AAMC biology and biochem
dyad chromosome
a replicated chromosome that consists of 2 identical, sister chromatids joined at the centromere. (double chromosome, looks like an X)
motor end plate (neuromuscular junction)
a specific part of the sarcolemma that contains acetylcholine (ACh) receptors
archaic
ancient; old-fashioned Her archaic Commodore computer could not run the latest software. Synonyms: ancient; antediluvian; antique; bygone; dated; dowdy; fusty; obsolete; old-fashioned; outdated; outmoded; passé; prehistoric; stale; superannuated; superseded; vintage
lipid rafts
cholesterol filled microdomains for signal transduction and endocytosis Lipid rafts are subdomains of the plasma membrane that contain high concentrations of cholesterol and glycosphingolipids
pragmatic
dealing with things sensibly and realistically in a way that is based on practical rather than theoretical considerations
Endosomes and Lysosomes
early endosomes mature into late endosomes which mature into lysosomes due to the time it takes to lower pH
aerie(aery,eyrie,eyry)
elevated nest
Alternative splicing occurs only in
eukaryotes
Cells of the immune system
1. Made up of leukocytes a) Granulocytes (Neutrophils, eosinophils, basophils) b) Lymphocytes (T-Cells, B-Cells, natural killer cells) c) Monocytes (precursor to macrophage) d) Dendritic cells (antigen presenting cells; APC) 2. Produced in the bone marrow
Hermits
A person who lives away from society for religious reasons
Gyrase
A prokaryotic enzyme used to twist teh single circular chromosome of prokaryotes upon itself to form supercois. Supercoiling helps to compact prokaryotic DNa and make it sturdier. topoisomerase in prokaryotes
genome size
The total amount of DNA contained within one copy of a single genome
adhesion proteins
Adhesion proteins are membrane proteins that form junctions between adjacent cells.
Aldosterone ions
Aldosterone increases reabsorption of NaCl and HCO3- and other salts while increasing secretion of K+ and H+ (from the Kaplan Biology book in Ch.5). If there's a deficiency of aldosterone, then less NaCl and bicarbonate will be reabsorbed, and less K+ would be secreted in urine, resulting in lower NaCl and bicarbonate while a higher K+ and H+ in the serum
pyruvate structure
C3H4O3 CH3-C(=O)-C(=O)-OH
ca pump and neuron
Ca2+ pumps actively transport Ca2+ out of a neuron. However, intracellular Ca2+ primarily controls neurotransmitter release, not the resting potential of a neuron.
Dehydrogenase
Catalyzes oxidation-reduction reactions (e.g., pyruvate dehydrogenase).
bacteria have double stranded
DNA and plasmid
Mature erythrocytes doesn't contain
DNA or mitochondria or nucleus
Which DNA polymerase do the DNA replication ?
DNA polymerase III 3prime to 5 prime exonuclease activity
ischemia consequences
Decreased delivery of energy (e.g., glucose) Damage to blood vessels Vasomotor paralysis Vasoconstriction Changes in blood Desaturation Clotting Sludging
Ectoderm
Epidermis of the skin and its derivates including sweat glands and hair follicles Epithelial lining of mouth and anus Cornea and lens of the eye Nervous system Sensory receptors in epidermis Adrenal medulla Tooth enamel Epithelium of pineal and pituitary glands
tertiary and quaternary structure interaction
Further folding into a unique three-dimensional shape that may be globular or fibrous Secondary structure held together by Hydrogen bonds of the backbone atoms, tertiary held together by covalent and/or non-covalent bonds between side chain atoms. Quaternary structure is side chain interactions (again could be covalent and/or non-covalent) but these are between the side chains of different polypeptides (can only happen if the protein is multiple subunits). disulfide in tertiary
Fibroblasts
In connective tissue, cells that secrete the proteins of the fibers.
transmembrane proteins
Integral proteins that span the membrane Involved in the transport of selected ions and molecules across the plasma membrane Able to affect membrane permeability
Vikings
Invaders of Europe that came from Scandinavia
catalytic efficiency
Kcat/Km
Anion exchange chromatography
Negative proteins stick to positive beads, only positive proteins go through
Y-linked inheritance
Only males have Y chromosomes, passed from fathers to sons, all Y-linked traits are expressed
Penitents
People who admit their sins, are truly sorry for having sinned, and wish to be restored to the good graces of God and the Church.
The LXCXE motif is described as being "highly conserved." Certain parts of the genome, such as that encoding the LXCXE motif, are highly conserved because they are: A. vital to an organism's survival. B. chemically incapable of mutation. C. stored in vesicles for later secretion. D. stored in cell compartments where they are unlikely to be secreted.
Solution: The correct answer is A. From an evolutionary point of view, DNA sequences that are vital to an organism's life are conserved among species. Mutations can affect any portion of DNA. DNA is stored in the nucleus, not in vesicles. The fragment is stored in the nucleus and it is unlikely to be secreted, however this is not the definition of "highly conserved." Instead, genomic conservation refers to maintaining sequence similarity across species.
Fibroblasts
Spindle-shaped cells that form connective tissue proper
How to increase force of contraction
Stimulating a larger number of cells or motor units
stroke
Sudden loss of consciousness, sensation, and voluntary motion caused by rupture or obstruction (as by a clot) of a blood vessel of the brain.
tetrad chromosomes
Tetrad consist of two homologous chromosomes, each with two chromatids. Thus, a tetrad consists of 4 chromatids, two of which belong to each chromosome called sister chromatids which are joined at the centromere.
vili atrophy
Villi atrophy results in a decrease in the surface area of the small intestine, leading to a decrease in nutrient absorption. I
What bond is cleaved by IN during the first reaction of integration? A P-C B P-P C P-H D P-O
What bond is cleaved by IN during the first reaction of integration? Jack Westin Advanced Solution: No tricky verbiage here, we can answer this question just by knowing about the bond that's cleaved. We'll revisit the passage, but a quick glance at our answer choices shows we're going to have phosphorus bound to another atom. We'll get details about the cleavage taking place, and then we'll jump into specifics using our general knowledge.We have part of the passage here. I want to focus on the middle of the second paragraph. It says Upon binding, IN catalyzes cleavage of a GT dinucleotide from each 3PRIME vDNA terminus. This is the cleavage we're focused on. The cleavage of the GT dinucleotide. How do two nucleotides bond in the first place? DNA is polymerized when the 3′ hydroxyl group of the nucleotide, attacks a 5′ phosphate group to create a phosphodiester bond. When we look at a phosphate group, the only thing the phosphorus is attached to is 4 oxygen atoms. That means our ideal answer is going to be phosphorus bonded to oxygen. In our breakdown we went through the formation of the dinucleotide and the bonds formed by phosphorus in the phosphate group. Phosphorus only binds oxygen, not carbon, hydrogen, or another phosphorus. That makes it easy to eliminate answer choices A-C. We're left with our best answer, the only answer that matches our breakdown, answer choice D: the bond that is cleaved is between phosphorus and oxygen. A P-C B P-P C P-H D P-O How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. Solution: The correct answer is D. This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is D because in the first reaction of integration, integrase cleaves a dinucleotide from the 3 end of a strand of viral DNA. This involves the cleavage of a P-O bond. It is a Knowledge of Scientific Concepts and Principles question because you are asked to recall the structure of DNA.
Osteocytes
a bone cell, formed when an osteoblast becomes embedded in the matrix it has secreted.
Tip to tail method
a method for finding a resultant vector in which the tail of one vector is drawn from the tip of another vector
chinks
a narrow opening
high heat capacity
absorbs and releases large amounts of heat before changing temperature
Fungi reproduce
asexually and are eukaryotes
eluded
avoided or escaped by cleverness or speed
Large hydrophobic aromatic molecules can pass the membrane through diffusion
because they are planar
Endosomal pathway
begins at endocytosis where, following internalization of extracellular materials into vesicles, the vesicles containing these materials mature to early endosomes, late endosomes, and finally, lysosomes
Lysosomes
cell organelle filled with enzymes needed to break down certain materials in the cell Hydrolytic enzymes within it do the breakdown of macromolecules and as well as material that has been taken into the cell by phagocytosis low pH than cytoplasm
hewn
cut or shaped with hard blows of a heavy cutting instrument like an ax or chisel
SDS-PAGE
denatures the proteins and masks the native charge so that the comparison of size is more accurate, but the functional protein cannot be recaptured from the gel non reducing don't break down disulfide bonds reducing break down disulfide bonds
commensal
describes an organism that lives symbiotically with a host; this host neither benefits nor suffers from the association
monocyte function
develop into macrophages in tissues and phagocytize pathogens or debris
ovarian cells and osteoclasts
epithelial and connective tissue respectively
If the GAPDH gene is continuously expressed, where is it most likely found?
euchromatin
Cochlear implants
for nerve deafness not literally though, sensorineural hearing loss Nerve deafness occurs when the cochlea(typically the hair cells on the basillar membrane) or the auditory nerve (nerve 8 I think) become damaged. Cochlear implants directly electrically stimulate the basillar membrane bypassing the need for hair cells needing to shift and depolarize. Cochlear implants can only fix the cochlea sensorineural hearing loss and cant do dick if your nerve is botched.
arched
having a curved shape at the top
chaperone proteins
help other proteins fold correctly
age-related diseases
illness connected with getting old
A classic hallmark of GPCR receptor is
increase in Camp levels
tumor necrosis factor
inflammation, apoptosis, act together to promote response
jest
joke, prank
The ornithine decarboxylase reaction has been studied extensively by biomedical researchers. The most likely reason for the interest of these researchers is that the reaction is: A. a good source of ornithine. B. an early event in cell division. C. unique to mammals. D. one in which a chiral reactant is converted to an achiral product. Solution
kinda late to the party but my reasoning was D) chiral reactant -> achiral product is not all that rare & exciting even if D) it was a really exciting thing, that would be much more interesting to chemists, whereas biologists would care more about B)cell division (e.g. maybe connects to cancer, etc) My reasoning was that the idea that it's an early event in mitosis might be significant in cancer research (e.g. inhibiting ornithine decarboxylase as a way to prevent tumor cells from passing the metaphase/anaphase checkpoint) but idk
wicker
material made of thin twigs or branches that bend easily
Regulatory T cells
may function in preventing autoimmune reactions control the T-cell response
structural isomers (constitutional isomers)
molecules that have the same molecular formula but differ in the way the atoms are connected
nucleoside
nitrogenous base + sugar (no phosphate group)
Autsomal Dominant
no generations are missed, both mother and father are causing the disease to all children
palindromic sequence
nucleic acid sequence on double-stranded DNA or RNA where reading the 5' to 3' forward on one strand matches the sequence reading backward 5' to 3' on the complementary strand
pillaging
plundering, looting, destroying property by violence
lymphatic system functions
recovery of excess tissue fluid, detection of pathogens, production of immune cells, defense against disease
Lyophilized
the process of rapidly freezing a substance at extremely low temperatures and the dehydrating the substance in a high vacuum
exocrine secretions of pancreas
trypsinogen, chymotrypsinogen, lipase, amylase, bicarb a secretion deposited into a body cavity through a duct
Kcat
turnover number (molecules catalyzed per second in optimal conditions) Vmax / [E]
precariousness
uncertainty; danger
feedforward regulation
~ Enzymes regulated by intermediates that precede them
initial filtration step in the glomerulus of the mammalian kidney occurs primarily by:
There are three pressures that work together to regulate filtration in the glomerulus: glomerular capillary pressure, capsular hydrostatic pressure, and blood colloid osmotic pressure. The glomerular capillary pressure will force filtrate from a capillary into Bowman's capsule; the other two forces promote movement of the filtrate in the opposite direction.
If we have inflammation what will happen to the blood vessels
They will dilate to increase blood flow and leukocytes to the site
According to the passage, platelets are LEAST likely to contain transmembrane serotonin transporters
This is incorrect. The passage states that platelets are carrying serotonin that has been synthesized outside the platelets; therefore, there must be a mechanism for transporting serotonin into platelets. Serotonin would be transported into platelets by transmembrane transporters. This option is very likely to be true; therefore, it is not the correct answer.
Succinate dehydrogenase complex
Transfers electrons from succinate (via bound FADH2) to oxidized ubiquinone complex 2
ducts
Tubes that lead from a gland to a target organ
cation exchange chromatography
Type of adsorption chromatography based on ionic interactions between a negatively-charged stationary phase and positively-charged protein.
The following graph plots a muscle power curve superimposed on a curve showing the relationship between the force a muscle generates against a load and the velocity at which the muscle shortens Assuming the duration of contraction is the same, which point on the power curve corresponds to the LEAST amount of work being done by the muscle?
When time is same then Power = work because fxv time is same then fxd so with the lowest power would be when F is zero or v is zero
Which statement best accounts for the hereditary transmission of SDH-linked paraganglioma in a parent specific manner? SDH is: A. an imprinted gene. B. a Y-linked gene. C. an X-linked gene. D. a tumor suppressor gene. Genetically, this mutation is transmitted in the germ line and segregates with a benign form of neuroendocrine tumor known as paraganglioma, but only after paternal, not maternal, transmission of the mutated allele.
Which statement best accounts for the hereditary transmission of SDH-linked paraganglioma in a parent specific manner? SDH is: Jack Westin Advanced Solution: The second paragraph is the key to understanding the hereditary transmission of SDH-linked paraganglioma: according to the passage it only segregates with the SDH-linked paraganglioma after paternal transmission of the mutant allele. There is no indication that males and females are unequally affected and the mutation segregates exclusively with the paternal lineage so something specific to the fathers must be at play. A. an imprinted gene. Jack Westin Advanced Solution: Imprinted genes are parent-specific based on inherited epigenetic modifications. For example, if the gene is methylated and "off" in the parent, it will be "off" in the offspring. If the transmission is exclusive to the paternal lineage and affects both sexes, an imprinted gene would be a great way to account for the differential transmission of the SDH-linked paraganglioma. B. a Y-linked gene. Jack Westin Advanced Solution: A Y-linked gene would only affect the male offspring. The passage gives us no reason to believe only male offspring are affected. C. an X-linked gene. Jack Westin Advanced Solution: An X-linked gene could affect both males and females, but it would disproportionately affect males and would not be exclusive to paternal transmission. D. a tumor suppressor gene. Jack Westin Advanced Solution: Tumor suppressor genes are not transmitted in a parent-specific manner. Only answer choice A correctly identifies a parent-specific transmission that could explain transmission exclusively via the paternal lineage. "but only after paternal, not maternal, transmission of the mutated allele" this implies that both males and females can carry the mutated allele, but the disease comes from only male transmission of the mutated allele, so female transmission does not cause a mutation. This is because imprinted genes are epigenetically modified and are only imprinted on one allele. In this case, the female allele is presumably silenced. Don't worry, I also got this Q wrong when I took the FL. It is some semantic bull if you ask me, but that is where they get the answer from. It means both sexes can inherit the mutated gene from either the father or the mother and will have the same genotype. However, the phenotype of offspring will be affected only if the mutated gene is inherited from the father (only after paternal, not maternal, transmission). The mother mutated gene has been silenced epigentically. An imprinted gene is one that is expressed in a parent-specific manner, so like the passage is saying in this case, only paternal. This was definitely a tricky question that made me pause for awhile, but the key in the passage is that it makes reference to the mutated allele being possessed both paternally and maternally (" ... but only after paternal, not maternal, transmission of the mutated allele), so if it were a Y-linked inheritance pattern, the allele wouldn't be expressed by the mother. So because the mutated gene can only be transmitted paternally and not maternally, it is indicative of an imprinted gene.
pyrrole ring
4 membered ring containing nitrogen as a part of heme group
Basophils
A circulating leukocyte that produces histamine. Account for 0%-1% of WBCs in the blood.
Fibroblasts
A fibroblast is a type of cell that contributes to the formation of connective tissue, a fibrous cellular material that supports and connects other tissues or organs in the body. Fibroblasts secrete collagen proteins that help maintain the structural framework of tissues.
Eremitical
A form of monasticism that began in Egypt in the third century in which individuals and small groups withdrew from cities and organized society to seek God through prayer
X-linked dominant inheritance
All females of the affected father are diseased. Affected mother can pass on the disease to both male and female offspring. don't skip generation males are affected by mothers
Monocytes classification
An agranular leukocyte that is able to migrate into tissues and transform into a macrophage.
Can Glycoproteins cross the cell membrane?
As a general rule no. They are too large and charged to simply not be repelled by the lipid plasma membrane. But there are lots of other selective mechanisms to get across this plasma membrane barrier. They can bind to receptors on the surface and that might trigger a secondary message if a hormone or cytokine. Or the receptor for the glycoprotein might trigger endocytosis
B cells
Cells manufactured in the bone marrow create antibodies for isolating and destroying invading bacteria and viruses.
Chemotaxis
Chemotaxis in microbiology refers to the migration of cells toward attractant chemicals or away from repellents. Virtually, every motile organism exhibits some type of chemotaxis.
Esinophils
White blood cell (Granulocyte) that defend against potozoan and helminth parasites. Stain red and may play a role in allergies.
Endoderm
Epithelial lining of digestive tract Epithelial lining of respiratory system Lining of urethra, urinary bladders and reproductive system Liver Pancreas Thymus Thyroid and Parathyroid glands
5 types of electron carriers in the ETC:
Flavin Rings - 2 electrons Iron-Sulphur Centers (Fe-S) - 1 electron Ubiquitin - 2 electrons Cytochrome C - 1 electron Copper Centers (Cu) - 1 electron
microtubule organizing center (MTOC)
General term for any structure (e.g., centrosome and basal body) that organizes microtubules in cells.
Inhibitor
Ki is independent of inhibitor concentration, which should be a constant 60.3uM from Table 1 since it is an inherent property of the inhibitor.
Mesoderm
Notochord Skeletal system Muscular layer of stomach and intestine Excretory system Circulatory and lymphatic system Reproductive system except germ cells Dermis of skin Lining of body cavity Adrenal cortex Kidney Ureter
post-transcriptional modification
Post transcriptional modifications are ones that are on the mRNA before it is translated. They can also occur at the same time as transcription and pretty much transform pre-mRNA into mature mRNA. Modifications include: addition of the 5' cap, poly A tail and splicing. The cap and poly A tail are added co-transcriptionally, but still count as post transcriptional. Splicing is what gives rise to isoforms. Depending on how you arrange the introns/exons you get different proteins by alternative splicing. They usually have pretty similar properties/functions.
Post-trasnlational modification
Post translational are modifications that occur that happen after the protein has already been translated. Post translational modifications are usually groups being added to the protein by covalent bonds or cleavage of protein. Examples of the groups are phosphorlation, glycosylation and ubquitination. An example of cleavage is with insulin where the hormone is activated by cleavage of a section of the protein. Post translational modifications don't usually give completely different proteins, but deal with activation/inactivation or enhancing the protein's function.
Convergent Evolution (Analogy)
Process by which unrelated organisms independently evolve similarities when adapting to similar environments
Protozoa
Protozoa are a group of single-celled eukaryotes, either free-living or parasitic, that feed on organic matter such as other microorganisms or organic tissues and debris.
Restriction enzymes
Restriction enzymes are special proteins that recognize a specific SEQUENCE and cuts the DNA at that site. So, if it cleaves at TATA and you have CCCAAATATA after your left with CCCAAA.
Retrograde transport
Retrograde transport means things are moving toward the nucleus, and this is accomplished by dynein. This pathway would look like:Retrograde transport = Golgi -> COP I vesicle -> ER
An ester is prepared by the method of direct esterification using an esterase enzyme as a catalyst. Which of the following modifications will NOT appreciably increase the final yield of ester? A. Using 2 times as much enzyme B. Using 2 moles of RCOOH instead of 1 mole C. Using 2 moles of RCH2OH instead of 1 mole D. Removing RCOOCH2R from the reaction mixture as it is formed
Solution: The correct answer is A. This modification will increase the rate of formation, not the final yield of ester. Doubling the concentration of RCOOH will increase the yield of ester by Le Châtelier's principle. Doubling the concentration of RCH2OH will increase the yield of ester by Le Châtelier's principle. By Le Châtelier's principle, removing the ester product as it forms will drive the reaction to form more ester.
The viruses that encode either FNDC5 or control protein transfer genetic material via: A. transformation. B. transduction. C. contamination. D. conjugation.
Solution: The correct answer is B. Transformation is the process that transfers genetic material from the environment into bacteria. Transduction is the process by which nucleic acids are transferred from viruses to cells. Contamination is the act of making a solution or sample impure. Conjugation is the exchange of nucleic acids between bacteria.
A major obstacle to obtaining useful energy from a nuclear fusion reactor is containment of the fuel at the very high temperatures required for fusion. The reason such high temperatures are required is to: A. eliminate the strong nuclear force. B. remove electrical charge from reactants. C. decrease the density of the fuel. D. enable reactants to approach within range of the strong nuclear force.
Solution: The correct answer is D. The strong nuclear force cannot be eliminated by increasing the temperature because this force manifests whenever nucleons are present and within a range of picometers. The electrical charge is an intrinsic property of matter that is independent of temperature. Decreasing the density of the fuel is detrimental because the probability of fusion increases with the increase in fuel density because the attractive component of the strong nuclear force acts at short distances indicative of high density for the fuel material. The probability of fusion increases with the decrease in the average distance between fuel particles that enables attractive nuclear forces to overcome the repelling nuclear forces acting at medium and long distances. An increase in the temperature is equivalent to an increase in the root-mean-square speed of the fuel particles that will travel the average distance between fuel reactants in smaller times. The associated increase in the kinetic energy of the particles relative to the center of mass of the nuclear fuel system essentially correlates with a decrease in the electrostatic potential barrier that repels particles of the same electric charge. In turn, this increases the probability of particles to undergo the tunnel effect by penetrating the electrostatic barrier. These combined effects enable reactants to approach within range of the strong nuclear force. Eliminate B - charge is conserved regardless of temperature. Eliminate A - the strong nuclear force is what binds a nucleus together - fusion would want a strong nuclear force to hold together protons and neutrons from different nuclei. Eliminate C - the higher density the material, the closer the nuclei are to each other, something that is desirable for fusion since it increases likelihood of reactions. Choose D and recognize that strong nuclear forces are what bind nuclei together. Combining reactants at the subatomic level (which is what happens in fusion reactions) requires a ton of energy, because the strong nuclear force causes the atoms to repel each other. High temperatures indicate high amounts of heat energy, which increases the movement velocity of the reactants and ultimately the force with which they collide. If they collide hard enough, they can overcome the SNF and fuse together. The answer that most closely matches this explanation is D 3 ReplyShare level 2No_Physics_8481OP·14 days ago Thank you and congrats on the score! 2 ReplyShare Continue this thread level 1Bball5uper5tar·14 days ago I honestly thought of it in terms of spontaneity of the reaction. Think abt Delta G = Delta H - TDeltaS High Temp overcomes initially positive delta H to become a spontaneous process (Negative Delta G). Honestly may be completely flawed reasoning but this is how I thought of it 2 ReplyShare level 1Sweaty_Ask322·16 hr. ago I thought of it as Nuclear (+) charges trying to fuse together as they repel each other and so what the higher temp. does is increase the KE of the system so that the distance between the molecules shortens.
