypotonic solution, water will move into cell and cause it to swell and burst