Approximately how many moles of Kr+ are contained in the laser tube at 0°C and 1 atm? A. 3 x 10-7 B. 2 x 10-6 C. 4 x 10-5 D. 5 x 10-4
Solution: The correct answer is D. This number of moles of gas occupies only 0.00672 mL at STP. This number of moles of gas occupies only 0.0448 mL at STP. This number of moles of gas only occupies 0.90 mL at STP. There is 1 mole of gas in a volume of 22.4 L = 2.24 x 10 cm3 at STP; there are approximately 5 x 10-4 moles in 11 cm3.
If oligonucleotides such as mRNA were not degraded rapidly by intracellular agents, which of the following processes would be most affected?
The coordination of cell differentiation during development
Centrioles
a minute cylindrical organelle near the nucleus in animal cells, occurring in pairs and involved in the development of spindle fibers in cell division.
patron
a person who gives financial or other support to a person, organization, cause, or activity
Conjugation in bacteria
The process of conjugation involves production of a special conjugation pilus (sex pilus) by one bacterium and transfer through it of DNA to another bacterium. It requires special genes for the pilus and these are usually present on a plasmid, a separate extragenomic strand of DNA not incorporated into the bacterium's own DNA. This plasmid is referred to as the fertility or F factor. Conjugation is a feature of Gram-negative bacteria. It confers the advantages of sexual reproduction on the bacterium. The plasmid benefits by being able to move from one host bacterium to another through the conjugation pilus.
Henderson-Hasselbalch equation
pH = pKa + log [A-]/[HA]
intermediate filaments
permanent rather than microfilament/ tubules structural support for the cell resist mechanical stress can be compared to springs inside the mattress
Serotonin polar or non polar
polar because of amine and alcohol group per AAMC so can't ross the membrane and so its receptors would be embedded in the cell membrane
reclusive
preferring to live in isolation
autosomal dominant inheritance DO THESE PLEASE
presence of certain genes that means there is a 100% chance of the person eventually getting the disease
ubiquitous
present, appearing, or found everywhere: his ubiquitous influence was felt by all the family | cowboy hats are ubiquitous among the male singers.
divergent evolution (adaptive radiation)
process by which a single species or a small group of species evolves into several different forms that live in different ways
Cell Differentiation / Specialization
process of making cells "different" or "special" in function For example: some cells are muscle cells, some are bone cells **differential gene expression - genes turned "on" or "off"
example of exocrine glands
salivary glands, mammary glands, chief cells in stomach, liver, and pancreas through ducts
histone acetylation
the attachment of acetyl groups (-COCH3) to certain amino acids of histone proteins, the chromatin becomes less compact, and the DNA is accessible for transcription Histone acetylation is part of gene regulation. Acetylation removes the positive charge on the histone, decreasing the interaction with the DNA and converting it from beterochromatin to euchromatin so that it can readily be transcribed. It has been shown that ROS can increase histone acetylation, leading to increased gene expression.
primary immune response
the initial adaptive immune response to an antigen, which appears after a lag of about 10-17 days
polar bodies
the polar bodies do not have less organelles or genetic material; it's just a smaller amount of cytoplasm that does not allow the cell (the polar body) to survive.
mature RNA sequence for a protein that is secreted or that will locate to the cell membrane will contain
the signal sequence. This portion of the mRNA is located in the 5′ region and will signal to the ribosome that translation needs to be continued in the rough endoplasmic reticulum. Transmembrane proteins must be shuttled through the ER/Golgi to form a vesicle that fuses with the membrane and places AT1 in the transmembrane space. Once translation begins in the cytoplasm, the protein must have a signal sequence that allows it to bind to the ER and continue on the pathway to the membrane.
Quantitative PCR (qPCR)
uses fluorescent probe to monitor the amplification process It monitors the amplification of a targeted DNA molecule during the PCR, not at its end, as in conventional
Beta-branched Amino Acids
valine, isoleucine, threonine
Ingestion of a foreign particle
when a macrophage ingests foreign material, the material initially becomes trapped in a phagosome. The phagosome then fuses with a lysosome to form a phagolysosome. Inside the phagolysosome, enzymes digest the foreign object. Of the cell structures listed, the labeled carbohydrate is most likely to be microscopically visualized within a lysosome (phagolysosome).
Synapsis (meiosis)
when the homologous chromosomes come together and pair up with each other in prophase I
proximal convuluted tubule
where is 65% of the filtrate is reabsorbed in the nephron? https://www.khanacademy.org/test-prep/mcat/organ-systems/the-renal-system/a/tubular-reabsorption-article
anarchic
with no controlling rules or principles to give order
short chain fatty acids
less than 6 carbons
epithelial cells functions
protection, absorption, filtration, secretion skin cells that cover the outside of the body and line the internal surfaces of organs
imparted
to give a particular quality to something
Transferase
transferases are responsible for transferring of functional groups between molecules.
Endonuclease
"Endo" inside "nuclease" enzyme that cleaves nucleotides. Thus, it is an enzyme that breaks apart the SINGLE nucleotides in dna backbone starting in the MIDDLE of the strand.
transcription
'm not sure.. but the NF-kB pathway creates IL-6 and IL-10. In order to make these interleukins, they must be transcribed/expressed. Their genes can be expressed upon the binding of the cp65 and cRel transcription factors, since transcription factors are a big part of transcription. Having the binding sites of the transcription factors on the promoter allows for the transcription factors of the NF-kB pathway to bind and thus, allow for gene expression gene expression is enhanced by certain proteins binding to the promoter region. If you look at the nucleus depicted in figure 1, you'll see that a p65/cReI protein dimer binds to DNA and causes transcription of iL-6 and iL10.
Ubiquination of proteins
1) Ub is attached to a place where there is a lysine side chain 2) This attachment signals need for that to be degraded - Sometimes multiple Ubs are attached 3) Ubiquinated protein goes into protesome 4) Ub is removed/released + protein degrades to peptides 5) Peptide is further broken down to Aa's Aa's go on to participate in biosynthesis, be catabolized into carbon skeletons and amino groups Proteasomes are different from lysosomes mainly due to the fact that proteasomes are just free floating in the cytoplasm while lysosome are vesicles in the cell. But both of them have very similar functions. You're probably thinking of lysosomal proteases. In regards to your question, I believe ubiquitination leads to the proteasome coming to the degrade the protein. Protease and not proteasome are found in lysosome.
Lytic granules are generally released from CTLs when the T-cell receptors on these cells bind specifically to: A. viral antigens presented on the surface of virus-infected cells. B. growth factors secreted by helper T lymphocytes. C. B-cell receptors on activated B lymphocytes. D. constant regions of secreted antibodies.
A especially if its a virus no anti bodies just kill them
Pentose Phosphate Pathway
A metabolic process that produces NADPH and ribose 5-phosphate for nucleotide synthesis.
non-competitive inhibitor
A substance that reduces the activity of an enzyme by binding to a location remote from the active site, changing its conformation so that it no longer binds to the substrate bind the enzyme and enzyme substrate complex with the same affinity
non-competitive inhibitor
A substance that reduces the activity of an enzyme by binding to a location remote from the active site, changing its conformation so that it no longer binds to the substrate.
polymerase chain reaction (PCR)
A technique for amplifying DNA in vitro by incubating with special primers, DNA polymerase molecules, and nucleotides. You need polymerase, dNTPS, primers, target DNA The steps are denaturation annealing amplifying 2^n for number of cycles
ligand-gated ion channel
A transmembrane protein containing a pore that opens or closes as it changes shape in response to a signaling molecule (ligand), allowing or blocking the flow of specific ions; also called an ionotropic receptor. an influx of Na across the motor end plate resulting in depolarization of muscle fiber membrane in respond to acetylcholine
steroid
A type of lipid characterized by a carbon skeleton consisting of four rings with various functional groups attached.
Which organelles have phospholipid bilayer?
All membranous eukaryotic cell organelles have the common feature of a phospholipid bilayer, although the proteins differ in each case
archangel
An angel of high rank.
Warfarin structure
Anticoagulant non polar
Ionizable Amino Acids
Arg, Lys and His: -NH3+ Asp and Glu: -COOH Tyr: -OH Cys: -SH DR CHEKY
Best primers for PCR
Best primers for PCR have a high GC content and CG bases in 5′ and 3′.
To be an effective therapy, an antisense gene that is incorporated into a genome that contains the target gene must be:
If you want the antisense RNA to have a therapeutic effect, it would need to be expressed whenever the target mRNA that it's meant to block is to be expressed. Thus the target gene and the antisense would have to be regulated in a similar manner. I.e same promoter Just because a gene is physically near another doesn't mean if one is expressed the other will be. Gene expression is based on a handful of regulatory processes.
unmortared
Mortar is a workable paste which hardens to bind building blocks such as stones, bricks, and concrete masonry units, to fill and seal the irregular gaps between them, spread the weight of them evenly, and sometimes to add decorative colors or patterns to masonry walls Not secured by mortar. The unmortared wall of bricks would be easy to dismantle, merely requiring backs strong enough to lift the bricks and enough time to do it.
Large amounts of protein are found in the urine of a patient. Based on this information, which portion of the nephron is most likely malfunctioning? A. Collecting duct B. Distal tubule C. Glomerulus D. Loop of Henle
Large amounts of protein are found in the urine of a patient. Based on this information, which portion of the nephron is most likely malfunctioning? A. Collecting duct Jack Westin Advanced Solution: The collecting duct is responsible for concentrating the filtrate and future urine via aquaporin-mediated water reabsorption and secretion. No proteins are reabsorbed at this location. B. Distal tubule Jack Westin Advanced Solution: The distal convoluted tubule is primarily responsible for additional salt (NaCl), bicarbonate and water reabsorption, and proton, potassium and ammonia secretion; no proteins here. C. Glomerulus Jack Westin Advanced Solution: The glomerulus normally prevents large molecules, including proteins, from making it into the filtrate. If the glomerulus is compromised or damaged, the proteins from the blood can make their way into the filtrate and produce proteinuria or protein in the urine. This is the correct answer. D. Loop of Henle Jack Westin Advanced Solution: The loop of Henle and its countercurrent multiplier allow for the reabsorption of water in the descending limb and the reabsorption of salt in the ascending limb. Proteins are neither secreted nor reabsorbed in the loop of Henle. The glomerulus is the only answer choice that regulates the movement of proteins in the nephron.
sacked
Looted or plundered a captured city or town; fired
Neutrophils
Most abundant white blood cell., The most abundant type of white blood cell. Phagocytic and tend to self-destruct as they destroy foreign invaders, limiting their life span to a few days.
X-linked recessive inheritance
No male to male transmission (mutation is only on the X chromosome). Son of heterozygous mothers have a 1/2 chance of being affected. unaffected parents affected children only mother is causing
Overexpression of which enzyme is likely to result in increased levels of HIF? Jack Westin Advanced Solution: A. Succinyl decarboxylase B. Succinyl-CoA synthetase C. Succinate dehydrogenase D. Succinate carboxylase
Overexpression of which enzyme is likely to result in increased levels of HIF? Jack Westin Advanced Solution: 39. According to the passage, succinate competitively inhibits HIF hydroxylase which normally targets HIF for degradation. An enzyme that increases the levels of succinate will inhibit the degradation of HIF by HIF hydroxylase and result in increased levels of HIF. A. Succinyl decarboxylase Jack Westin Advanced Solution: Decarboxylating succinyl-CoA, the precursor to succinate, will produce an entirely different molecule with only three carbons; succinate has four carbons. B. Succinyl-CoA synthetase Jack Westin Advanced Solution: This is the correct answer. Succinyl-CoA synthetase is responsible for the synthesis of succinate from succinyl-CoA. Overexpression of succinyl-CoA synthetase will increase succinate levels, increase the inhibition of HIF hydroxylase thereby decreasing the degradation of HIF and ultimately result in increased levels of HIF. C. Succinate dehydrogenase Jack Westin Advanced Solution: Succinate dehydrogenase catalyzes the conversion of succinate to produce fumarate. Test day tip: the first term in the name of an enzyme is often its substrate. This is not a hard and fast rule but can be useful nonetheless. The overexpression of an enzyme that uses succinate as its substrate would decrease the levels of succinate as it is converted to another compound ultimately decreasing HIF levels. D. Succinate carboxylase Jack Westin Advanced Solution: Like the enzyme above, succinate carboxylase would catalyze a reaction in which succinate is a reactant, or substrate. If an enzyme that uses succinate as its substrate were overexpressed, succinate levels would go down and HIF degradation would rise. Only succinyl-CoA synthetase would cause an increase in HIF levels.
monastic
Relating to or resembling a monastery (where monks or nuns live), esp. by being quiet, secluded, contemplative, strict, and/or lacking luxuries
DNA polymerase I
Removes RNA nucleotides of primer from 5' end and replaces them with DNA nucleotides exonuclease 5 prime to 3 prime
secondary immune response
The adaptive immune response provoked by a second exposure to an antigen. It differs from the primary response by starting sooner and building more quickly.
distal convulted tubule and collecting duct
The distal convoluted tubule and collecting ducts are then largely responsible for reabsorbing water as required to produce urine at a concentration that maintains body fluid homeostasis.
The terminal electron acceptor in the metabolic pathway responsible for the chemical changes observed when Culture A was electrically stimulated is:
The muscle cells in Culture A use lactic acid fermentation to provide the energy for the contractions that result from electrical stimulation. In this process, NADH reduces pyruvate to produce lactate. Therefore, pyruvate serves as the electron acceptor in the production of lactate.
peripheral proteins
The proteins of a membrane that are not embedded in the lipid bilayer; they are appendages loosely bound to the surface of the membrane. Peripheral membrane proteins are found on the outside and inside surfaces of membranes, attached either to integral proteins or to phospholipids. Unlike integral membrane proteins, peripheral membrane proteins do not stick into the hydrophobic core of the membrane, and they tend to be more loosely attached.
Convents
The residences of religious women who are bound together by vows to a religious life like nuns
Which amino acid can form disulfide bonds (covalent bonds)?
Think of the structure of methionine and cystine. Cystine has S-H Methionine S-CH3. I like to think because of the Ch3 it would not be able to participate in disulfide bonds
transcription factors
Transcription factors are proteins that help turn specific genes "on" or "off" by binding to nearby DNA. Transcription factors that are activators boost a gene's transcription. Repressors decrease transcription. Groups of transcription factor binding sites called enhancers and silencerscan turn a gene on/off in specific parts of the body. Transcription factors allow cells to perform logic operations and combine different sources of information to "decide" whether to express a gene.
Glucosamine
a substance produced naturally in the body; often used as a supplement to maintain cartilage in the joints
prions infect
animals- slow acting, cause brain disorder (mad cow disease)
dwindeled
diminish gradually in size, amount, or strength
Kallman syndrome
hypothalamic --> NO GnRH delayed puberty, microphallus, cryptorchidism...secondary amenorrhea in females can test pit fxn with METYRAPONE..inhibits cortisol and should see a rise an ACTH
The Cori Cycle converts:
lactate into pyruvate
How does methylation work?
protein synthesis is blocked and the gene is prevented from being expressed (methylation groups are like switches) DNA methylation works by adding a chemical group to DNA. Typically, this group is added to specific places on the DNA, where it blocks the proteins that attach to DNA to "read" the gene. This chemical group can be removed through a process called demethylation.
Clathrin
protein that coats the inward-facing surface of the plasma membrane and assists in the formation of specialized structures, like coated pits, for phagocytosis form vesicles for intracellular transporting
inhibitor constant (Ki)
reduces concentration of ES complex and overall reaction velocity Ki measures the affinity of the inhibitor for the enzyme. The question says that when there's no substrate, Ki is 54 uM. In other words, the equilibrium constant for the reaction EI <--> E + I is 54 uM When you add enough substrate that all enzyme is in the form ES, and THEN add inhibitor (to form ESI), you get a different binding constant: 76 uM. That means the equilibrium constant for the reaction ESI <--> ES + I is 76 uM Ki works like any dissociation constant: larger number means lower affinity. Therefore the inhibitor has a lower affinity for ES than for just E alone. If you remember from inhibitors, a mixed inhibitor binds to both ES and E, but usually with unequal affinity. If the inhibitor prefers E, then it is more like a competitive inhibitor than an uncompetitive inhibitor (that's what it's a "mix" of- competitive and uncompetitive), and mixed inhibitors increase the Km. So this must be a mixed inhibitor that increases the Km, since it prefers E over ES. If Ki had not changed when substrate was added, it would be a noncompetitive inhibitor, binding E and ES with equal affinity. If Ki had been lower when substrate was present, it would have been more like an uncompetitive inhibitor than a competitive one, and Km would have decreased (because uncompetitive inhibitors decrease Km). By the way, B and D are actually two ways of saying the same thing, since a noncompetitive inhibitor is just a special case of a mixed inhibitor in which Km does not change.
Isomerase
responsible for rearranging of the structure of molecules.
Imidazole
side chain of histidine aromatic ring with two nitrogens pka of side chain is 6 - so at a physiological pH the side chain is neutral amide then will be protonated at a higher pH
The primary muscle type of the uterus (B), arteries and veins (C), and the small intestine (D) is which muscle
smooth
T cells (T lymphocytes)
specialized white blood cells that receive markers in the thymus, are responsible for cellular immunity and assist with humoral immunity Unlike T-cells and macrophages, B-cells don't kill viruses themselves. In the Viral Attack story, the B-cell sweeps up the leftover viruses after the T-cell attack. Actually, B-cells are as important as T-cells and are much more than just a final clean-up crew. They make important molecules called antibodies.
Layers of the epidermis (superficial to deep)
stratum corneum, stratum lucidum, stratum granulosum, stratum spinosum, stratum basale
connective tissue functions
support and protect (bone/cartilage) transport materials (blood) energy storage (adipose) defense (lymph)
Where in the human male reproductive system do the gametes become motile and capable of fertilization?
testis epididimys urethra prostate glands epidimys
Exonuclease
"Exo" outside "nuclease" enzyme that cleaves nucleotides.. is an enzyme that breaks apart the SINGLE nucleotides in the dna backbone starting at the ENDS (3' or 5').
quaint
(adj.) odd or old-fashioned in a pleasing way; clever, ingenious; skillfully made
sense and antisense strands
-antisense strands are the DNA strand that ARE transcribed (complementary to eventual RNA sequence) -Sense strand is the strand that is NOT transcribed (identical to the RNA sequence - except the DNA strand has T and not U) I remember it as 5-3=2 (which is a positive + strand) "PLUS, CODING just makes SENSE" [+, coding, sense, anti-template] 3-5= -2 (negative, so - strand) And opposite of the previous ones [-, non-coding, anti-sense, template] n genetics, a sense strand, or coding strand, is the segment of double-stranded DNA running from 5' to 3' that is complementary to the antisense strand of DNA, which runs from 3' to 5'. The sense strand is the strand of DNA that has the same sequence as the mRNA, which takes the antisense strand as its template during transcription, and eventually undergoes (typically, not always) translation into a protein.
Metal affinity chromatography
-protein is artificially produced by inserting its gene into cells, gene can be modified to include extra His residues (his-tag) -His tag proteins bind tightly to Ni2+ chelating resin -Hi-tagged protein is eluted by adding imidazole to the buffer, which out competes His-tag and protein no longer binds to column
ion exchange chromatography
-stationary phase is made of either negatively or positively charged beads (attract & bind compounds that have opposite charge) -salt is added to elute proteins stuck to column
Huckel's rule for aromaticity
4n+2 is the number of pi electrons so when I put different values of n from 1 I see a pattern if number of electrons are not according to that pattern then it is not aromatic
sulfhydryl group
A functional group consisting of a sulfur atom bonded to a hydrogen atom (—SH). A sulfhydryl group (also called "thiol group") consists of a sulfur atom with two lone pairs, bonded to hydrogen. The sulfhydryl group is ubiquitous in our body and mostly found in the oxidized form as disulfide linkages. The disulfide linkages contribute to the tertiary and quaternary structures of proteins.
Microtubules
A hollow rod composed of tubulin proteins that makes up part of the cytoskeleton in all eukaryotic cells and is found in cilia and flagella.
feedback inhibition (negative feedback)
A metabolic pathway is switched off by the inhibitory binding of its end product to an enzyme that acts early in the pathway.
endomembrane system
A network of membranes inside and around a eukaryotic cell, related either through direct physical contact or by the transfer of membranous vesicles. The endomembrane system (endo- = "within") is a group of membranes and organelles in eukaryotic cells that works together to modify, package, and transport lipids and proteins. It includes a variety of organelles, such as the nuclear envelope and lysosomes, which you may already know, and the endoplasmic reticulum and Golgi apparatus, which we will cover shortly.
motor proteins
A protein that interacts with cytoskeletal elements and other cell components, producing movement of the whole cell or parts of the cell. Motor proteins are the driving force behind muscle contraction and are responsible for the active transport of most proteins and vesicles in the cytoplasm. They are a class of molecular motors that are able to move along the surface of a suitable substrate, powered by the hydrolysis of ATP. There are three superfamilies of cytoskeletal motor proteins. Myosin motors act upon actin filaments to generate cell surface contractions and other morphological changes, as well as vesicle motility, cytoplasmic streaming, and muscle cell contraction. The kinesin and dynein microtubule-based motor superfamilies move vesicles and organelles within cells, cause the beating of flagella and cilia, and act within the mitotic and meiotic spindles to segregate replicated chromosomes uses ATP so active transport Kinesin kicks out dyenin drags in
natural killer cells
A type of white blood cell that can kill tumor cells and virus-infected cells; an important component of innate immunity.
Which statement is LEAST supported by the experimental data? A. A high cgr ratio correlates with increased excretion of reduced digoxin. B. The cgr operon is functionally more active in the FAA 1-3-56 strain than in the DSM2243 strain of E. lenta. C. A low cgr ratio correlates with reduced digoxin inactivation. D. The abundance of the cgr operon in gut flora may predict the bioavailability of digoxin in individual patients.
A. A high cgr ratio correlates with increased excretion of reduced digoxin. Jack Westin Advanced Solution: This is supported by the experimental data. Group 1 in Table 1 corresponds to the DSM2243 strain and group 2 corresponds to the FAA 1-3-56 strain. The DSM2243 strain has a higher cgr ratio. In Fig. 1, the DSM2243 strain produces greater reduction of digoxin compared to FAA 1-3-56 strain. Therefore the higher cgr ratio is correlated with greater digoxin reduction and excretion B. The cgr operon is functionally more active in the FAA 1-3-56 strain than in the DSM2243 strain of E. lenta. Jack Westin Advanced Solution: Per Fig. 1, the DSM2243 strain has a higher percentage of reduction than the FAA 1-3-56 strain, the opposite of this answer choice. This must be the correct answer to this LEAST supported question. C. A low cgr ratio correlates with reduced digoxin inactivation. Jack Westin Advanced Solution: As shown above, the FAA 1-3-56 strain has a lower cgr ratio and the same strain has lower reduction (and lower digoxin inactivation) than the DSM2243 strain. This answer choice is supported by the experimental data. D. The abundance of the cgr operon in gut flora may predict the bioavailability of digoxin in individual patients. Jack Westin Advanced Solution: The abundance of the cgr operon (cgr ratio) is correlated with the reduction and inactivation of digoxin (% reduction in Fig. 2). This means that the abundance of the cgr operon could predict the bioavailability (approximated by the percentage that remains active) of digoxin. This is supported by the experimental data; answer choice B is the correct answ hey, i do not know this you still need this answer but for anyone else that might Table 1 shows that the more cgr units you have, the more Digoxin reduction that will be observed. From the passage, we know that Digoxin reduction inactivates it and transforms it to another compound, notably dihydrodigoxin. so what option D is saying essentially is that, the amount of cgr units one has may dictate how much available Digoxin units there are available in the body.. which is true cuz the more cgr, the less Digoxin is available since they are being reduced and changed to another compound.
According to the passage, relative to slow-twitch fibers, fast-twitch fibers are likely to exhibit which property? A. Greater Ca2+-pumping capacity B. Increased capillary density C. More mitochondria D. Higher levels of oxygen-binding proteins
A. Greater Ca2+-pumping capacity Jack Westin Advanced Solution: A. Greater Ca2+-pumping capacity would improve contractile strength which corresponds to fast-twitch fibers. Remember that the movement and subsequent binding of calcium to troponin is what allows for the binding of actin and the myosin heads, ultimately producing contraction. Also, it does not directly correlate with increased oxidative capacity which would be a marker for slow-twitch fibers. This is a great answer choice. B. Increased capillary density Jack Westin Advanced Solution: B. Increased capillary density would promote blood flow and oxygen delivery to the slow-twitch muscle fibers, thereby improving oxidative capacity. The question stem asks for a property of fast-twitch fibers so this is incorrect. C. More mitochondria Jack Westin Advanced Solution: C. More mitochondria correspond to greater oxidative capacity for slow-twitch muscle fibers. D. Higher levels of oxygen-binding proteins Jack Westin Advanced Solution: D. Like answer choices B and C, this answer choice would improve oxidative capacity for ATP production in slow-twitch fibers. Only answer choice A would be a property of fast-twitch muscle fibers.
Which type of catalytic activity is most likely missing from cFLIP? A. Oxidoreductase activity B. Lyase activity C. Isomerase activity D. Hydrolase activity says it has protease activity
A. Oxidoreductase activity Jack Westin Advanced Solution: A. Unlike caspase-8, cFLIP is catalytically inactive, so it is missing the catalytic activity that caspase-8 possesses. According to the first paragraph, caspase-8 is a protease. Proteases cleave peptide bonds and use water to do so, meaning they possess hydrolase activity. Oxidoreductases change the oxidation state of substrates via reduction and oxidation reactions. B. Lyase activity Jack Westin Advanced Solution: B. Lyases cleave bonds via mechanisms other than water. Peptide bonds are cleaved using water so while tempting, this answer choice is not the best answer. C. Isomerase activity Jack Westin Advanced Solution: C. Isomerases catalyze isomerization reactions and alter the chemical structure of its substrates while maintaining the same chemical composition. D. Hydrolase activity Jack Westin Advanced Solution: D. Hydrolases cleave bonds using water, like proteases cleave peptide bonds using water. This is the correct answer.
Ach in muscles
Acetylcholine is released at the neuromuscular junction where it binds to receptors on the muscle cells and, depending on the type of muscle cell, causes depolarization or hyperpolarization of the cell membrane. In skeletal muscles, acetylcholine binds to its receptors, which leads to depolarization of the muscle cell membrane and muscle contraction. Ach has both inhibitory and excititory properties via the muscarinic and nicotinic recpeters, respectively. ACh does so many things! It's a neurotransmitter that acts on the presynaptic ganglions for both the peripheral nervous system (PNS) and central nervous system (CNS) and post synaptic ganglions for the PNS to activate muscles and support cognitive functions. Here's a quick diagram to show you what I mean: https://media1.britannica.com/eb-media/48/8248-004-9927D90E.jpg In the PNS ACh is released at the neuromuscular junction to activate muscles (It binds to nicotinic receptors on the end plate and causes Calcium ions to rush into the cell, which induces a muscle contraction). It also plays a major role in the autonomic nervous system - it induces GI movements for the "Rest and Digest" reflex of the parasympathetic nervous system. If you've heard of "Curare" it inhibits these nicotinic ACh receptors to induce paralysis. Here's a diagram: https://image.slidesharecdn.com/neuromuscularjunction-121105173344-phpapp02/95/neuromuscular-junction-12-638.jpg?cb=1352139237 In the CNS, the nicotinic receptors/ ACh pathway between the cerebral cortex and hippocampus supports memory functions. Acetylcholinesterase inhibitors (like Memantine) inhibit the degradation of ACh in the synaptic cleft of these neurons and help treat Alzheimers disease.
Which cells harvested from adult mice were most likely used as the highly proliferative benchmark in the experiment that generated the data shown in Figure 3?
Adipocytes- tend to get fatter/they're kinda like transitional cells in that they expand to accommodate more stuff. Probably average proliferation rates. Cardiac muscle cells - muscle cells don't divide. These cells fuse together. Getting swole = bigger cells/more blood, not more cells. These are highly specialized cells, and many highly specialized cells lose replicative abilities (as well as many other 'normal function.' Nerve cells - like muscle cells, highly specialized and don't divide. Another example: oocytes. Highly specialized, doesn't proliferate. Intestinal epithelial cells: these cells are constantly exposed to pathogens and rough environments. It is likely that they are constantly being degraded and replaced, given their strenuous role. They likely must rapidly proliferate to replace infected cells or used up cells or cells that have lysed in the moving/kinetic environment they work in. 20 ReplyShareReportSave
No drowsiness was initially felt by the alcoholic because the previous abuse of alcohol had: A. denatured the cP-450. B. inhibited the cP-450. C. reduced the cP-450. D. induced the cP-450. Solution
Ah this is mentioned in the passage as increasing drug metabolism but doesn't go into much detail other than that, however it does talk about breaking down barbituates. cp450 is part of a family of CYPs is commonly known in the pharma world as what ****s your small molecule drugs up. It basically takes anything hydrophobic and makes it hydrophilic to be excreted via your kidneys by sticking hydroxyl groups on everything. This is what the passage meant if you really want to go into detail on cp450 (CYPs). so if you have increased drug metabolism from cp450, you excrete the drug faster. You excrete the drug faster by filtering it out of your blood, the less you feel the effects of it. By ingesting alcohol in the past (it's like you purposefully poisoned yourself) you're inducing the liver to make more cp450. So the basal [cp450] is increased. Solution: The correct answer is D.This is a Biology question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is D, because the passage states that cP-450 is inducible and thus previous abuse of alcohol, a cellular toxin, results in an increased concentration of cP-450. This leads to rapid barbiturate metabolism by cP-450 and the absence of initial drowsiness. This is a Scientific Reasoning and Problem Solving question because it requires you to propose an explanation for the absence of the initial drowsiness.
secondary structure of protein
Alpha-helix and beta-pleated sheets, result of hydrogen bonding between primary structures Secondary structure is formally defined by the pattern of hydrogen bonds between the amino hydrogen and carboxyl oxygen atoms in the peptide backbone.
Enzyme assay
An enzyme assay is the name given to any laboratory technique that measures enzyme activity within a sample. Enzyme assays can be used for a variety of purposes, which include identifying the presence of an enzyme, investigation of specific enzyme kinetics or the activity of inhibition within a sample.
Epitope and antibody
An epitope binds one specific antibody. Therefore, a molecule with several epitopes will bind several distinct antibody molecules. The presence of several antibodies on the surface of an antigen is expected to elicit a stronger immune response than if fewer antibodies were present. The table shows that molecules with higher molecular weights are associated with higher numbers of epitopes, which will bind a greater number of antibodies. These molecules are therefore expected to elicit a more robust immune response.
passive immunity vs active immunity
An individual does not produce his or her own antibodies, but rather receives them directly from another source, such as mother to infant through breast milk
AITA is a single-stranded RNA virus. For genomic replication to occur, the virus must package its own RNA polymerase into the virion because eukaryotic host cells: lack an enzyme that can synthesize RNA from RNA. lack an enzyme that can synthesize DNA from RNA. lack an enzyme that can synthesize RNA from DNA. utilize RNA that is structurally different from viral RNA.
Answer A is correct. Eukaryotic cells do not generate RNA from RNA, and therefore eukaryotic cells do not have enzymes that perform this function. To make mRNA for either protein synthesis or new genomic material, RNA viruses such as the influenza virus must possess an RNA-dependent RNA polymerase, as suggested by answer choice A. Eukaryotes also lack an enzyme that can synthesize DNA from an RNA molecule, which could make Answer B tempting. However, the influenza virus never uses DNA in its replication cycle (as noted in Paragraph 1), eliminating Answer B. Answer C is incorrect because eukaryotes do possess enzymes that synthesize RNA from DNA. Answer choice D is also incorrect because nucleic acids from eukaryotes and viruses are believed to be chemically identical.
The Compound 4 precipitate is pale yellow, while Compound 3 is white. This color difference occurs because Compound 4: contains delocalized π electrons whereas Compound 3 does not. reflects light because it is more planar than Compound 3. has a more extensive conjugated system than does Compound 3. is not capable of hydrogen bonding, as is Compound 3.
Answer C is correct. Answer A is incorrect because both compounds contain aromatic rings with delocalized electrons. The planarity of a molecule by itself does not give rise to color; for example, benzene is colorless. This eliminates Answer B. The ability to hydrogen bond does not contribute to absorption in the visible region; for example, water is colorless. This makes Answer D incorrect. Compound 4 is a conjugated enone with an extensive delocalized network of electrons stretching from the carbonyl through an interconnecting alkene and into the ring system. The large number of conjugated double bonds allows Compound 4 to absorb light in the visible region. Answer C is therefore correct.
Ammonia is an example of a Bronsted-Lowry base. Through which process does ammonia raise the pH of biological solutions? Ammonia is a strong base, so small concentrations cause large increases in pH. Ammonia releases NH2-, which is analogous in size to the hydroxide ion. By abstracting protons from water, ammonia causes an increase in [OH-]. By abstracting protons from water, ammonia causes an increase in [H+].
Answer C is correct. Bronsted Lowry bases by definition accept protons. In aqueous solutions, they can accept a proton from water leaving behind hydroxide. If hydroxide increases, pH goes up. Choice A is incorrect, this describes strong bases such as KOH and NaOH. Choice B can be ruled out because the hydrogens of ammonia are not readily dissociable; therefore, this does not occur. Choice D states that by removing H+, the [H+] increases, which is impossible.
Which finding from a subsequent study would most support the passage conclusion that chemokines are necessary for myeloid cell migration? Compared to WT mice injected with histamine, MCP-1-/- mice not injected with histamine recruit fewer myeloid cells to the blood stream. Compared to WT mice injected with histamine, MCP-1+/+ mice injected with histamine recruit more myeloid cells to the injection site. WT mice injected with histamine recruit myeloid cells to the injection site, but MCP-1-/- mice injected with histamine do not. WT mice injected with histamine recruit myeloid cells to the injection site, but MCP-1+/+ mice injected with saline do not.
Answer C is correct. The passage tells us that MCP-1 is a chemokine. Although not entirely necessary to answer this question, it may help to know that histamine induces inflammation. Choice C is the correct and strongest answer because it shows that the chemokine (MCP-1) is necessary for the myeloid cells to be recruited to the inflammation site. Answer A should be rejected because two variables are being altered, the presence or absence of MCP-1 (WT vs. MCP-1-/-), and a histamine injection vs. no histamine injection. Basic scientific protocol requires that only one treatment be varied at a time. Otherwise, it cannot be determined which of the multiple differences caused the observed result. Answer B should be rejected because a WT mouse is expected to have the genotype MCP-1+/+, suggesting that the two treatment groups being compared in this answer choice are identical with respect to chemokine expression. Answer D is incorrect because it isolates the presence or absence (saline) of the histamine injection, not the presence or absence of the chemokine.
If all three of the molecules shown are mixed together in a single solution, which species will participate in hydrogen bonding? I. CH4 II. CH2O III. CH3OH I only III only II and III only I, II, and III
Answer C is the correct answer. Hydrogen bonds are formed between electronegative atoms such as oxygen, and a hydrogen atom that is covalently bound to an electronegative atom (F, O, or N). Methanol (III) would be the only molecule capable of forming hydrogen bonds with itself, but because they are all mixed together, the carbonyl oxygen in formaldehyde can act as a hydrogen bond acceptor (but not a hydrogen bond donor). This makes options II and III both correct and Answer C the best choice. Methane would not be miscible with formaldehyde and methanol and is not capable of hydrogen bonding under any circumstance because it lacks the necessary electronegative atom.
NADH exhibits a characteristic UV absorbance of 340 nm. This absorbance is the result of photons causing the: vibration of NADH sigma (σ) bonds. cleavage of NADH sigma (σ) bonds. cleavage of NADH pi (π) bonds. excitation of NADH pi (π) electrons.
Answer D is correct. NADH absorbs electromagnetic radiation in the UV region as the result of the excitation of π electrons. Answer A is incorrect because the vibration of bonds occurs in the infrared (IR) region; this accounts for IR spectroscopy, but not for UV absorbance. Answers B and C are both incorrect because the cleavage of bonds is not associated with UV absorption. Therefore, Answer D is the best choice.
Which of the following is the MOST likely explanation for the greater severity of Duchenne muscular dystrophy compared to Becker muscular dystrophy?Duchenne MD involves a frame-shift mutation, while Becker MD does not. Becker MD involves a frame-shift mutation, while Duchenne MD does not. Duchenne MD causes the deletion of bases in multiples of three, while Becker MD does not. Duchenne MD is a somatic mutation, while Becker MD is a germ-line mutation.
Answer choice A is correct. Frame-shift mutations are the most deleterious type of mutation to the final gene product, as every amino acid downstream of a frame-shift mutation is likely to be incorrect and, in addition, premature STOP codons will often truncate the final protein. Thus, of the choices listed, Answer A is the mostly likely explanation for the greater severity of Duchenne MD. Answer B is incorrect because it suggests that the more serious mutation causes the less serious disease condition. Answer C is incorrect because deletions of exactly three bases (or multiples of 3) do not affect the reading frame. One or more amino acids may be absent in the gene product, but the other amino acids will remain unchanged. Answer C suggests that the non-shifted reading frame causes Duchenne MD, the more severe form, while the shifted reading frame causes Becker MD, the less severe form. This logic is exactly backwards. Answer D is incorrect because if Duchenne MD was a somatic mutation it would not be heritable.
Passive immunity could result from: I. receiving a vaccine. II. contracting a viral or bacterial infection. III. receiving an injection of antibodies. III only I and II only II and III only I, II, and III
Answer choice A is correct. Passive immunity is immunity generated when one receives antibodies from some outside source. The immune system of the person receiving the passive immunity is never directly activated. This can occur when a person is injected with antibodies developed against an antigen in another person or host, as in Statement III. A more common example is the antibodies passed from mother to child in mother's milk. The mother's immune system produces the antibodies and passes them to the child, but the child's immune system is never activated against the antigens to which those antibodies bind. Statements I and II are both false because either a vaccine or an infection activate the host's immune system and it generates its own antibodies. Because Statement II is false, Answers B, C and D must be rejected.
Researchers used the following assay to investigate whether the TRAF6 protein is required for the dissociation of IκB from NF-κB. The impact of TRAF6 on two kinases of the IκB-NF-κB dissociation pathway was measured. β-actin was included in the assay in order to: ensure the same quantity of protein was loaded for all samples. ensure all samples were transfected with the same amount of virus. determine if the virus also causes degradation of β-actin. determine if the virus also causes overexpression of β-actin.
Answer choice A is correct. The assay shown is a Western blot, and β-actin was used as a loading control. In a Western blot, researchers are usually trying to measure the expression of a protein, or the amount of a protein product. Students should recall that the proteins in the original tissue sample must first be separated by electrophoresis. If, during this step, an unequal amount of sample is loaded into each lane, the resulting Western blot could show a smaller signal in a given lane, NOT because there was less of the protein being studied, but simply because less sample was loaded. For this reason, a structural protein such as β-actin (that is not affected by any of the treatments) is used as a control. This allows confirmation on the final blot that all lanes contained equal samples. Loading controls also allow researchers to verify that proteins were transferred equally from the gel to the blotting membrane. This makes Answer A correct. If the researchers wanted to ensure the same levels of HTLV-1 were in all samples, they would need to add an HTLV-1 antibody, eliminating Answer B. Based on passage information, and the fact that β-actin is never mentioned, it is highly unlikely that β-actin is involved in this process or being investigated by the researchers. Nothing in the passage suggests the virus affects actin molecules. Even if it did, to examine the impact of the virus on β-actin (degradation or overexpression) the researchers would need to compare b-actin in HTLV-1 cells to β-actin in a control cell line such as Jurkat. This is not seen in the Western blot. Rather, TRAF6 is being compared to a control. Therefore, Answers C and D are both incorrect.
A researcher who needs to separate proteins according to size should use which method? SDS-PAGE PAGE only Isoelectric focusing ELISA
Answer choice A is correct. The treatment of proteins with SDS, a detergent, equalizes the charge to mass ratio, removing the factor of charge from their electrophoretic mobility. Under that condition, they separate according to size. Choice B is PAGE alone. Under these circumstances, the proteins' electrophoretic mobility is partly determined by charge. Isoelectric focusing, Choice C, separates proteins based on their isoelectric point, so this choice should be eliminated. Choice D can be eliminated as the ELISA method relies on immunological techniques to identify the presence of a particular protein. All four laboratory techniques are commonly addressed in biochemistry and to a lesser degree in biology and organic chemistry. Ironically, when originally released, the Altius MCAT-2015 practice materials were criticized by many students for covering these techniques, SDS-PAGE and ELISA being absent from every other MCAT prep resource. We stood firm and told our students that they must know these topics. Many months later, when the AAMC released their first scored full-length practice exam and the AAMC Section Banks, our foresight was proven to be accurate. These topics are tested multiple times on recently-released AAMC materials. In the AAMC Section Bank alone you can find them on Passage 7, as well as stand-alones #52, #69, and #70. This is only one of dozens of cases in which our experienced team of PhD authors PREDICTED an AAMC trend only to see that prediction realized on future materials. Rather than get defensive when you miss a challenging question, decide to trust us. Master the question types we present and you won't face unnecessary surprises or frustrations come test day.
Sweat pores in the skin release water and various other substances onto the surface of the skin. What physical property of water allows sweating to reduce body temperature? Lower density in the solid phase High heat capacity Low heat capacity High surface tension
Answer choice B is correct. Water has a number of properties that make it uniquely suited to serve as "the solvent of life." Most of these are due to the strong intermolecular forces among water molecules and its polarity. The one that is most relevant to cooling the body is water's high heat capacity—when water molecules acquire enough energy to evaporate from the surface of the skin, they have absorbed a relatively large amount of heat and therefore, take away a large amount of heat. This is known as evaporative cooling. Answers A and D describe properties of water, but are not relevant to evaporative cooling. Answer C is incorrect as water has a high, not a low, heat capacity.
Ftt is an obligate intracellular bacterium. An antibody response can still be effective against Ftt because antibodies: enhance the rate of phagocytosis of Ftt. bind to Ftt plasma B-cells, stimulating antibody production. bind to Ftt and prevent it from entering the cell. cross the Ftt cell membrane and prevent translation of Ftt proteins.
Answer choice C is correct. The primary function of antibodies is to activate complement, enhance phagocytosis, and neutralize infectious agents by preventing binding to the receptor on the target cell. Although answer choice A is correct in that antibodies can enhance the phagocytosis of bacteria into the cell, this would not be effective given that Ftt are already obligate intracellular bacteria. Ftt is a prokaryotic cell, therefore it does not have its own immune system, or plasma B-cells that produce antibodies—making choice B incorrect. The best answer is that the antibodies will neutralize the infection by preventing F. tularensis from interacting with its cellular receptor and thereby preventing its entry into the host cell. Answer choice D is incorrect because antibodies are not widely known to cross the cell membrane and have no known function in preventing translation. This is even less likely for a bacterium with a cell wall.
The researchers selected naïve mice because they were: too young to have fully functional immune systems. a different species that does not have macrophages. not injected intraperitoneally with pristane. injected intraperitoneally with pristane.
Answer choice C is correct; Naïve may be an unfamiliar term to some examinees, but its meaning here can be inferred from the figure, since the treatment groups were "Naïve" and "Pristane," suggesting the naïve group did not receive pristane. Following the rules of experimental science, the comparison to the treatment should be identical in every variable except the treatment. Thus, choice C is the correct answer, since the only difference should be in the presence or absence of pristane. For this reason, Answer choice A should be rejected as mice of the same age should be used. Answer B can be eliminated for the same reason: all mice should also be of the same species. Answer choice D is false as the experimental mice were all injected with pristane, but the naïve control mice were not.
What is the most probable impact of a point mutation that substitutes the terminal guanine of a 5' exon with an alternate nucleotide? The mRNA will be prematurely terminated. The protein will be nonfunctional due to an amino acid change. The intron will not be spliced from the mRNA transcript. There will be no effect, because this is a silent mutation.
Answer choice C is the correct answer. A guanine at the 5' position is required for RNA splicing. If a different nucleotide were present, the process would not function and the intron would not be removed. This makes choice C the best answer. This will not create a stop codon, making choice A incorrect. Answer B may be tempting because failure of the splicing mechanism may well result in a non-functional protein. However, given both answers, B and C, answer C is a more complete and accurate description that emphasizes the primary importance of guanine in the splicing mechanism. This is particularly important given that the importance of a guanine substitution was specifically referenced in the stem. Examinees should expect AAMC questions to feature distractors that may be true, or could be true, but are not as complete, or as correct, as the best answer. Answer D must be incorrect because, as already described, failure of exon splicing will have multiple deleterious effects.
During normoxia, HIF-1α is predominantly located in which part of the cell? Nucleus Mitochondria Cytoplasm Lysosome
Answer choice C is the correct answer. The passage states that HIF-1α is degraded by the proteasome under non-hypoxic conditions, which is a cytoplasmic process. It is possible that there could be a small amount of HIF-1α that makes its way into the nucleus or that is residual from a previous hypoxic condition but the predominant location of HIF-1α would be the cytoplasm. Answer A is incorrect because the nucleus is the destination to which HIF-1α is translocated during hypoxia, and the question stem asks about its location during normoxia. Answer B is incorrect because functional cellular proteins other than those directly involved in mitochondrial transport or metabolism would not be found inside of the mitochondria. Answer D is incorrect because HIF-1α is a protein and would be degraded inside of a lysosome.
A student hypothesizes that NhhA would make a good vaccine target. Is the student's hypothesis reasonable? No, because bacterial infections cannot be prevented using vaccines. No, because vaccines are ineffective against proteins that signal apoptosis. Yes, because NhhA is a soluble cytosolic protein. Yes, because NhhA is located on the surface of the bacteria.
Answer choice D is correct; good vaccine targets for bacteria and viruses should be membrane proteins, so that antibodies in the bloodstream can access the protein and form an immune response to them. Cytoplasmic proteins would not be accessible by the immune system, so vaccines designed using them would not trigger an immune response. Answer choice A is incorrect because vaccines can be developed against bacteria. Answer B is incorrect because even though it is true that NhhA triggers apoptosis, apoptosis is irrelevant to vaccine design. A vaccine against a pro-apoptotic protein can still be effective, assuming the protein has other qualities that make it a good vaccine candidate. Answer C is incorrect because NhhA is not a soluble protein. It is a membrane protein, meaning that it must have a significant hydrophobic region.
Which expression gives the probability that a mother and father will have four children, two girls and two boys, in any birth order? (½)4 4 (½) 2 (½)4 6 (½)4
Answer choice D is the correct answer. Each birth is an independent event. For each "birth event" the probability of having a boy is ½ and the probability of having a girl is also ½. This type of problem would be more simple had the question asked "What is the probability of having four children, all of them boys?" The probability of having four boys in a row is (1/2)4 or 1/16. However, the problem changes when the question states that the births can appear in any order. In this case, one must first predict all of the possible birth orders by which these parents could have two boys and two girls: GGBB, GBGB, GBBG, BBGG, BGBG, BGGB It may be worthwhile for the examinee to try for a moment to think of a seventh or eighth possible birth order—it will become apparent that these six permutations represent all of the possible orders for having two girls and two boys. Next, one must calculate the possibility of each birth order. Because the probabilities of having a boy or a girl are both ½, the probability of the first order is ½ x ½ x ½ x ½ = (½)4 = 1/16. Notice that the math will be identical for the other five birth orders. The final key to this question is to recognize that six birth orders are six different events, ANY of which could satisfy the probability requirement of two boys + two girls, in any order. When multiple events are possible, and any of those events satisfy the question, the probabilities of each event must be ADDED. To clarify, think of a simple example. If you draw a ball from a bowl every morning and the bowl contains 33% red balls, 33% blue balls, and 33% green balls, the chance of drawing a green ball is clearly 1/3. But what if the question is "What is the probability of drawing either a green OR a blue ball?" It is now apparent that 66% of the balls in the bowl satisfy this requirement, so your odds are 2/3. Notice that this is the same as adding the probability of drawing a green ball to the probability of drawing a blue ball: 1/3 + 1/3 = 2/3. In the same way, we must add the probabilities of these two parents having two girls and two boys in any of these six possible birth orders. We established previously that the probability of one birth order is 1/16. Therefore, the chance of any of them happening is: 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6(1/16). This is equivalent to Answer D: 6(1/2)4. This proof eliminates Answers A, B and C.
WT mice have which genotype? RORγt-/- RORγt+/- RORγt-/+ RORγt+/+
Answer choice D is the correct answer. Identifying the correct answer requires some understanding of the use of induced mutations and of the standard way in which researchers represent those mutations. With the AAMC now using excerpts from peer reviewed journals for almost every passage, and with the emphasis on "reasoning about the design and execution of research," an understanding of basic research terms and protocols will be expected. Researchers often utilize study subjects such as mice into which they have induced various mutations. The term "wild-type" refers to an organism that is unaltered and has the normal, unaffected genotype. Plus (+) and minus (-) signs are used to represent the presence or absence of a normal gene or gene product. WT organisms do not carry the mutation, so they are represented as +/+. A +/- symbol represents a heterozygote, which may still have some functional protein, but will experience reduced expression compared to WT. Following this system, a -/- symbol indicates an individual with two mutant genes and therefore no functional protein expression. Answer D is the correct answer because wild-type mice should lack all mutations and therefore are assigned the RORγt+/+ designation. Answers B and C should be rejected because they both have the same meaning—reversing the order of the plus and minus signs does not change anything. Answer A is a double mutant with no functional protein—the exact opposite of a wild-type mouse. NOTE: You will also see plus and minus signs used in many academic journal figures to represent the presence or absence of individual drugs, compounds, or other treatments during individual experimental trials. A plus sign indicates the compound was present and a minus sign indicates it was absent.
Which steps involved in the contraction of a skeletal muscle require binding and/or hydrolysis of ATP? I. Dissociation of myosin head from actin filament II. Attachment of myosin head to actin filament III. Conformational change that moves actin and myosin filaments relative to one another IV. Binding of troponin to actin filament V. Release of calcium from the sarcoplasmic reticulum VI. Reuptake of calcium into the sarcoplasm
Answer: I, III, and VI only It seems they're taking a broad-stroke approach to this problem. While I agree with you that the actual power-stroke event occurs when ADP + Pi are released, this can only happen when ATP was bound to the myosin head and subsequently hydrolyzed. So in all intents and purpose, the power stroke is dependent on the attachment and hydrolysis of ATP, albeit somewhat indirectly.Now, that leads us to the task of internalizing this mode of thinking and looking back at the answer choices. In this general framework of choosing anything that could be a product or event, occurring immediately or not from the actual binding/hydrolysis of ATP, only II sees enticing. However! ATP binding allows the head to be cocked, but, as the explanation states, it's not ATP that allows it bind - instead, it's calcium and it's role in removing the troponin/tropomyosin complex.And lastly, how could you have gotten this right? A good lesson from this question and a motto I work by when attacking MCAT questions is to start very general. Don't restrict your way of thinking to one particular area without consulting the question and answer choices with an open mind. For anyone still looking this up 5 years later, I found this video to be super helpful for a visual because I got this wrong. From this video, it looks like the hydrolysis of ATP to ADP and PI re-cocks the myosin head into its high energy conformation such that once the binding site is exposed on actin the myosin head attaches. It is not the hydrolysis of ATP that CAUSES the attachment of the myosin head to actin. The hydrolysis of ATP to ADP and Pi ONLY re-cocks the myosin head so it CAN attach. What causes the attachment is the binding of calcium to troponin and the subsequent conformational change of tropomyosin, and the dissociation of only phosphate from the myosin head (ADP dissociates after Pi, during the power stroke). ie. Ca2+ is released from sarcoplasmic reticulum and binds to troponin Tropomyosin undergoes conformational change to expose myosin head binding Myosin head (that was bound to ADP and Pi) dissociates only phosphate and attaches to actin ADP dissociates, causing myosin and actin to move relative to one another (power stroke) ATP binds the myosin head moving it to a low energy conformation ATP is hydrolyzed to ADP and pi and myosin head re-cocks but does not yet attach to the actin filament Cycle repeats, when a new ca2+ binds and myosin releases phosphate only so head can attach Thus, hydrolysis' of ATP is involved in re-cocking, not the actual attachment.
Antergrade transport
Anterograde transport means things are moving out of the cell, and is accomplished by kinesin. This would follow Anterograde transport = ER -> COP II vesicle -> Golgi Retrograde transport = Golgi -> COP I vesicle -> ER
Which conclusion is most reasonable given the MO diagrams for benzene and cyclobutadiene? Electrons in antibonding MOs increase a molecule's energy. Nonbonding MOs are lower in energy than the atomic orbitals from which they form. In aromatic molecules, nonbonding MOs are lower in energy than bonding MOs. Partially filled antibonding MOs increase a molecule's stability.
Antibonding MOs are shown in figures 1 and 2 to be higher in energy than the bonding MOs. Electrons in these orbitals increase a molecule's energy. Answer A is therefore correct. The MO diagram for cyclobutadiene shows the nonbonding MOs at the same energy level as the atomic p orbitals. This is a characteristic of non-bonding orbitals; non-bonding MOs are orbitals that have essentially the same energy in the molecule as they do in the atomic system. Therefore, they don't contribute to the bond order. Answer B is thus incorrect. Bonding MOs are lower in energy than the atomic orbitals from which they form. Nonbonding orbitals have the same energy. These considerations make Answer C incorrect. Antibonding orbitals are the highest energy MOs. Electrons in these orbitals increase the energy of a molecule and decrease its stability, whether partially or completely filled, so Answer D is incorrect.
Archea
Archaea are a group of micro-organisms that are similar to, but evolutionarily distinct from bacteria. Many archaea have been found living in extreme environments, for example at high pressures, salt concentrations or temperatures. These types of organisms are called extremophiles.
Chloramphenicol did NOT inhibit translation in E. coli cells containing the cat6xbs expression plasmid. What experimental parameter could be changed in order to affect translation inhibition?
As stated in the passage, the cat transcript encodes chloramphenicol acetyltransferase. Without even being familiar with this protein, the reader should be able to tell by looking at the name of the protein that it is an enzyme (-ase) that causes chloramphenicol to undergo a chemical change (acetyl transfer). This renders chloramphenicol ineffective and confers antibiotic resistance to the cell regardless of the chloramphenicol concentration or incubation time. A moderate increase in the temperature should increase the translation efficiency. Thus, use of an alternate antibiotic is the only reasonable choice of those presented. Thus, D is the correct response.
As one step in the estimation of the efficiency of neuronal induction, scientists calculated the average number of induced cells present in 30 randomly selected 20× visual fields. Which change to this particular aspect of the experimental protocol would increase the accuracy of the estimates of efficiency? Increase the magnification of oculars used to define the field of view Increase the number of visual fields counted per petri dish Select visual fields from the central portion of the petri dish Use the presence of green fluorescence to identify cells appropriate for quantification
B Because they were taking a random sample of cells to determine how many were successfully induced (aka GFP positive). If they screened for GFP, then they would be selecting only cells that were induced, giving them a 100% ratio which is not accurate. It's all in the wording. I think because the gfp is only on in induced cells so you are only counting induced cells/field instead of induced cells/total cells per field Basically by having more visual fields you can more confidently say that you didn't just randomly choose fields that had a number of induced cells that wasn't representative of the whole sample. So by having more fields your assumption about the whole sample becomes more accurate. Increasing the magnification would probably make estimates less accurate.
Despite the effects of diabetes, the brains of diabetic patients still receive adequate nourishment. This is most likely because the brain uses: A. less glucose than do other body tissues. B. insulin-independent transporters for the uptake of glucose. C. fatty acids for energy instead of glucose. D. insulin-dependent transporters for the uptake of glucose. B
Based on the passage, diabetes affected individuals will use proteins and lipids as a source of glucose, thus increasing, rather than decreasing, their appetite. High levels of glucose are a symptom of diabetes. This is due to the fact that kidneys can reabsorb glucose up to 320 mg/min (tubular maximum). The levels of glucose in the filtrate that are in excess will pass into urine, leading to a sweet taste. Unexplained weight loss might occur as the body uses proteins and lipids as a source of energy. Feeling of fatigue might occur as the body uses amino acids and lipids as a source of energy.
Based on the passage, the overexpression of which protein is most likely associated with GM6112 cells becoming cancerous? A. cFLIP B. Caspase-8 C. FADD D. Fas
Based on the passage, the overexpression of which protein is most likely associated with GM6112 cells becoming cancerous? Jack Westin Advanced Solution: Most of the proteins described in the passage promote apoptosis. Cancerous cells are able to avoid apoptosis so the correct answer will either not promote apoptosis and/or actively prevent apoptosis. A. cFLIP Jack Westin Advanced Solution: A. ccFLIP lacks the catalytic activity to activate the FADD pathway which, when activated, promotes apoptosis. Therefore, cFLIP does not promote apoptosis and because it binds at the same site as the catalytically active caspase-8, prevents the activation of the FADD pathway and decreases the amount of apoptosis that occurs. B. Caspase-8 Jack Westin Advanced Solution: B. Caspase-8 is catalytically active and activates the pro-apoptotic FADD pathway. Overexpression would increase apoptosis. C. FADD Jack Westin Advanced Solution: C. FADD promotes apoptosis, the opposite of what we're looking for in cancerous cells. D. Fas Jack Westin Advanced Solution: D. The second sentence of the passage states that Fas stimulation recruits FADD which is a death domain protein. Like answer choices B and C, this is the opposite of what we expect. Cancerous cells are able to avoid apoptosis but this answer choice promotes cell death. Inhibition of apoptosis is a characteristic of cancer cells. So, if cFLIP is inhibiting apoptosis and is being overexpressed, we will have cancerous GM6112. The overexpression of cFLIP would prevent the activation of caspase-8, which would lessen the apoptotic response. You also said it in your post, "cFLIP... would prevent cell death." We want apoptosis in cancerous cells.
Assume that a certain species with sex chromosomes R and S exists such that RR individuals develop as males and RS individuals develop as females. Which of the following mechanisms would most likely compensate for the potential imbalance of sex-chromosome gene products between males and females of this species? A. Inactivation of one R chromosome in males Jack Westin Advanced Solution: A. In this question stem, males have two copies of the R chromosome and females have one R and one S chromosome. The imbalance stems from the different numbers of R and S chromosomes. To compensate for the potential imbalance, either the males will need to have one less active R chromosome, have a form of the active S chromosome, or females would have to have another active R chromosome. It is unlikely that one of the sexes would gain an additional chromosome, but inactivating the "extra" R chromosome in males would mean each sex only has one active R chromosome, which would compensate for the imbalance. B. Doubling transcription from the S chromosome in females Jack Westin Advanced Solution: B. Doubling the transcription of the S chromosome in females would be like having another active S chromosome in females. However, males would still have no S chromosomes so the imbalance would be magnified, not compensated. C. Inactivation of the R chromosome in females Jack Westin Advanced Solution: C. Inactivating the only R chromosome in females would be like removing the R chromosome and would worsen the imbalance; males would have two functional R chromosomes and females would have none. D. Doubling transcription from the R chromosomes in males Jack Westin Advanced Solution: D. Like answer choice B worsened the imbalance, doubling the transcription of the R chromosome in males would be the equivalent of having more active R chromosomes and would worsen the imbalance because women would still only have one. Answer choice A is the only answer choice that compensates for the imbalance instead of worsening it.
But how would inactivating the X chromosome help with an imbalance? If anything there would be more of an imbalance then because it would result in potential offspring XX, XY, X, & Y and Y cannot even survive so it's just XX, XY, and X which is further imbalanced 1 ReplyShare level 2dill3377-2·5 yr. ago·edited 5 yr. ago The inactivation occurs after offspring, so XX and XY will always give 50/50 XX and XY. Let's just say 4 children were born, two females and two males, the two females would total X each and the males XY each. The imbalance is due to how much genetic information is held in each part. A human X chromosome holds 804 genes and a Y chromosome holds 63. So a female has 804 and a male has 867 genes. While this is still a small imbalance, it's significantly less than if the female kept both X chromosomes, totaling 1608 genes versus the males 867. Doubling transcription in this case would mean 804 genes from the X and 126 from the Y (63x2), totaling only 930, still nowhere near the 1608 that the female with two activated X chromosomes has. Edit: I believe it's called lyonization in humans. I got this question right by comparing it with what happens in humans. Females have a XX chromosome and inactivate one of them, this is essentially just the opposite situation. Adding to this, females do this because the "X" chromosome holds significantly more genetic material than the Y chromosome does. Hence the imbalance of genetic information. Essentially, all males RR would become just R and all the females RS would stay RS, remember, only males are inactivating their R chromosome and the S is only a tiny bit of information that adds to the overall.
The bglF transcript is known to have a short half-life within the cytosol. What mechanism is most likely responsible for the transport of this transcript to the cytoplasmic membrane once it is synthesized? diffuse across the cytoplasm transport via attachment to the mitotic spindle active transport via cytoskeletal filaments transport from the endoplasmic reticulum in vesicles
C 1st not because it would take a long time 2nd not because its a prokaryotes who does binary fission same for the last one It doesn't diffuse across filaments it gets carried by a motor protein (Kinesin) along the filament from the nucleus to the membrane. This would be preferable to just sending the transcript into the cytoplasm and waiting for it to reach the membrane by chance because it would degrade before it would reach the membrane. Active transport along a filament provides a direct route from the nucleus to the membrane. Ah I see the confusion. Motor proteins moving along the filaments is active because it actively uses ATP! prokaryotes don't have a nucleus and no mitosis but have motor proteins
Which of the following energy conversions best describes what takes place in a battery-powered resistive circuit when the current is flowing?
Cant think of any examples I've seen on other exams, but you need to tackle these problems piece by piece. If you're consistently getting them wrong, it probably means you dont have a solid understanding of the different types of energy. This question can be solved pretty intuitively though. Batteries are made of chemicals (like nickel and cadmium), and chemical energy is the energy stored in bonds, so chemical energy is the answer here. Current then flows down the wires. Current is the movement of charge, so this sounds a lot like electrical energy. Lastly, resistors cause drops in potential, and that excess energy is lost as heat -> thermal energy.
Certain viruses contain RNA as their genetic material. One of the ways these RNA viruses replicate themselves is to: A. code for or carry a transcriptase that copies viral RNA. B. infect microorganisms possessing RNA as their genetic material. C. alter the host cell's polymerase in order to synthesize progeny viral RNA from the viral RNA template. D. stimulate the transcription of specific sequences of the host's DNA, which, in turn, direct the assembly of viral particles.
Certain viruses contain RNA as their genetic material. One of the ways these RNA viruses replicate themselves is to: A. code for or carry a transcriptase that copies viral RNA. Jack Westin Advanced Solution: This is correct. RNA viruses contain transcriptases, namely reverse transcriptase, that can use the viral RNA transcript to make cDNA or complementary DNA that can be recognized by the host for replication. Another name for reverse transcriptase that you might see on test day is RNA dependent DNA polymerase because it depends on RNA and will produce DNA. B. infect microorganisms possessing RNA as their genetic material. Jack Westin Advanced Solution: RNA viruses may only possess RNA as their genetic material, but their hosts contain DNA. Remember that the central dogma of biology for living organisms is DNA -> RNA -> protein. C. alter the host cell's polymerase in order to synthesize progeny viral RNA from the viral RNA template. Jack Westin Advanced Solution: RNA viruses code for or carry their own polymerases that recognize the RNA template to produce DNA. They do not alter the host's own polymerases. D. stimulate the transcription of specific sequences of the host's DNA, which, in turn, direct the assembly of viral particles. Jack Westin Advanced Solution: The key here is that these viruses contain RNA as their genetic material; how can the RNA instructions be used for viral replication? Even if the viral RNA stimulated the transcription of host genes, the problem of converting the RNA genetic material into clear instructions would remain. On the other hand, answer choice A and the presence of reverse transcriptase would allow for the viral RNA instructions to be converted to DNA instructions that the host can use to replicate the virus.
Which step of the SAM pathway is most likely catalyzed by Pfs? SAH + H2O → homocysteine + adenosine B12-CH3 + homocysteine → B12 + methionine methionine + ATP → SAM + 3Pi SAM + precursor → SAH + methylated product
Choice A is correct. Pfs is an S-adenosylhomocysteine (SAH) nucleosidase, meaning that it cleaves off a nucleoside (a DNA or RNA base attached to a sugar; essentially a nucleotide without phosphate groups) from a homocysteine. This reaction is depicted as the lower-most reaction in the circular pathway shown in Figure 1. Answer A shows a conversion from S-adenosylhomocysteine to homocysteine, which results from the removal of adenosine, a nucleoside. This matches the name of the enzyme and the logic of Figure 1, making A the best choice. Answer choice B is a conversion from homocysteine to methionine, which is a methylation reaction (addition of -CH3). Answer choice C is the addition of a nucleoside, adenosine, to create S-adenosylmethionine from methionine (SAM instead of SAH). Answer choice D is a demethylation reaction (removal of a methyl group).
Within the ADH active site, the pKa values of cys46, cys174, and his67 are: decreased due to charge destabilization of their protonated forms by Zn2+. decreased due to charge stabilization of their protonated forms by Zn2+. increased due to charge stabilization of their protonated forms by Zn2+. increased due to charge destabilization of their protonated forms by Zn2+.
Choice A is correct. The coordination of a cation such as Zn2+ to a neutral Lewis base increases the positive charge and destabilizes the Lewis base. This significantly increases the acidity of any protons attached to the cationic atom of the Lewis base and decreases their associated pKa values. Answer A is the correct choice because Zn2+ destabilizes the protonated form of cysteine or histidine, making the proton more acidic (lower pKa). Answer B is incorrect because it is the deprotonated form, not the protonated form, that would be stabilized by coordination to Zn2+. Answer C is incorrect because the pKa is decreased, not increased, and also because the protonated form is destabilized, not stabilized. Answer D is incorrect because the pKa is decreased, not increased.
Replacing the topoIb catalytic tyrosine with which amino acid best mimics wild-type topoIb, while eliminating catalysis? Phenylalanine Threonine Tryptophan Alanine
Choice A is the best answer. Recall that Tyrosine is a polar aromatic amino acid that contains a hydroxyl group attached to an aromatic ring. Phenylalanine and tryptophan are the only options that include a bulky ring structure similar to the tyrosine side chain, meaning neither threonine nor alanine provide adequate mass or hydrophobicity to mimic wild-type topoIb. Thus, neither B nor D can be the correct answer. Tryptophan, however, carries two rings, making the side-chain structure too large to properly mimic the tyrosine side-chain. Phenylalanine is nearly identical to tyrosine, but lacks the primary alcohol used in catalysis. These facts make Answer A better than Answer choice C.
For a condition exhibiting X-linked dominant inheritance, an affected male and a normal female will have: affected daughters. unaffected daughters. both affected and unaffected sons. affected sons.
Choice A is the correct answer. The genotypes for the father and mother are X'Y and XX, where X' represents an X chromosome with the affected allele (the allele for which one copy alone will cause the disease for this X-linked dominant condition). Daughters of these two individuals must get one unaffected X from their mother (that is all she has), and must get one affected allele (X') from their father: thus, all daughters must be X'X (affected). This finding supports Answer A. If the inheritance pattern were X-linked recessive, these daughters would be unaffected carriers--because they have only one copy of the disease-causing allele. Answer B is false because an unaffected daughter from these parents actually indicates X-linked recessive. Any child that is a son must have received a Y chromosome from his father. Given that the mother is normal, sons of these two parents NEVER receive an affected allele and therefore all sons are unaffected, making answer D false. Answer C is true in part (unaffected sons), but because it states that some sons will be affected, it must also be eliminated.
Researchers used extracts made from total plant organ biomass rather than the purified active ingredient. If the active ingredient was 1.0 % of the total plant biomass, the Ef values reported by the researchers in this study: overestimated the effectiveness of the plant compounds. underestimated the effectiveness of the plant compounds. overestimated the effectiveness of ciprofloxacin. underestimated the virulence of the microorganisms.
Choice A is the correct answer. The key to answering this problem is to recognize that mg/mL is 1000 fold more concentrated than µg/mL. The concentration of plant extract is 2.28 mg/mL and the concentration of the ciprofloxacin is 16.6µg/mL; To compare these directly, they must have the same units. The plant extract concentration (total biomass) in µg/mL is 2280. According to the stem, the active ingredient is only 1.0 % of this, or 22.8 µg/mL. So, we have 22.8 µg/mL active plant ingredient, but only 16.6 µg/mL ciproflaxin. Thus, when the researches added 30 µL of both treatments to the agar plate, they were actually adding more of the active ingredient from the medicinal plants than they were ciproflaxin. This may be counterintuitive because the stem states that only a tiny portion (1%) of the plant extract was actually active ingredients. However, when expressed in the same units, it is clear that even 1% of the more concentrated plant extract is still more than the concentration of the ciproflaxin. This suggests that Answer A is true, the researchers overestimated the effectiveness of the plant extracts because they were treating the agar plates with more plant active ingredient than they were ciproflaxin. Answer B can be eliminated because the scientists overestimated the effectiveness of the plant compounds, they did not underestimate them. Answer choice C is incorrect because Ef values do not report on the effectiveness of ciproflaxin; they report on the relative effectiveness of another compound compared to ciproflaxin (per Equation 1). If anything, the researchers could have underestimated the potency of ciproflaxin because they were adding less of it than they were the active plant ingredients. This is an example of a difficult MCAT question. If the examinee does not compare the units of each dosage as given in the passage, he or she would be very unlikely to choose Answer A. Notably, however, this is the kind of "difficulty" the AAMC will present. Almost universally, status quo prep companies misunderstand MCAT difficulty—thinking that difficult questions test more advanced science knowledge or require complex calculations. Difficult MCAT questions do NOT usually test difficult topics. They test basic science topics, but in a more challenging way. Difficult AAMC questions will require you to either a) think more conceptually or critically than you may be used to, or b) require greater care and precision (as in noting the units for this question before choosing an answer). In this way, the AAMC rewards the best critical thinkers without rewarding rote memorization of obscure or advanced scienc
For most materials, the maximum static friction is: arger than the kinetic friction force. smaller than the kinetic friction force. equal to the kinetic friction force. equal to the gravitational force.
Choice A is the correct answer. Under normal situations, when a horizontal force is applied to an object the static friction force increases to its maximum limit, and then the object moves. The maximum limit is given by f = µsN. Compare this to the normal kinetic friction force experienced when the object is moving, which is f = µkN. The two expressions are very similar, but since µs > µk for essentially all known surfaces, the maximum static friction force will always be greater than the kinetic friction force. Choice A is therefore correct, and this eliminates choices B and C. Choice D could be a tempting option since the friction force is indeed related to the gravitational force mg. Under normal conditions, and if the object is on a horizontal surface, the maximum static friction force is equal to µN = µmg. However, µmg = mg only in the hypothetical circumstance where µ = 1, so choice D is also incorrect.
Enveloped viruses do not infect cells surrounded by a cell wall. Given this information, an enveloped RNA virus is most likely to infect which host range? Plants Animals Bacteria Fungi
Choice B is correct. Enveloped viruses can only infect organisms that do not have cell walls. Animals are the only organisms on the list that do not have a wall surrounding their cells. Plants (Answer A) have a cellulose cell wall, bacteria (Answer C) have a peptidoglycan cell wall, and fungi (Answer D) have a chitin cell wall.
Which mutation would represent an oncogenic activation of Ras if the normal Ras mRNA sequence is the one shown below? 5'-CGUAAUGUUGAUCGUCAGGCCCUGGUCGUUUUUUCACCCACUGUAAAAA-3' 5'-CGUAAUGUUUAUCGUCAGGCCCUGGUCGUUUUUUCACCCACUGUAAAAA-3' 5'-CGUAAUGUUGAUCGUCAGGCCCUGGUCGUUUUUUCACACACUGUAAAAA-3' 5'-CGUAAUGUUGAUCGUCAGGCCCUGGUCGUUUUUUCACACCACUGUAAAAA-3' 5'-CGUAAUGUUGAUCGUCAGGCCCUGGUCGUUUUUUUACCCACUGUAAAAA-3'
Choice B is correct. It is stated in the final sentence of the passage that the oncogenic forms of Ras have point mutations in the 12th or 13th codons. When counting codons, or reading mRNA transcripts generally, it is important to look for the start codon, AUG, first. This is where translation will begin and it is also the first codon. Answer choice B has a point mutation in the 12th codon, making B the correct answer choice. Answer choice A has a point mutation in the second codon, so it is incorrect. Answer choice C has an insertion in the 12th codon, which causes a frameshift mutation, not a point mutation. Answer choice D has a mutation 38 nucleotides in, which might appear to be the 12th codon if the student started counting codons at the 5' end instead of at the start codon. There is no question that this will be a difficult challenge for most students. The science is relatively basic, but the long nucleotide sequence has a high intimidation factor, and careful analysis of each answer option may take much longer than the average question. We include questions such as these to prepare you for similar situations almost certain to show up on test day. We've heard a few horror stories of "huge nucleotide sequences" on the real MCAT, but our hope is not to prepare you only for analyzing long nucleotides. We hope you'll take home a bigger-picture message from this practice question. Truly difficult questions are actually quite rare. Our algorithm calculates an average of 3-6 questions of this caliber per section, so that's about what we include in our practice tests. However, whether it is a surprisingly long nucleotide sequence, a passage with more figures than you've ever seen before, or a graph with a dozen different variables graphed on the same axis, the take-home message is the same. Stay calm. Even the difficult questions are usually based on basic science or logic. If you need to, mark the question and come back to it only if you have time to do so later. Whatever you do, don't let one silly question intimidate you and influence your confidence on the rest of the exam!
What is Vmax for the SidA-catalyzed reaction in the presence of 13 µM celastrol? 0.25 mol/min 0.5 mol/min 1.0 mol/min 2.0 mol/min
Choice B is correct. The y-intercept of a double reciprocal plot is equal to 1/ VmaxIn Figure 3, the y-intercept is 2.0 (mol/min)-1. Therefore, 2.0 (mol/min)-1 = 1 /Vmax. Solving for Vmax gives 0.5 mol/min. Answer B is therefore the correct Vmax. Answer A would be correct if the y-intercept had been 4.0 (mol/min)-1. Answer C corresponds to a y-intercept of 1.0 (mol/min)-1, and Answer D corresponds to a y-intercept of 0.5 (mol/min)-1.
When other biomolecules are unavailable, acetoacetic acid can be used as a fuel source in all of the following organs EXCEPT the: heart. liver. brain. muscles.
Choice B is the correct answer. Acetoacetic acid is a ketone body. Ketone bodies can be used as fuel in the brain, the heart, and muscles (so Answers A, C, and D are incorrect), but they cannot be used in the liver because it lacks the necessary enzymes to convert ketone bodies to ATP. This is because ketone bodies are produced in the liver when blood glucose is low.
Which experimental result from the study of transgenic rats pre-disposed to AD would most strongly support the theory that a reduction in the number of dendritic spines causes AD? The number of dendritic spines on cortical neurons is inversely correlated with the severity of AD symptoms. Increasing the dosage of a drug that promotes dendritic spine formation is associated with a significant decrease in AD symptoms. Symptom-free rats with a reduced number of dendritic spines on their cortical neurons eventually develop AD. A drug that inhibits dendritic spine formation on cortical neurons is associated with a significant increase in AD symptoms.
Choice B is the correct answer. All of the answers provide some level of support for the conclusion that dendrite morphology is an integral part of AD, but only two of them suggest (but do not prove) causality. Answers A and C are strictly correlations and provide no evidence of causality. Answers B and D provide the strongest evidence because they are not strictly correlative. Answer B is a stronger choice than Answer D because it implies that not only does the increase in dendritic spines result in decreased AD symptoms, but it also suggests a dosage effect: more drug (and therefore more dendritic spines) results in even fewer AD symptoms. A dosage effect is stronger evidence of causation than a simple association between treatment and outcome. Association = two things are related Correlation = two things are related, but the basis of the relationship is unknown. Aka, when one changes, the other also changes, but it's not necessarily directional and other things can be influencing the relationship Causation = two things are related in a directional manner such that changing one results in a change in the other. Correlation and causation both imply association.
Researchers wondered if ResV could improve age-related decreases in heart function to mimic the level of function observed in young mice. To explore this possibility, they administered high doses of ResV to old mice. For this study, researchers most likely used mice from which group as a control? YC OC CR ResV
Choice B is the correct answer. For the study described in the stem, the investigators would examine old mice given high doses of ResV and compare their heart function to a control group of old mice who did not receive ResV. This makes the old control mice mentioned in the passage (OC) the best control. Answer A, the young control group (YC) may be tempting, but in actuality it could not act as a control for old mice receiving ResV. Comparing the treatment group to young mice receiving or not receiving ResV would introduce two manipulated variables. Sound experimental design requires that only one variable be manipulated at a time. Otherwise, when examining the results, the researchers could not determine whether the observed outcomes were the result of the mice being different ages, or the mice receiving or not receiving ResV. Answer C is implausible because this would also introduce a second manipulated variable, calorie restriction. Answer D cannot be a control because the treatment group is already being treated with ResV. The control needs to be mice identical in nearly every way to the treatment group EXCEPT that they are NOT dosed with ResV.
Given a genome content of 50% AT, and assuming random nucleotide variability, the probability of the AUUUA sequence is one in every: 25 bases. 45 bases. 2.0 x 105 bases. 4.0 x 105 bases.
Choice B is the correct answer. If a genome is 50 % A/T, then it is also 50 % G/C. At each position there is therefore a 1 in 4 (25%) chance of having the necessary base out of the four possible bases. Since the sequence is five nucleotides long, one must multiply the individual probabilities: (¼)(¼)(¼)(¼)(¼) = 1/1024. Thus, the probability of finding AUUUA is one in every 1024, or 45 bases. Answer A may be selected if the student thought there was a ½ chance of a given base at each position, but this is incorrect because there are four possible bases. Answer choices C and D are permutations of Answers A and B given in scientific notation and both are numbers far too large.
Which finding, if true, most weakens the theory that the human appendix is an evolutionary remnant without a current functional purpose? Many people develop appendicitis, which is the infection, inflammation, and potential rupture of the appendix. Many people suffer from diarrhea following appendectomy, which is the surgical removal of the appendix. Patients with a rare genetic disease do not have an appendix and these patients have no apparent symptoms. An ancient primate used the appendix to store cellulose enzymes, but this trait is no longer present in primates.
Choice B is the correct answer. The question asks for a finding that would WEAKEN the concept that the appendix is an evolutionary remnant without a current functional purpose. Answer B is correct because if patients suffer from diarrhea after an appendectomy this is an indication that the loss of the appendix may have resulted in the loss of some digestive function. This is certainly not the only possible explanation for diarrhea after appendectomy, but it is a small piece of possible evidence that would be contrary to the theory outlined in the stem. This makes B the best answer because none of the other options are logical or plausible. Answer A is false because it is irrelevant to the question. Many people develop meningitis too, and this does not suggest that the brain has no functional purpose. Answer C is false because if the lack of an appendix causes no symptoms, this supports the theory rather than weakening it. Answer D is false because, if ancient primates utilized the appendix for a digestive purpose, but that trait has now disappeared, this supports the concept that in evolutionary history the appendix did have a function, but further evolution has replaced that function and now it has no current physiological purpose.
The histones released by ischemic neurons were primarily found in which cellular location prior to the ischemic trauma? Dendrites Soma Synaptic end plate Axon
Choice B is the correct answer. This question requires recognizing the function of a histone and combining that knowledge with an understanding of the structure of a neuron. Histones are the proteins around which DNA is wound in its condensed conformation. In an intact cell (including neurons), histones will be primarily located in the nucleus, associated with the DNA they stabilize. In a neuron, the nucleus is found in the soma, or cell body, of the neuron, making B the only plausible answer. Because DNA is not present in the axon, synaptic end plate (distal part of the axon at the synapse), or the dendrites (projections off of the soma), histones will not be found in these locations to any significant degree, eliminating Answers A, C and D.
When the PAM in Figure 1 is activated (contracted), the PAM pulls to the left on the force sensor and the force sensor pulls to the right on the PAM. What is the relationship between these two forces? FPAM on sensor < Fsensor on PAM FPAM on sensor = Fsensor on PAM FPAM on sensor > Fsensor on PAM FPAM on sensor = Fsensor on PAM = 0
Choice B is the correct answer; the two forces must be equal. This is a test of Newton's Third Law, which says that the force of object 1 on object 2 is ALWAYS equal and opposite to the force of object 2 on object 1. It doesn't matter if the two objects are stationary, moving to the left or to the right, or even accelerating to the left or right... these partner forces are always equal. Choices A and C are therefore incorrect because the two forces are equal. Choice D is false because the PAM creates a force when it is actuated, so the force cannot equal zero.
Prions are infectious agents composed entirely of protein. In human hosts, prions cannot: be inherited or coded for by a gene. convert normal proteins into pathogenic proteins. cause pathology outside the central nervous system. be acquired through food consumption.
Choice C is correct. Although prion infections are usually associated with the consumption of contaminated meat (Answer D), they can also be inherited and encoded for in a person's DNA (Answer A). Prions convert normal proteins of the brain into pathogenic ones (Answer B). This conversion is exponential and takes years to decades to become pathological. The correct answer is option C because prion pathology is only observed in the central nervous system of the infected person or animal.
The synthesis of glucagon in pancreatic α-cells is most closely associated with which cellular organelle? Nucleus Lysosome Endoplasmic reticulum Mitochondria
Choice C is correct. Because glucagon is a secretory protein, it will be translated at the rough endoplasmic reticulum, with the growing protein chain oriented into the ER lumen as translation proceeds. Initially, the mature mRNA will be bound by a cytosolic ribosome, but a signaling sequence will pause translation and cause translocation to the rough ER. This makes Answer C correct. The recognition that glucagon is a secretory protein is necessary for this question, but does not require prior memorization. Students should recognize glucagon as a hormone that regulates blood sugar and know that it is synthesized for the entire body in the pancreas; therefore it must be secreted by the α-cells that produce it. Answer A may be tempting because the gene coding for glucagon would be found in the nucleus, but synthesis refers to translation of the mRNA transcript to form the actual protein, so choice A is incorrect. Answer B is incorrect, and is also implausible because lysosomes contain proteolytic enzymes. Answer D is incorrect because, although mitochondria do synthesize a few proteins for their own use, they do not synthesize secretory proteins for use outside the cell. The overwhelming majority of all mitochondrial proteins are coded for by the nuclear genome.
Epinephrine contains an acidic amine hydrogen described by the following equilibrium: R2NH + H+ ↔ R2NH2+ (pKa = 8.55). When epinephrine is released into the bloodstream during a "fight or flight" response, most epinephrine molecules will: (Note: pH of blood = 7.4) be uncharged. carry a 1- charge. carry a 1+ charge. be zwitterionic.
Choice C is correct. Epinephrine is an amine base. In solutions with a pH < pKa, the equilibrium shifts in favor of protonation. This places the proton on the amine nitrogen, giving it a 1+ charge. The equilibrium shown in the question stem is helpful in visualizing this conversion. Answer A is incorrect because epinephrine is charged. Answer B is incorrect because the charge is 1+ not 1-. Answer D is not correct because epinephrine does not have a functional group that carries a 1- charge to neutralize the 1+ charge.
The fluorescent light emitted from the β-isoindole fluorophore must: be infrared light. be ultraviolet light. have a wavelength greater than 420 nm. have a wavelength less than 420 nm.
Choice C is correct. Fluorescent light has to be of lower energy (longer wavelength) than the source light used to cause the excitation. The passage indicates that the light incident on the tube was 420 nm, so the light fluoresced must be of wavelength longer than 420 nm (lower energy). Answer A is unlikely to occur because infrared light is on the opposite end of the visible spectrum, in excess of 700 nm. Regardless, it is not true that it "must" occur as the stem requires; emission could occur at any wavelength greater than 420 nm. Answer B is much closer to the likely wavelength of the emitted light, but it is still impossible because 420 nm is already above the ultraviolet region and the wavelength is getting longer, not shorter. Answer D is false because it would indicate a higher energy wave of shorter wavelength.
If Experiment 2 was repeated with a mutant form of Rac1 that is continuously activated, HIF-1α mRNA levels would: decrease substantially during hypoxia. decrease substantially during normoxia. increase substantially during hypoxia. increase substantially during normoxia.
Choice C is correct. In Experiment 2, a deactivating mutation of Rac1 (Rac1T17N) results in a loss of activation of mRNA expression for HIF-1α. This indicates that Rac1 activity is necessary for activation of HIF-1α transcription. Therefore, it is likely that an activating mutation of Rac1 might further increase transcription of HIF-1α during hypoxia. Choice A is incorrect because HIF-1α is transcribed in response to hypoxia; a mutation in an upstream protein is unlikely to result in a decrease in transcription under a hypoxic state. If that protein is necessary for transcription, and the mutation renders the protein non-functional, then no change might be seen, but a decrease is highly unlikely. Choice B is similarly incorrect, because an activating mutation in a necessary modulator is even less unlikely to decrease HIF-1α transcription under normoxia if hypoxia is the necessary trigger for upregulation. Choice D is incorrect because while it may be possible that a highly active form of Rac1 might increase basal levels of HIF-1α transcription (because based on the results in Figure 2, Rac1 plays a role in upregulating HIF-1α transcription), it is unlikely to do so substantially under normal conditions, especially given the trend shown in the Western blot. Hypoxia is the trigger that stimulates upregulation of HIF-1α transcription, so C is a better answer.
In the enzyme-catalyzed conversion of L-Ornithine to N5-Hydroxyl-L-Ornithine: L-Ornithine is oxidized and NADPH is reduced. L-Ornithine is reduced and NADPH is oxidized. L-Ornithine is oxidized and NADPH is oxidized. L-Ornithine is reduced and NADPH is reduced.
Choice C is correct. The nitrogen atom of L-Ornithine is in a -3 oxidation state. The nitrogen atom of N5-Hydroxyl-L-Ornithine is in a -1 oxidation state. NADPH is the reduced form of the Flavin. NADP+ is the oxidized form. Therefore, L-Ornithine is oxidized, as well as NADPH, and the correct choice is Answer C. Answers B and D can be eliminated because they report L-Ornithine as being reduced. Answer A is eliminated because it falsely states that NADPH is reduced. O2 is the molecule that is reduced in this reaction
If a woman who is a heterozygous carrier for Duchenne MD has a son with a normal man, what is the probability this son will suffer from Duchenne MD? 0% 25% 50% 100%
Choice C is correct. The passage tells us that Duchenne MD is X-linked and recessive. The easiest way to answer this problem is construct a Punnet square, as shown below, in which Xn and Xm represent an X chromosome with a normal and a mutant allele, respectively. The Punnet square reveals that any sons born to this couple will have a 50% chance of being affected by Duchenne MD, which is answer choice C. This also eliminates answer choices A, B and D. However, it is important to recognize that the question asks about the chance for sons to inherit the disease, not all children. Answer choice B is the correct answer for the probability of any offspring being affected. Also notice that among this couple's daughters, there is a 50% chance of being a carrier, but zero chance of being affected.
If a segment of DNA complementary to the Cldn6 primer was transcribed into mRNA, would a start codon be found in the transcript? Yes, because there would be an ATG sequence. No, because there would be no ATG sequence. Yes, because there would be an AUG sequence. No, because there would be no AUG sequence.
Choice C is correct. The primer ATG GCC TCT ACT GGT CTG CA would result in a complimentary sequence of DNA reading TAC CGG AGA TGA CCA GAC GT which would transcribe back into mRNA with the sequence: AUG GCC UCU ACU GGU CUG CA. The first three bases are the start codon AUG so, yes, there is a start codon in the transcript. This makes answer choice C correct and eliminates answer choice D. Answers A and B can be eliminated immediately because the thymine base (T) is not found in RNA.
The following bacterial cells were grown in a nutritive environment. Following 24 hours of incubation, which bacterial population will include the greatest number of bacteria? 10 cells with a doubling time of 8 hours 100 cells with a doubling time of 12 hours 1 cell with a doubling time of 2 hours 1000 cells with a doubling time of 24 hours
Choice C is correct. This question stem concerns the exponential rate of bacterial growth. Answer C is the correct answer because in a 24-hour period a population that doubles once every two hours will double 12 times. This will be 4096 cells in 24 hours. The cells in Answer A will double 3 times in 24 hours, producing 80 cells. The cells in Answer B will only double twice in a 24-hour period, so the 100 cells will double to 200 and then to 400 cells. Lastly, the cells in Answer D will only double once in a 24-hour period resulting in 2000 cells. Because Answer C produces the most cells, it is the correct answer. In many ways, this question is conceptually similar to a half-life problem in reverse. You could think of it as a double-life problem; an application of bacterial "doubling" rather than radioactive "halving."
Quorum sensing in S. aureus most likely does NOT include communication between which two entities? Modulatory proteins secreted by one bacterium and transcription factors in another bacterium. One S. aureus bacterium and other S. aureus bacteria growing within the same colony. Quorum protein AI-2 secreted by one bacterium and chemoreceptor neurons in another bacterium. Paracrine signaling molecules secreted by one bacterium and the cell membrane of another bacterium.
Choice C is correct. This question stem includes a qualifier, so we are looking for the answer that is NOT a plausible way in which quorum sensing could occur between S. aureus bacteria. Answer C is impossible; being prokaryotes, bacteria lack organized cellular organelles, much less developed nervous systems with specialized chemoreceptors. This makes C the correct answer because it is clearly NOT how quorum sensing functions. Answers A and D both use slightly different terms to describe how quorum sensing most likely occurs. The passage states that a protein (AI-2) is a signaling molecule produced by bacteria for this purpose. We are also told that the process is paracrine-like and involves stimulus-response sets between bacteria. Therefore, Answer D, secretion of a signaling molecule, followed by binding of that ligand to a receptor on a neighboring bacterium, is highly plausible. Answer A uses very similar logic, but introduces transcription factors. This is plausible as well because the passage indicates that bacteria use this process to "coordinate gene expression." Finally, Answer B simply states that S. aureus bacteria in the same colony communicate with one another, which is central to the definition of quorum sensing.
What level of protein structure is disrupted by the point mutation in the C-terminus of NhhA? I. Primary II. Secondary III. Tertiary IV. Quaternary III only I and II only I and IV only I, II, and III only
Choice C is the correct answer. Because the primary structure is defined as the amino acid sequence, a point mutation will always change primary structure as long as the mutation results in substitution of an amino acid (Roman Numeral I). The only time a mutation would not alter the primary sequence is when the mutation resulted in a related codon that still coded for the same amino acid. That is not possible in this case because the passage tells us that the mutation reduced trimerization—so there must have been a change to the protein. This change must have impacted the quaternary structure because the passage indicates that the point mutation disrupts the trimer formation of NhhA, and trimerization is an example of quaternary structure. Answers A, B, and D are all incorrect because none of them include quaternary structure (Roman Numeral IV). Also, it is possible that this point mutation does not alter the secondary and tertiary structure of NhhA. The primary structure can be changed such that an amino acid is substituted but the region still remains formed into a beta sheet or alpha helix, and the tertiary structure may still remain the same (eliminating Roman Numerals II and III). That one amino acid could be the binding point for the trimer such that the tertiary structures of the native protein and the mutant protein are indistinguishable, but trimer formation is impossible in the mutant form.
The final electron acceptor during alcohol fermentation in S. cerivisiae is acetaldehyde. Is this process analogous to anaerobic respiration in humans? Yes, because acetaldehyde is the final electron acceptor. Yes, because oxygen is the final electron acceptor. No, because pyruvate is the final electron acceptor. No, because oxygen is the final electron acceptor.
Choice C is the correct answer. In humans, when insufficient oxygen is available (as in muscles during extended exercise), or when the necessary cellular equipment is lacking (as in red blood cells), glucose undergoes anaerobic fermentation with pyruvate acting as the final electron acceptor. Many students confuse this with lactate, which is formed in the final step, but it is the reactant molecule, pyruvate, that is reduced and is therefore labeled as the final electron acceptor. Similarly, oxygen is identified as the final electron acceptor in aerobic respiration, even though water is the molecule formed in the final step. Answers A and B are false because yeast fermentation and human anaerobic fermentation utilize different final electron acceptors. Answer D is false because oxygen is the final electron acceptor in human AEROBIC respiration and the question asks about anaerobic respiration.
The germ cells of a newly discovered diploid organism divide without the DNA replication step that normally occurs prior to meiosis. For this organism, which type of chromosome is found at the metaphase plate during Meiosis I? Tetrads, consisting of two pairs of sister chromatids Dyads, consisting of a single pair of sister chromatids Monads, consisting of a single chromatid Bivalents, consisting of four pairs of sister chromatids
Choice C is the correct answer. Prior to Meiosis I, under normal circumstances, all chromosomes (46 in humans, consisting of 23 homologous pairs) are replicated. This results in the formation of dyads, duplicated chromosomes that each consist of two strands of DNA; one strand individually is referred to as a chromatid; and the two chromatids are joined by a single centromere. When two homologous dyads pair up during meiosis I, the complex as a whole is called a tetrad (4 strands of DNA). However, in the organism described in the stem, DNA replication does not occur prior to meiosis. Therefore, for a diploid organism with 2n = 46, there will still be 46 chromosomes at the start of meiosis, but each chromosome will contain only one strand of DNA. There will be no sister chromatids present, during either Meiosis I or Meiosis II. This greatly simplifies the question, eliminating answers A, B and D because they all refer to sister chromatids. Bivalent is another word for tetrad, and both terms are exactly equivalent in meaning. Tetrads/bivalents consist of two pairs of sister chromatids (4 DNA strands), NOT four pairs (8 DNA strands) as is suggested by Answer D. A monad is a chromosome consisting of a single DNA strand.
One mole of which lipid will produce the most ATP upon complete oxidative metabolism? Glycophospholipids Sphingolipids Triacylglycerides Steroids
Choice C is the correct answer. Triacylglycerides are formed when three fatty acids are attached to a glycerol backbone. The other lipid classes listed have 2 or fewer equivalents of fatty acid per mole. Fatty acids produce the most ATP/carbon of any lipid. Answers A and B are not correct since these are structural lipids used for forming membranes and contain 2 or fewer fatty acid chains. Answer D is not correct; steroids are both structural components in the membrane and precursors for a variety of hormones, but do not produce significant amounts of ATP when metabolized.
Suppose airborne transmission of H7N9 is verified from chickens to humans, but not to ferrets, even though ferrets have tracheal architecture similar to humans. If infected humans experience symptoms of H7N9, which explanation is most plausible? Human tracheal architecture is rapidly evolving to meet the needs of the virus. Human tracheal architecture is more closely related to birds than to mammals. A mutation in the viral HA protein allows it to bind a receptor in the human trachea. A mutation in the ferret SA protein prevents the HA protein from binding its receptor.
Choice C is the most plausible explanation for infection of humans with a novel influenza virus strain. Influenza virus strains are highly adaptable due to their rapid mutation rate and lack of effective DNA repair mechanisms. Influenza pandemics in the past have been caused by strains that obtained an evolutionary advantage via mutations that allow binding to a different host receptor; this is the scenario described by Answer choice C. There is also a hint in the passage, where it is mentioned that it is a viral HA mutation that allows the virus to switch from avian to human hosts. Choice A is incorrect because the human trachea is not changing and would not evolve to help a virus. Be careful to never assign intent to the organisms or genes in an evolutionary scenario. The MCAT often tempts examinees with wrong answers that makes it sound as if a virus, gene, or organism "intentionally" mutated, or that it evolved as a reaction to its surroundings, or that it evolved "in order to obtain" a desirable advantage. All mutations are random. Evolution occurs by sheer happenstance when an accidental mutation happens to convey an adaptive advantage. Choice B is not plausible because humans are mammals and therefore share more traits with other mammals than with birds. Choice D can be eliminated because although mutating ferrets might obtain an evolutionary advantage, it is highly unlikely that all ferrets will mutate in such a way that they are no longer infected by influenza virus. Further, an important concept for MCAT-2015 is the principle that viruses undergo rapid mutation because they lack many of the DNA repair mechanisms found in higher-order organisms. Therefore, the likelihood of viral mutation >> likelihood of mammalian mutation.
A disruption in the expression of Cldn6 is LEAST likely to affect which layer of the epidermis? Stratum corneum Stratum granulosum Stratum spinosum Stratum basale
Choice D is correct. According to the passage, the effects of altered Cldn6 expression are observed in the suprabasal layers of the epithelium, which are the uppermost layers, above the stratum basale. The suprabasal layers consist of the stratum corneum (answer A), the stratum granulosum (B), and the stratum spinosum (C). However, specific memorized knowledge is not necessary to answer the question. The term "suprabasal" used in the passage suggests layers above the basal layer, and the figure provided in the stem makes it clear how each layer is ordered. Answer choice D, the stratum basale, as its name implies, is the basal layer of the skin and therefore least likely to be affected by a condition that impacts suprabasal epithelium.
The Hardy-Weinberg equation is shown. Assuming q represents the recessive allele in a hypothetical population, which variable can be determined using a population survey? p² + 2pq + q² = 1 p p2 2pq q2
Choice D is correct. Assume the dominant allele is A and the recessive allele is a. The sum of the frequency of each genotype in a population must equal one, therefore: AA homozygotes + Aa heterozygotes + aa homozygotes = 1. The stem states that q is the frequency of the recessive allele, so p must be the frequency of the dominant allele. Thus, the frequencies of gametes and progeny can be determined from a simple Punnet Square: p2 = the frequency of individuals of genotype AA, 2pq = the frequency of individuals of genotype Aa, and q2 = the frequency of individuals of genotype a (p2 + 2pq + q2 = 1). The value q2 then, is the frequency of homozygous recessive individuals in the population, which can be easily counted or observed because they will show the recessive phenotype; this is answer choice D. Choices B and C cannot be counted directly, as it is impossible to distinguish between homozygous dominant individuals and heterozygotes for a gene exhibiting simple dominance. Answer choice A, which is p, can be determined from p2, but through calculation, not from a population survey.
The primary mRNA transcript for dystrophin contains 300,000 fewer nucleotides than the DMD gene from which it is transcribed. The primary mRNA transcript is shorter because: introns have been deleted. exons have been deleted. the poly-A tail was not transcribed. regulatory sequences were not transcribed.
Choice D is correct. Genes contains large regulatory sequences, such as the promoter region, that are not transcribed. However, these regulatory sequences are still considered part of the gene they help to regulate. As a result, the primary transcript will always be shorter than the gene itself. This matches best with answer choice D. Answer choices A and B can be eliminated because both introns and exons are part of the primary transcript. Introns will be removed during post-transcriptional processing, prior to translation. Choice C is incorrect because the poly-A tail is added onto the primary transcript during post-transcriptional processing. In other words, the poly-A tail would actually add nucleotides to the primary mRNA transcript.
Where X represents the conjugate base, which of the acids listed is expected to have the smallest H-X bond dissociation energy? Acid A, pKa = 4.10 Acid B, pKa = 4.75 Acid C, pKb = 9.63 Acid D, pKb = 10.90
Choice D is correct. Solving this question requires understanding several concepts and definitions. First, the stronger the bond is between an acidic hydrogen and its conjugate base, the weaker that acid will be. Strong acids are expected to have relatively weaker H-X bonds (smaller bond dissociation energies) than weaker acids (larger bond dissociation energies). Second, a strong acid is defined by the extent to which it dissociates in water. A measure of its dissociation is its Ka, a type of equilibrium constant. pKa is defined as: pKa = -log Ka. The relationship between Ka and Kb is defined by the dissociation of water: Kw = 1 x 10-14 = KaKb. It therefore follows that pKw = 14 = pKa + pKb. This relationship is important so that the pKb values given in the answer choices can be converted to pKa values for direct comparison. Therefore, the pKa values in the list are: A) pKa = 4.10 (given); B) pKa = 4.75 (given); C) pKa = 14 - 9.63 = 4.37; D) pKa = 14 - 10.90 = 3.10. Since choice D has the smallest pKavalue, it is the strongest acid and will be expected to have the weakest H-X bond. Much of this information is presented as background info for the student. If an examinee easily recognizes that C and D can be converted to pKa values, this question is as simple as picking the answer with the lowest pKa.
Carotenoids are brightly colored biological pigments. Which carotenoid will absorb the longest wavelength of visible light?
Choice D is correct. The longer the extended conjugated double bond system, the longer the wavelength of visible light that will be absorbed. Answer choice A, b-Carotene, has 11 conjugated double bonds; Choice B, lutein, has 10; Choice C, violaxanthin, has 9; Answer D, astaxanthin, has 13 conjugated double bonds, including the C=O in the ring structure. Structures A, B, and C have a lower number than structure D, and thus Answer D is correct.
The average protein is composed of approximately 500 amino acids. This suggests that the average mature mRNA polymer contains approximately: 500 base pairs. 500 nucleotides. 1500 nucleotides. >1500 nucleotides.
Choice D is correct. The stem indicates that the average length of a protein is about 500 amino acids. This requires the translation of approximately 500 codons. However, each codon is composed of three nucleotides, so approximately 1500 nucleotides are required to form 500 codons. These facts alone would suggest Answer C, however, a mature mRNA molecule also features untranslated regions on both the 5' and the 3' ends, as well as a poly A tail. These are variable in length, but allow one to definitively conclude that if the average number of amino acids in a protein is 500, then the average number of nucleotides in mature mRNA must be quite a bit larger than 1500. This makes choice D the correct answer. Answer A can be eliminated because the mature mRNA is composed of a single strand of nucleotides, not base pairs. Answer B can be eliminated as 500 nucleotides would not produce a protein with 500 amino acids, since three nucleotides make up a single codon.
The ability of the BBB to regulate flux into and out of the CNS relies heavily on tight junctions. Tight junctions are important in the BBB because: hey permit cell-to-cell communication between cells in the CNS and those in the PNS. the adherens protein is overexpressed at the BBB to ensure tight associations between cells. they create a chemical barrier that inhibits all active transport across the BBB membrane. they seal off extracellular pathways, requiring passage through highly-selective cell membranes.
Choice D is correct. There are several kinds of cell junctions in animals, and this question requires, in part, recognizing the correct description of one type of cell junction, the tight junction (TJ). The BBB is devoted to controlling passage of molecules and ions into the brain. TJs seal off the extracellular pathway and require substances to cross the cell membrane, where highly-selective control can be exerted for which molecules do or do not pass through the barrier. Choice A is a description of gap junctions, which would not exert control over transport. It also references communication between CNS and PNS, which would occur through synapses, not gap junctions. Choice B is a description of adherens junctions which stabilize the cells, but have no clear effect on selective transport. Choice C does not describe any kind of cell junction, as no example of a "chemical barrier" between cells is currently known.
S. aureus has a generation time of 20 min. If biofilm density is proportional to colony size, which value gives the optical density (OD560) of the biofilm formed by the SBY2 colony one hour after the measurements taken for Figure 2? 0.3 0.8 6.0 16.0
Choice D is correct. This answer requires understanding two concepts. First, if the generation time is 20 min, the number of bacteria will double three times in one hour. Second, if the optical density is proportional to the colony size (which may or may not be true in practice), then the OD560 value will increase linearly with the colony size. At the time point measured in Figure 2, SBY2 was associated with an optical density of 2.0. Twenty minutes later (one doubling time for the bacteria), this will have doubled to 4.0. Forty minutes from the original time point it will have doubled again to 8.0, and one hour from the original time point it will have doubled a third time, to 16.0, or Answer D. Answer C is the result of multiplying 2.0 by three instead of doubling it three times. Answers A and B use the same logic as choices C and D, but are misapplied to the SBY3 colony instead of the SBY2 colony.
Following ingestion of a meal high in simple carbohydrates, the primary metabolic process at work is: glycolysis gluconeogenesis glycogenolysis glycogenesis
Choice D is the correct answer. Glycogenesis is the process of building glycogen to store glucose for the future. Answer A is one of the ways to consume glucose from the blood stream. Glycolysis only consumes a small amount of glucose prior to the pathway being shut down because the cell has sufficient energy. Because glycolysis consumes such a small amount of glucose prior to shutting down, it is not a major player in the balancing of blood sugar. Glycogen can grow to any size and therefore Answer D plays a much larger role in balancing blood sugar. Answers B and C are not correct since both of these processes will result in increased blood sugar since glucose is released from the cell.
A student attempts to isolate a solid organic drug compound with a known melting point range of 148-151 °F. An impure sample of this drug is likely to have which melting point range? 148-150 °F 150-155 °F 145-147 °F 144-150 °F
Choice D is the correct answer. Impurities in a solid both depress the melting point, and broaden the range over which the solid melts. Impurities never increase the temperature at which the solid melts, so Answer B is eliminated. Answer A suggests that the melting point stays approximately the same, but the melting point range becomes more narrow. Just the opposite is caused by an impurity, so Answer A must be rejected. Answer C is false because, although the melting range is depressed as expected, the melting point range is more narrow than the range for the pure solid given in the stem. Answer D is correct because the melting point is depressed and the range over which the solid melts is broadened. As is often the case, the lower end of the melting point decreases significantly, while the upper end of the range changes very little, resulting in a net broadening of the melting range. so think impurities melting point depression and boiling point elevation since impurities want the substance to never leave the liquid phase
In its native form, β-A is a soluble protein with extensive α-helical domains. Misfolding of β-A causes an α-helix-to-β-sheet transition that decreases solubility. β-A aggregation is likely the result of which interaction between β-sheet motifs on misfolded β-A proteins ? Charge-charge attractions Charge-charge repulsions Hydrophilic interactions Hydrophobic interactions
Choice D is the correct answer. Soluble proteins have their hydrophobic regions sequestered on the inside of the folded protein, away from the hydrophilic cytosol. When the pleated β-sheet forms, hydrophobic groups are exposed to the polar cytosol and are therefore attracted to the sheets of other unfolded proteins. A favorable energy change is therefore associated with aggregation. None of the other choices correctly attribute the aggregation pattern to the hydrophobic regions in the unfolded protein. Answers A, B, and C all list the interaction as hydrophilic or charged. To the degree that misfolding exposes more polar or charged groups to the cytosol, this would increase β-A solubility. Therefore, Answers A, B and C should be eliminated.
Based on the amount of KLF6 mRNA expressed after treatment with actinomycin D (Figure 2), cell lines derived from liver carcinomas exhibit: inhibited transcription of KLF6 mRNA. enhanced transcription of KLF6 mRNA. inhibited degradation of KLF6 mRNA. enhanced degradation of KLF6 mRNA.
Choice D is the correct answer. The passage states that transcription occurs normally in cancerous cells. However, Actinomycin D blocks transcription. Because transcription is blocked, answers A and B can be eliminated. After the addition of actinomycin, the amount of each type of mRNA remaining at a given time point is determined by the mRNA half-life, or the rate at which mRNA is degraded. The half-life of KLF6 mRNA is much shorter in both liver carcinoma cell lines. Put another way, the amount of KLF6 mRNA declines more rapidly in these cell lines. This rules out Answer C. Answer D is the only choice remaining, but also makes logical sense since the cancerous cell lines have the smallest amount of mRNA at the final time point—indicating mRNA degrades fastest in these cells.
Silkworms excrete a structural protein in which protein chains are organized and stabilized by which repetitive structural motif? -Helices Disulfide bonds Salt bridges β-pleated sheets
Choice D is the correct answer; silk is formed through b-pleated sheet secondary structure within the silk protein. Any helical structure, disulfide bonds, or salt-bridges play a minor role in stabilizing the structure of silk. Silk is presented in almost every textbook as an example of b-pleated sheet secondary structure. That being said, this does require specific prior knowledge that is difficult to infer or arrive at through a process of elimination. We have included this question because we see exactly this trend on a very small number of AAMC questions; usually only one question per exam section or about 2%. If you happen to remember the secondary structure of silk, this would be a very easy question. That is usually the pattern when the AAMC uses similar questions. If you remember the factoid, it's easy. If you don't, it's very frustrating. Keep in mind, however, that 1) these are quite rare, 2) they are unlikely to impact your score significantly, and 3) thanks to the Altius full-length exams, you have been exposed to this possibility so it won't surprise you. If you encounter something like this and don't recall the factoid, do your best to rule out distractors, make an educated guess, and move on. For all you know, it may be a beta question that doesn't even count!
Researchers have discovered the Pfs gene in P. aeruginosa, a bacteria that causes opportunistic infections in patients with Cystic Fibrosis (CF). If CF is an autosomal recessive condition, is it reasonable to conclude that treatment with a small interfering RNA (siRNA) that blocks translation of the Pfs gene transcript will cure a patient with CF? Yes, blocking Pfs will prevent biofilm formation by P. aeruginosa and therefore prevent CF. Yes, blocking Pfs will prevent colonization by P. aeruginosa and therefore prevent CF. No, siRNA molecules cannot prevent gene activation in vivo, so CF cannot be prevented. No, siRNA inhibition of the Pfs gene transcript will prevent biofilm formation, but not CF.
Cystic fibrosis is not caused by P. aeruginosa, it is an opportunistic infection that requires CF for its pathology. The stem clearly states that CF is a genetic condition. Autosomal recessive conditions are not caused by bacterial infections; they result from inheritance patterns. In this case, the bacterial infection may make the situation worse, but it is a separate process and preventing the bacterial infection can never change the genes of the person with CF. Therefore, blocking any pathways in P. aeruginosa would not result in a cure for cystic fibrosis. This eliminates Answers A and B. Short-interfering RNA molecules (siRNAs), as their name implies, interfere with normal mRNA function and have been used successfully in vivo for blocking gene function. Students should be aware of siRNAs and what they do, but this information isn't required to answer this question. Finally, if the siRNA were directed against the Pfs mRNA, the biofilm formation by P. aeruginosa will be inhibited, but CF will NOT be cured, which is answer choice D.
Meselon and Stahl Experiment Results
DNA replication occurs in a semiconservative way and, after two rounds of bacterial growth, there will be bacteria containing only 14N and bacteria containing the same amount of 15N and 14N. Thus, the genome mass will be 5.4 fg for the bacteria that only contain 14N, and 5.45 fg for the bacteria that contain equal levels of 14N and 15N.
Dihydrodigoxin levels in the culture medium of E. lenta cells would most likely be highest in which data set? A. 2 B. 4 C. 6 D. 8
Dihydrodigoxin levels in the culture medium of E. lenta cells would most likely be highest in which data set? Jack Westin Advanced Solution: Dihydrodigoxin is the inactivated metabolite produced by the reduction of digoxin. This means that the highest level of dihydrodigoxin will be present when there was a lot of digoxin for the E. lenta cells to metabolize. The second paragraph notes that cgr1 and cgr2 are upregulated in E. lenta when digoxin is present. The greater the presence of digoxin and the greater metabolism of digoxin, the greater the cgr1 and cgr2 expression and the higher the levels of the dihydrodigoxin metabolite respectively. A. 2 Jack Westin Advanced Solution: Data set 2 has the greatest cgr2 expression which means it would have the highest dihydrodigoxin levels. B. 4 Jack Westin Advanced Solution: Data set 4 has about half the cgr2 expression of data set 2 so the dihydrodigoxin levels will be lower. C. 6 Jack Westin Advanced Solution: Data set 6 has comparable cgr2 expression to data set 4. It will also have lower dihydrodigoxin levels than data set 2. D. 8 Jack Westin Advanced Solution: Data set 8 has the lowest levels of cgr2 expression out of all of the answer choices; it will have the lowest levels of dihydrodigoxin.
The CD spectroscopy signal that was used to generate the data in Figure 1 arises from the chirality of the: A. α carbon. B. amide nitrogen. C. carbonyl carbon. D. β carbon.
Eliminate B/D - they are achiral. Eliminate D - large majority of beta carbons are achiral. The difference in measured chirality in denaturation experiments is attributed to exposure of chiral backbone alpha carbons that would otherwise be unable to polarize light due to being buried in the protein. Solution: The correct answer is A. In a protein, each amino acid residue, except Gly, has a chiral α carbon. Amide nitrogen is achiral. Carbonyl carbon is achiral. With the exception of Ile and Thr,the β carbon in amino acid residues is achiral. CD spectra arise from the massively greater presence of α carbon chirality.
What is different between two enantiomers?
Enantiomers have identical chemical and physical properties and are indistinguishable from each other except for the direction of rotation of the plane of polarized light. They are described as optically active.
Uncompetitive Inhibitors bind to
Enzyme Substrate complex at an allosteric site Uncompetitive inhibition occurs when an inhibitor binds to an allosteric site of a enzyme, but only when the substrate is already bound to the active site. In other words, an uncompetitive inhibitor can only bind to the enzyme-substrate complex.
Protozoans characteristics
Eukaryotic, single-celled; they move in a variety of means including flagella, cilia, and amoeboid motion Complex inside, many make a tough survival stage called the CYST, which is often the stage that they transmit disease, the remainder of the time they are in a form known as a Trophozoite Usually found in water or in moist areas, some are photosynthetic, many are predators, some absorb nutrients
Samples from various time points of the proteolysis of TPMTwt were subjected to SDS-PAGE under reducing conditions. Which figure best depicts the expected appearance of the gel? (Note: The arrow indicates the movement of the protein through the gel.)
For proteolysis, the protein is expected to be broken up over time. Since smaller fragments will travel further in an SDS-PAGE, eliminate A and B because they indicate that the protein is consolidating/oligomerizing over time. Eliminate C because the top band stays the same darkness/concentration - proteolysis would deplete the initial substrate. D correctly shows both degradation into smaller fragments and depletion of the larger protein chunks over time. Solution: The correct answer is D. This gel depicts protein aggregation, not proteolysis, over time. The molecular weights of the bands are increasing rather than decreasing. This gel shows protein aggregation, instead of proteolysis, and does not even decrease the amount of starting protein. Although the increase in the number of lower molecular weight bands with time fits proteolysis, the original protein band at the highest molecular weight should diminish over time. As befits proteolysis, the number of lower molecular weight bands with time increases and the original protein band at the highest molecular weight diminishes with time.
secondary active transport
Form of active transport which does not use ATP as an energy source; rather, transport is coupled to ion diffusion down a concentration gradient established by primary active transport. The electrochemical gradients set up by primary active transport store energy, which can be released as the ions move back down their gradients. Secondary active transport uses the energy stored in these gradients to move other substances against their own gradients. As an example, let's suppose we have a high concentration of sodium ions in the extracellular space (thanks to the hard work of the sodium-potassium pump). If a route such as a channel or carrier protein is open, sodium ions will move down their concentration gradient and return to the interior of the cell. In secondary active transport, the movement of the sodium ions down their gradient is coupled to the uphill transport of other substances by a shared carrier protein (a cotransporter). For instance, in the figure below, a carrier protein lets sodium ions move down their gradient, but simultaneously brings a glucose molecule up its gradient and into the cell. The carrier protein uses the energy of the sodium gradient to drive the transport of glucose molecules.
Different junctions
Gap junctions: allow rapid ion current between cells, making them common in synchronized excitable tissue (most notably, cardiomyocytes and smooth muscle). Made of 2 connexons. Desmosomes: allow quick patches to hold together cells with high turnover rates (GI tract lining and skin). "Spot welds." Tight (occluding) junctions: hold cells together and regulate passage of stuff through cells. Associated with certain types of epithelia such as skin, blood-brain barrier, and renal cells. Hemidesmosomes: attach the basement membrane of the epidermis to the dermis.
endocrine secretions of pancreas
Glucagon - Alpha Cells Insulin - Beta Cells Somatostatin - Delta Cells secretion of a hormone from an endocrine gland which is transported in the blood to a target organ
glyceraldehyde-3-phosphate dehydrogenase (GAPDH)
Glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate
Which of the following best describes the chemical energy that is derived from the Krebs cycle? Energy is produced in the forms of: A. ATP, which directly supplies energy for many cellular processes, and NAD+, which supplies energy for the electron transport chain. B. NAD+, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain. C. ATP, which directly supplies energy for many cellular processes, and NADH, which supplies energy for the electron transport chain. D. NADH, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain.
I don't know if this helps but I view GTP as ATP for the sake of the MCAT and TCA cycle. Thus the TCA cycle provides ATP (GTP) which directly supplies energy for cellular processes. And NADH which supplies energy to the ETC. Which of the following best describes the chemical energy that is derived from the Krebs cycle? Energy is produced in the forms of: A. ATP, which directly supplies energy for many cellular processes, and NAD+, which supplies energy for the electron transport chain. Jack Westin Advanced Solution: NAD+ is a substrate and not a product of the Krebs cycle. Recall that NAD+ is the oxidized form of NADH and that the reduced form NADH is what serves as the electron carrier that feeds into the electron transport chain to produce energy. B. NAD+, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain. Jack Westin Advanced Solution: As noted above, NAD+ is not a source of energy and is not a direct product of the Krebs cycle. C. ATP, which directly supplies energy for many cellular processes, and NADH, which supplies energy for the electron transport chain. Jack Westin Advanced Solution: This is a solid answer choice. Both ATP and NADH are products of the Krebs cycle and are direct and indirect suppliers of energy respectively. ATP can be used directly in many cellular processes while NADH donates its electrons to the electron transport chain where a series of redox reactions creates a proton gradient that is used for the production of ATP. D. NADH, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain. Jack Westin Advanced Solution: Both NADH and ATP are products of the Krebs cycle. However, NADH does not directly supply energy; NADH feeds its electrons into the electron transport chain to produce ATP which then serves as a direct energy supplier. Answer choice C remains the best answer. See below for an overview of the connection between the Krebs cycle and the ETC.
Which other cellular components are likely to be located near the lacY6xbs transcript in the cell membrane? proteins and glycolipids glycolipids and sterols sterols and phospholipids phospholipids and proteins
I had to think a lot but the thing is they said lac Y is in E coli and so Ecoli is a prokaryote and we know lac Y is located near the cell membrane so what are the things found in the prokaryote cell membrane D glycolipids are made in golgi and prokaryotes don't have membrane-bound organelles sterols like cholesterol are made in ER and so prokaryotes don't have that E. coli gram neg membranes are roughly 75% protein and 25% phospholipid, by mass. Eukaryotic cell membranes: contain phospholipids, glycoproteins, and cholesterol prokaryotic cell membranes: contain proteins and phospholipids (cell walls can contain peptidoglycan) The question is asking about the cell membrane, while what you are talking about with gram negative bacteria is the cell wall. Prokaryotes have both a cell membrane and a cell wall which are two different things level 1 madelameee ·3 yr. ago You're right that LPS is a glycolipid, but it's considered part of the cell wall, not the cell membrane. The terminology is confusing since it's referred to as part of the "outer membrane," but anything outside of the inner layer (cell membrane) is considered separate.
Which of the following animal pairs best illustrates the outcome of convergent evolution? Jack Westin Advanced Solution: Convergent evolution occurs when fairly unrelated organisms evolve to meet similar environmental needs and is associated with analogous structures, like the wings of a bat and butterfly. The bat and butterfly are distant relatives but both needed the ability to fly in their environments. A. The dolphin and the shark Jack Westin Advanced Solution: A. Dolphins and sharks are distant relatives, specifically from different taxonomic families. They had to evolve similar characteristics in order to survive in similar environments, so this answer choice is looking good. B. The domestic sheep and the mountain goat Jack Westin Advanced Solution: B. Sheeps and mountain goats are a bit more closely related than dolphins and sharks and differ at the level of genus, so answer choice A is a better answer. C. The polar bear and the panda bear Jack Westin Advanced Solution: C. Polar bears and pandas belong to the same family, so like answer choice B, are more related than dolphins and sharks making it a poor answer choice. D. The light-colored and the dark-colored forms of the peppered moth Jack Westin Advanced Solution: D. These two are the most related out of all of the answer choices. For convergent evolution, we are looking for the least related animal pairing with similar traits making A the best answer. Now, you may be asking yourself, "am I supposed to know the family, genus and species of random animals for the MCAT?!" The answer is, thankfully, no. Without knowing the family and genus of each animal in these answer choices, you can probably tell that sheeps and goats are closely related, and that different types of bears are also closely related (probably why the MCAT test writer called it a panda bear instead of simply panda). You might also recognize that dolphins and sharks are less related; they have more visual and physical differences than some of the other answer choices. The MCAT is not about sitting down and memorizing a never-ending list of facts. Being able to critically reason your way through questions where you don't have the exact answer, such as this one, is an invaluable skill for test day. Keep working on it, you've got this!
I knew this because sharks and dolphins are the golden example of convergent evolution. And elephants and wholly mammoths (probably didn't spell this right lol) are the golden example of divergent evolution 3 ReplyShare level 2Foggybrainss·4 yr. ago So to be able to answer questions on finding the pair that best represents convergent or divergent evolution, is it then easiest to look at ancestral origin? Convergent evolution: 2 separate species that are converging b/c shared traits due to similar environmental factors Divergent evolution: 2 species, once came from 1 common origin, who now are diverging b/c less shared traits due to exposure to different environments. ex elephants live in hot africa + mammoths lived in cold arctic 3 ReplyShare The key here is that they have to have acquired similar characteristics due to similar environments. AAMC helps you by pointing out their different environments mountain goat vs domestic sheep Also polar bear is obviously in the cold weather while panda bears live in warmer weather Whereas you (hopefully) know dolphins and sharks live in the water Let me know if you still need clarification
What event will most likely occur if Protein X is inserted into the inner membrane of mitochondria? A.The citric acid cycle will cease to function. The electron transport chain will cease to function. C.The proton gradient across the inner membrane will dissipate. The pH of the intermembrane space will decrease.
I would say a complete understanding requires a more thorough analysis than that. Inserting Protein X into the Inner Membrane of the mitochondria (between matrix and IMS) has the effect of opening a channel of sorts, as evidenced by adding Protein X to the micelle and having it lose its fluorescence. Thus, you should realize that the ETC functions by pumping protons to the IMS and increasing the concentration there. Inserting a channel will only serve to prevent the concentration buildup in the IMS, because it dissipates the proton gradient (C). ETC would not cease to function because it can still do the transport and proton pumping, but the proton gradient is just continuously dissipating. pH of IMS would actually increase because a lower pH means more protons and because of the Protein X insertion there are FEWER protons in the IMS. Also your reasoning for A is off, the inner membrane surrounds the mitochondrial matrix, and Succinate Dehydrogenase is actually complex II of the ETC. Rather, you should know that no part of the CAC functions with gradients.
If HDAC inhibition by βOHB is a physiological response in living animals, the information in the passage indicates it is likely to occur when: A. there is sustained fatty acid oxidation. B. the pentose phosphate shunt is activated. C. there is increased gluconeogenesis. D. the Cori cycle is occurring. Butyrate is structurally related to β-hydroxybutyrate (βOHB), the major source of energy for mammals during prolonged exercise or starvation.
If HDAC inhibition by βOHB is a physiological response in living animals, the information in the passage indicates it is likely to occur when: A. there is sustained fatty acid oxidation. Jack Westin Advanced Solution: The second paragraph begins by saying that βOHB) is a major energy source when exercising or during starvation. When someone first starts fasting, glycogenolysis and gluconeogenesis are initiated to maintain blood glucose levels and provide energy. During prolonged fasting, such as when the body perceives starvation, fatty acid oxidation and ketogenesis are engaged to provide sustained energy and βOHB is produced. This means that βOHB will increase and HDACs will be inhibited, correctly addressing the question stem. B. the pentose phosphate shunt is activated. Jack Westin Advanced Solution: This will not increase the levels of βOHB. The pentose phosphate pathway is of note because no ATP is used or produced. This means that it wouldn't be especially useful during times of energy depletion the way that other energy-producing pathways are useful. See below for a review of the PPP. C. there is increased gluconeogenesis. Jack Westin Advanced Solution: While gluconeogenesis will be upregulated during fasting, it will not directly increase levels of βOHB unlike fatty acid oxidation and ketogenesis. D. the Cori cycle is occurring. Jack Westin Advanced Solution: Like many of the other answer choices, the Cori cycle does not produce βOHB. Only answer choice A directly affects βOHB levels in living organisms during prolonged fasting.
The muscle subtype represented by Culture C is LEAST likely to be characterized by:
If you understand that culture B is type I muscle fibers, then all you need to do is to identify culture A. Culture C will be the one left over. The things that give away Culture A's identity are its color and that it produces lactic acid upon stimulation. Muscle with oxidative capabilities is red because it contains myoglobin. Myoglobin is similar to hemoglobin in the blood, and it stores oxygen in Type I and Type IIa fibers. Type IIx does not use the Krebs Cycle or the ETC, and therefore does not need myoglobin. Therefore, it is white in appearance. In addition, the passage mentions that lactic acid forms immediately when Culture A is stimulated by Acetylcholine. Type IIa fibers would probably only use glycolysis and produce lactic acid once its oxygen store depleted. Culture B - has a lot of mitochondria (a lot of oxidative phosphorylation, meaning it requires oxygen and thus aerobic. It utilizes ATP very slowly meaning it uses power very slowly. And it can hold a contraction for a long time (over 30 minutes), this should be enough for you to know it is Fiber I. Culture A - produces a lot of lactic acid when stimulated, meaning it is getting energy from fermentation, an anaerobic process. It's a toss up obviously for IIa and IIx, but the fact that the pH drops so fast (under 2 minutes) means a lot of lactic acid was produced and dissociated to lactate and H+. So we can go ahead and assume that it is Fiber IIx. Culture C - must be Fiber IIa. It does have a (moderately) fast rate of contraction, Oxidative Capacity (oxidative respiration) and Glycolytic capacity (anaerobic respiration). Remember, Glycolytic capacity is a measure of the maximum rate of conversion of glucose to pyruvate or lactate that can be achieved acutely by a cell AKA anaerobic respiration. Finally, it has medium sized motor units.
Blood glucose levels in a healthy individual as a function of time are shown.
In response to changing glucose levels, which events occur at time points A and B? A. Hepatocytes secrete glucagon at point A, and the pancreatic alpha cells secrete insulin at point B. B. Pancreatic alpha cells secrete glucagon at point A, and pancreatic beta cells secrete insulin at point B. C. Pancreatic alpha cells secrete insulin at point A, and hepatocytes secrete glucagon at point B. D. Pancreatic beta cells secrete insulin at point A, and pancreatic alpha cells secrete glucagon at point B. Solution: The correct answer is D. Glucagon is secreted by pancreatic alpha cells, not the liver. Furthermore, insulin is secreted by pancreatic beta cells, not pancreatic alpha cells. High levels of circulating glucose stimulate insulin, not glucagon, secretion. Low levels of glucose stimulate glucagon, not insulin secretion. Pancreatic alpha cells secrete glucagon, not insulin. Hepatocytes do not secrete glucagon. Insulin is secreted in response to high blood glucose levels (point A) by pancreatic beta cells. Conversely, glucagon is secreted in response to low blood glucose levels (point B) from pancreatic alpha cells. At time A, insulin is secreted to lower the high blood glucose and at time B, glucagon is secreted to raise the low blood glucose (eliminate A/B). Pancreatic cells are responsible for secreting both (choose D).
Expression of the rHuEPO gene in E. coli bacteria produced an EPO protein that did not increase erythrocyte production when injected into humans. The most likely reason for this observation is that: Jack Westin Advanced Solution will be added here. A. prokaryotic ribosomes interpret the genetic code in a completely different manner than do eukaryotic ribosomes. B. E. coli cannot glycosylate EPO in the same way that it is glycosylated by eukaryotic cells. C. bacteria are unable to secrete eukaryotic proteins. D. only viruses contain the necessary cellular machinery to properly express recombinant proteins.
In terms of how the genetic code is interpreted, protein translation is the same for both prokaryotic and eukaryotic cells. Lack of EPO function could be caused by a failure of EPO to activate EPOR. Glycosylation is a process that is crucial for the structural conformation of the protein, in which structure is a main determinant of whether a particular ligand activates a receptor. Thus, differences in protein glycosylation could prevent EPO from properly activating EPOR. Many eukaryotic proteins are produced and secreted by bacteria. Production of eukaryotic proteins in bacteria is a process often used by researchers to better understand the function of a protein or to produce the protein in high levels. Viruses do not exclusively contain all cellular machinery necessary to express recombinant proteins. Bacteria do, as well. Prokaryotes read the genetic code in the same manner as eukaryotes (eliminate A). Bacteria are also often used to generate and secrete eukaryotic peptides (e.g. some insulin, eliminate C). Recombinant protein expression is not limited to viruses (eliminate D). B is reasonable, since post-translational modification such as glycosylation is generally limited to eukaryotes, and is often crucial for proper protein trafficking and functioning
Two hypotheses were proposed to explain the sudden decline in radioactivity that occurred after about 1200 s. Hypothesis 1 The decline is the result of random, nonspecific degradation of the protein by proteolytic enzymes. Hypothesis 2 The decline occurs when a precursor to Protein X is modified to produce a mature form of Protein X that is subsequently secreted by the cell. Question Which of the following graphs shows the radioactivity of the antibody-Protein X precipitate in the extracellular growth medium if Hypothesis 2 is correct?
In the experiment that they discuss in the passage, the researchers are measuring the amount of protein X inside the cells, which is how they got the graph in figure 1. Hypothesis B states that the decline in Figure 1 at 1200 secs is due to protein X being transported out of the cell (less protein X inside cell = decline). So if Hypothesis B is true, then beginning at 1200 secs, protein X should begin to be transported out of the cell and now you can detect protein X in the extracellular medium. The only answer where protein X begins to be detected in the extracellular medium beginning at 1200 secs is choice D. If none of that makes sense, you might be interpreting the passage incorrectly or not understanding the material.
In the reaction below, the activity of glutamine synthetase decreases as the concentration of tryptophan increases. The number of glutamine synthetase molecules in the cell remains constant regardless of the tryptophan concentration. These observations are consistent with a mechanism of: A. translational control. B. glutamine degradation. C. transcriptional control. D. feedback inhibition.
In the reaction below, the activity of glutamine synthetase decreases as the concentration of tryptophan increases. The number of glutamine synthetase molecules in the cell remains constant regardless of the tryptophan concentration. These observations are consistent with a mechanism of: A. translational control. Jack Westin Advanced Solution: Translational control would cause a change in the translation of the mRNA to protein and would change the number of enzymes produced and observed thereby contradicting the question stem. B. glutamine degradation. Jack Westin Advanced Solution: According to Le Chatelier's principle, glutamine degradation and the corresponding decrease in product concentration would shift the reaction forward to restore equilibrium. This is inconsistent with a decrease in the activity of glutamine synthetase. C. transcriptional control. Jack Westin Advanced Solution: Transcriptional control would alter the number of transcripts produced, which would affect the translation of the enzyme and like translational control, would change the number of enzymes produced. The question stem specifies that the number of enzyme molecules remains constant. D. feedback inhibition. Jack Westin Advanced Solution: Feedback inhibition relies on a component of the pathway affecting either earlier steps in the pathway (most types of feedback inhibition) or later (in the case of feedforward inhibition). Glutamine synthetase's activity is decreasing as the concentration of its product, tryptophan, increases. If tryptophan is inhibiting the enzyme, then the enzyme is under feedback inhibition. The product "feeds the signal back" to the enzyme. This is the correct answer.
muslce fiber types
It helps me to associate them with a particular activity Type 1 - think marathoning, you need lots of mitochondria to help utilize oxygen so AEROBIC respiration is used (high oxidative capacity) and it is very resistant to fatigue Type IIa - think swimming a lap as fast as you can, need mitochondria to help utilize oxygen, but not as much as Type 1. Also need to start producing energy ANAEROBICALLY because you are expending ATP faster than oxidative respiration could make it by itself (oxygen debt). So uses both anaerobic respiration (glycolysis) and ox-phos (oxidative) Type IIb - think jumping as high as you can or a one-rep max on the bench. Very short burst of energy that will need a quick source of ATP (glycolytic) and a lot of glycogen to supply it whenever, so will appear white. Less mitochondria (so low oxidative capacity), so will fatigue very easy when glycogen stores are depleted.
NPY amplified pituitary responses to GnRH by: A. 27%. B. 48%. C. 67%. D. 83%.
It's asking how much the rate of GnRH improved with the addition of NPY, so you put the improved number (7.03) over the initial number (4.74) to find the rate of improvement. Dividing 7.03/4.74 = 1.48, which means that it improved by .48 (x 100 = 48%).Always put the improved number over the initial number when it's asking how much something improved. A result of 1 would mean it didn't improve, but if there's decimals after the 1, that's the percent (if you multiply by 100) that the rate improved. Hope that makes sense! Because it's easier math. You could also have subtracted 7.03-4.47 = 2.29 Then done 2.29 / 4.74 (GnRH alone) to yield a 48 percent increase. The other way just avoids an extra calculation, but you need to know to drop the 1.00 value and read the 0.48.
The stereochemical designators α and β distinguish between: _
Jack Westin Advanced Solution: This is a standalone question that relies on knowing the difference between α and β designators. The designation is traditionally used to distinguish molecules with multiple chiral centers. Think of two carbohydrates that are virtually identical, except for their configuration on a single carbon. How do we distinguish between the two? Is there even a reason to distinguish between the two, or are they the same thing for all intents and purposes?There is certainly a reason! Epimers are stereoisomers that differ in the configuration of atoms attached to a chiral carbon. A big one we see is glucose and its epimers. The difference in the position of the hydroxyl at a chiral carbon creates these epimers. When glucose is in its cyclic form, a chiral center is generated at C-1. That carbon atom that forms the new chiral center (C-1) is the anomeric carbon. We can see two anomers below and how they differ in position at the anomeric carbon: Be careful with the verbiage here! We do need to distinguish between enantiomers, but that's with the R and S designations. This does not encompass the α and β designators we talked about in our breakdown of the question. A enantiomers at an epimeric carbon atom. B enantiomers at an anomeric carbon atom. Jack Westin Advanced Solution: Similar to answer choice A, but we at least mention the anomeric carbon in this case. This is better than answer choice A, but still not 100% consistent with the verbiage in our breakdown. C epimers at an anomeric carbon atom. Jack Westin Advanced Solution: This is consistent with what we came up with in the breakdown of the question. α and β designators are used to distinguish between epimers at an anomeric carbon atom. The visual above really helps demonstrate the difference between the two if it's unclear. D epimers at a non-anomeric carbon atom. Jack Westin Advanced Solution: This answer choice starts out strong until we get to "non-anomeric" which is not true about this question. Answer choice C matched our breakdown exactly, so we can stick with C as our correct answer. How do you know you are truly ready? That's where the Jack Westin MCAT Diagnostic Tool comes in. Solution: The correct answer is C. This is an Organic Chemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this item is C since the α versus β designation distinguishes between molecules with multiple chiral centers, but differ only in the configuration of the site known as the anomeric carbon atom. This is a Knowledge of Scientific Concepts and Principles question since you must recall and identify the distinction between α and β designations for molecules.
Mucous secretions in the respiratory tract inhibit microbial infections. These secretions are produced by which of the following tissue types found in the lungs? A. Smooth muscle B. Epithelial C. Nervous D. Connective
Mucous secretions in the respiratory tract inhibit microbial infections. These secretions are produced by which of the following tissue types found in the lungs? A. Smooth muscle Jack Westin Advanced Solution: Smooth muscle is muscle that is under involuntary control and forms the walls of hollow organs and blood vessels. The way I like to remember this is that smooth, hollow and blood all have two "o"s. Smooth muscle can be found in the lung airways, but it does not secrete mucus. For a more in-depth review of muscle structure and contraction, visit: https://jackwestin.com/resources/mcat-content/muscle-system/muscle-structure-and-control-of-contraction B. Epithelial Jack Westin Advanced Solution: Epithelial cells can be found in the luminal aspect of internal organs including the lungs. There are many types of epithelial cells: squamous, cuboidal, columnar, transitional and Goblet cells. Goblet cells in particular secrete mucus that can trap irritants and smaller foreign bodies. These Goblet epithelial cells can be found in the lungs to trap and inhibit microbial infections making this the correct answer. It should also be noted that the cilia in the bronchi will move the debris and mucus up and out of the lungs in a process called mucociliary clearance. C. Nervous Jack Westin Advanced Solution: The nervous system may provide signals to other endocrine and exocrine organs to secrete their respective hormones, but they themselves secrete neurotransmitters. Neither of these correspond to the secretion of mucus in the lungs. D. Connective Jack Westin Advanced Solution: Connective tissues include the blood, fat, bone and cartilage. None of these will secrete mucus. Answer choice B, epithelial cells, is the only one that mentions a cell type responsible for secreting mucus.
insulin sensitivity
My understanding of it just from intuition (so please correct me if I'm wrong) is if you have high insulin sensitivity, then that means your insulin receptors are working well, thus you would have a high tolerance to glucose, since if you have a spike in glucose you would want your body to release insulin so that your cells could take up that glucose and thus you would "tolerate" glucose. But if you have the opposite, insulin resistance, your insulin receptors would not be functioning property (similar to T2DM), thus if you had the same spike in glucose your cells would not be able to uptake the same amount of glucose and thus you would have low glucose tolerance. Insulin sensitivity refers to the ability for your cells to respond to increased insulin levels. Increased sensitivity: when insulin high, the body does a good job at receiving that signal, allowing for glucose to be imported into the cell to be used for glycolysis and glycogenesis. Decreased sensitivity (aka insulin resistance): body isn't as good at perceiving that signal, and thus requires a higher concentration of insulin before effectively up-taking and using that glucose in the blood. 3 ReplyShare level 2ExpensiveAd6014·1 yr. ago what they said. insulin resistance/decreased sensitivity can be identified by doing something like an insulin tolerance test. inject someone with insulin, see how much their blood sugar drops. if it doesn't drop very much after injection (over the course of a couple hours maybe), that would indicate insulin resistance/decreased insulin sensitivity.
After Sarah's accident, her attending physician detected the protein myoglobin in her urine. What type of injury is consistent with this observation? Broken bone Damaged muscle Damaged kidney
Myoglobin is found exclusively in the muscles. If it isn't there, that's a problem with the muscle. If somehow it does make it into the bloodstream, it is so large, that it should be reabsorbed at the kidney. So if it is in the urine, both systems must be damaged. When muscle is damaged, a protein called myoglobin is released into the bloodstream. It is then filtered out of the body by the kidneys. Myoglobin breaks down into substances that can damage kidney cells.
The rate of a typical enzymatic reaction is increased by which of the following changes? A. Decrease in a substrate concentration B. Increase in pH from 6.8 to 7.4 C. Increase in the energy of activation D. Increase in temperature from 20°C to 37°C
No, because the environment of the enzyme determines its ph optimum. Pepsin, the digestive enzyme in the stomach has a higher rate of activity at acidic ph's and b will be wrong so ph is dependent on which enzyme we are looking at. However temperature in general increases enzymatic reactions regardless of the situation albeit it doesn't exceed the melting temperature of the enzyme. Hope it make sense The rate of a typical enzymatic reaction is increased by which of the following changes? A. Decrease in a substrate concentration Jack Westin Advanced Solution: Decreasing the concentration of the substrate would decrease the rate at which the enzyme reacts, not increase it. B. Increase in pH from 6.8 to 7.4 Jack Westin Advanced Solution: The optimal pH at which an enzyme functions is dependent on the enzyme. Just think about the difference in pH between the stomach, the blood and the small intestine! At least one type of enzyme can be found in each location, highlighting the varied pHs at which enzymatic activity is optimized. C. Increase in the energy of activation Jack Westin Advanced Solution: Increasing the activation energy increases the energetic barrier for the reaction and would decrease the rate of the enzymatic reaction. D. Increase in temperature from 20°C to 37°C Jack Westin Advanced Solution: This is correct; enzymes are typically most active at a temperature of 37°C (average body temperature). If the temperature gets too hot or too cold, the enzymes will lose their enzymatic activity.
One consequence of advanced malnutrition is reduced amounts of plasma proteins in the blood. This condition would most likely cause the osmotic pressure of the blood to: A. decrease, resulting in a decrease of fluid in the body tissues. B. increase, resulting in a decrease of fluid in the body tissues. C. decrease, resulting in an increase of fluid in the body tissues. D. increase, resulting in an increase of fluid in the body tissues.
One consequence of advanced malnutrition is reduced amounts of plasma proteins in the blood. This condition would most likely cause the osmotic pressure of the blood to: Jack Westin Advanced Solution: Let's take a look at the two main pressures within capillaries, the hydrostatic and the osmotic pressures:In the arterial side of the capillary bed, the hydrostatic pressure is greater than the oncotic pressure and fluid moves out and into the interstitial space/tissues. However, on the venous side of the capillary, there is less fluid relative to the arterial side but the large plasma proteins that cannot cross the capillary wall are still present. The same amount of proteins with less fluid surrounding them will draw fluid back into the capillary as osmotic pressure in will be greater than hydrostatic pressure out. The net movement of fluid out of the arterial capillary and into the tissue due to hydrostatic pressure and the net movement of fluid into the venous capillary due to the osmotic pressure are for the most part balanced. However, there is some discrepancy between the two net pressures and not all of the fluid re-enters the capillaries. The lymphatics drain this fluid into the venous circulation to prevent edema. A. decrease, resulting in a decrease of fluid in the body tissues. Jack Westin Advanced Solution: The osmotic pressure is the pressure that draws fluid into the capillary. If there are less proteins in the blood, there will indeed be a drop in the osmotic pressure as there would be fewer solutes in the blood to pull in fluid. However, we noted that the osmotic pressure is responsible for bringing fluid out of the tissue and into the capillaries. If there is less osmotic pressure and less "pulling" of the fluid back into the capillary, there would be an increase in fluid remaining in the tissue, not a decrease. B. increase, resulting in a decrease of fluid in the body tissues. Jack Westin Advanced Solution: As noted above, the osmotic pressure would decrease with fewer proteins in the blood because the solute concentration that pulls fluid in will decrease. C. decrease, resulting in an increase of fluid in the body tissues. Jack Westin Advanced Solution: This is the correct answer. Fewer plasma proteins will decrease the osmotic pressure. This means less fluid will move out of the interstitial space/tissues and into the capillaries, resulting in more fluid remaining in the tissues. Pre-medical school fact: this mechanism is part of one of the two main hypotheses regarding the cause of edema! If the osmotic pressure drops too low, the fluid left in the tissues can exceed the lymphatic system's ability to drain the fluid back into the venous circulation and cause the accumulation of fluid in the interstitial space, most often the lower extremities. D. increase, resulting in an increase of fluid in the body tissues. Jack Westin Advanced Solution: Like answer choice B, this answer is incorrect because fewer proteins in the blood would mean a lower effective solute concentration and a lower osmotic pressure, not an increased osmotic pressure.
One potential complication of celiac disease is osteoporosis, which can occur as the body utilizes bone tissue to maintain adequate levels of calcium in the blood. The most effective way for the body to utilize bone tissue to increase blood calcium levels would be to simultaneously: A. increase osteoblast activity and decrease osteoclast activity. B. increase osteoblast activity and increase osteoclast activity. C. decrease osteoblast activity and decrease osteoclast activity. D. decrease osteoblast activity and increase osteoclast activity.
One potential complication of celiac disease is osteoporosis, which can occur as the body utilizes bone tissue to maintain adequate levels of calcium in the blood. The most effective way for the body to utilize bone tissue to increase blood calcium levels would be to simultaneously: Jack Westin Advanced Solution: The three key players in bone homeostasis that you should be familiar with on test day are osteoblasts, osteocytes and osteoclasts. OsteoBlasts are responsible for Building bones. They are responsible for the deposition of the calcium, collagen and protein matrix that make up bones. Osteocytes are osteoblasts that have become a part of the bone matrix and are involved in cell communication and mechanical sensing. These cells are recognizable due to their dendritic processes. OsteoClasts Consume or Chew bone; they break down and reabsorb the matrix. In order for the body to mobilize the calcium stores present in bones and increase blood calcium levels as effectively as possible, 1) bone growth should be downregulated to minimize calcium deposition and use, and 2) bone should be broken down to release and reabsorb calcium. A. increase osteoblast activity and decrease osteoclast activity. Jack Westin Advanced Solution: This would minimize blood or serum calcium levels, not maximize them. The activation of osteoblasts would promote the consumption of calcium and the inactivation of osteoclasts would decrease calcium release and reabsorption. B. increase osteoblast activity and increase osteoclast activity. Jack Westin Advanced Solution: While increasing osteoclast activity would promote the release of calcium, the increased osteoblast activity would increase the use of calcium in the deposition of new bone matrix. This will not maximize the increase in blood calcium. C. decrease osteoblast activity and decrease osteoclast activity. Jack Westin Advanced Solution: The first part of this answer choice is on the right track; decreasing osteoblast activity will slow the use of calcium thereby allowing more calcium to remain in the blood. However, decreased osteoclast activity will mean less bone is being broken down and less calcium is being released and reabsorbed. Osteoclast activity should be increased in order to mobilize calcium. D. decrease osteoblast activity and increase osteoclast activity. Jack Westin Advanced Solution: This is the correct answer because it is the most effective way that the body can use bone to increase serum calcium levels. The decrease in osteoblast activity means that less calcium is being deposited into the bone matrix so more remains in the blood, and increased osteoclast activity means more bone is being broken down and more calcium is released and reabsorbed into the blood. Both changes in activity will increase blood/serum calcium.
pyrrolidine ring
Organic ring structure containing one atom of nitrogen; linkage to another amino acid through this nitrogen favors formation of a linear, fibrous protein molecule
Overexpression of cFLIP in other cell types will most likely result in: A. decreased apoptosis in response to TNF. B. decreased apoptosis in response to Fas ligand. C. increased apoptosis in response to TNF. D. increased apoptosis in response to Fas ligand.
Overexpression of cFLIP in other cell types will most likely result in: Jack Westin Advanced Solution: The second paragraph states that cFLIP is structurally similar to caspase-8 but does not have catalytic activity. It also binds to FADD at the same (active) site as where caspase-8 binds. A. decreased apoptosis in response to TNF. Jack Westin Advanced Solution: A. According to the passage, cFLIP binds to FADD at the same site as the Fas ligand caspase-8. The passage also states that TNF signaling does not depend on FADD. cFLIP is part of an FADD-dependent pathway and TNF is a non-FADD-dependent pathway so this answer choice is incorrect. B. decreased apoptosis in response to Fas ligand. Jack Westin Advanced Solution: B. cFLIP lacks catalytic activity meaning it will not activate the apoptosis pathway when it binds to FADD at the same site as the catalytically active caspase-8. Since it binds at the same site, overexpression of cFLIP will mean less caspase-8 binds to FADD and less apoptosis will occur. C. increased apoptosis in response to TNF. Jack Westin Advanced Solution: C. As noted in the explanation for answer choice A, TNF works through a non-FADD-dependent pathway and cFLIP signaling is through the FADD pathway so this answer choice is incorrect. D. increased apoptosis in response to Fas ligand. Jack Westin Advanced Solution: D. As noted in the explanation for answer choice B, the correct answer, cFLIP is not catalytically active so it will not promote apoptosis. When it is overexpressed, there should be less apoptosis, not more.
Based on Reaction 1, when 1.0 atm of CO(g) completely reacts to form carbon suboxide at 550°C in a sealed container, what is the final pressure in the container? A. 0.00 atm B. 0.10 atm C. 0.25 atm D. 0.50 atm
PV=nRT. The chemical reaction shows that 4 molecules of CO react to form 2 total molecules of gas - since n is halved, then that means the pressure is halved. 0.5 x1atm = 0.5atm. Solution: The correct answer is D. There is still gas present in the container. Since pressure is directly proportional to moles of gas at constant V and T, this is a smaller pressure than the balanced equation allows. This result ignores the pressure contribution of one of the gases. Based on Reaction 1, 4 mol CO(g) forms 2 mol of gases. Because of the direct relationship between P and n at constant V and T, that means 1.0 atm CO(g) makes 0.50 atm of gases.
Which Roman numeral represents the CNS integration for a pain reflex arc? A. I B. II C. III D. IV
Pain reflex arcs integrate with the CNS at interneurons in the spinal cord - indicated by option II/B. Solution: The correct answer is B. Roman numeral I represents the dorsal root ganglion. Roman numeral II represents an interneuron, which serves as an intercommunication point for the afferent and efferent neurons within the CNS. Roman numeral III represents the ventral root. Roman numeral IV represents the effector muscle.
siRNAs (small interfering RNAs)
RNAs of similar size and functions as miRNAs that inhibit gene expression. siRNA requires perfect complementarity and will cleave it's target every time. miRNA can bind without being exactly complimentary but will only prevent translation. If the miRNA can bind perfectly it will also cleave the target
Reverse Transcriptase polymerase chain reaction
RT-PCR stands for reverse transcriptase PCR. How it works is, one extracts all the mRNA from a sample of cells, then exposes the mRNA to the reverse transcriptase enzyme, which converts the RNA into complementary DNA (cDNA). The cDNA can then be run through a PCR reaction to quantify the presence of particular mRNA sequences that have been transcribed in the cell. RT-PCR is actually used by a lot of labs these days, way more than Northern blot and Southern blot. Southern blot has been replaced by PCR for the most part, and Northern blot has been replaced by RT-PCR. Western blot is still a very popular technique though, as is ELISA.
Liver functions
Responsible for: The metabolism of fats, proteins, and carbohydrates. Excretion of bilirubin, cholesterol, hormones, and drugs. Enzyme activation. Storage of glycogen, vitamins, and minerals. Synthesis of plasma proteins, such as albumin, and clotting factors. Blood detoxification and purification. Bile production and secretion. cholesterol synthesis fat digestion Glucose metabolism blood clotting
Skin: Blood vessels
Skin blood vessels are capillaries, and so blood pressure is not controlled at the level of the capillaries, it's the arterioles where BP is controlled. At the level of skin blood levels, dilation/constriction functions to control heat radiation from body and nutreint supply to superficial body layers. So after a long time of constriction, the vessels dilate periodically to provide oxygenation to the superficial tissues. Passage won't help this question, this one needed some cold hard Physiology knowledge.
Autosomal Recessive Inheritance
Skips generations, usually seen in only 1 generation. 25% of offsprings from 2 carrier parents affected. Often due to enzyme deficiencies, more severe than dominant disorders; symptoms presents in childhood. affected children of unaffected parents, skips generations not just the males affected
Enantiomers can exhibit a difference in which chemical or physical property? A. Density B. Boiling point C. Smell D. IR spectrum
Smell is due to chemical receptors binding to specific molecules. Receptors themselves are chiral, so will bind specific chiral molecules. Other physical properties such as density, boiling point, and IR spectroscopy are independent of chirality. Solution: The correct answer is C. Enantiomers will not display different densities. The boiling points of enantiomers are identical, and they cannot be separated by distillation. Enantiomers have the same physical and chemical properties. They differ only in their three dimensional arrangement of atoms and their interactions with other chiral molecules. They can differ in their smell due to interacting differently with chiral odorant receptors. The IR spectra of enantiomers are identical when a normal light source is used. Circularly polarized light will potentially illustrate differences.
K and M two unlinked genes affecting hearing. Dominant K allele is necessary for hearing and dominant M allele causes deafness regardless of the other groups present. What fraction of offspring from parents with KkMm and Kkmm will most likely be deaf?
So I was reading over the answers and didn't really understand so I realized a way to explain it if you're more math brained. So the question, because the genes are UNLINKED, means that it is essentially asking what is the probability that we have "M" OR "kk" (not AND). So to solve these you use: p(A) + p(B) - p(A and B) probability of M is 1/2 probability of kk is 1/4 probability of both is just independent 1/2 * 1/4 = 1/8 1/2 + 1/4 - 1/8 = 4/8 + 2/8 - 1/8 = 6/8 - 1/8 = 5/8
The author's use of the word "resisters" gives the impression that the author wants to: A. discuss the migrants in a nonjudgmental way. B. emphasize that the migrants opposed authority. C. portray the migrants as unpatriotic citizens. D. suggest that the migrants' reasons for going to Canada were justified.
Solution: The correct answer is A. "Resister" implies, in a neutral way, the migrants' motivation for leaving the U.S. The nonjudgmental tone that is created by this word choice is implied early in the passage since resister is chosen over more familiar (but also negative) terms: "Just over 100 resisters, known more commonly as 'draft dodgers' and 'deserters,' came to Canada in 1964" (first paragraph). Given the context of the passage and the use of the term "resister" to specifically denote Vietnam war resisters, the migrants are shown to primarily resist or oppose involvement in the Vietnam War, not authority generally. The author uses a nonjudgmental tone throughout the passage that suggests the term "resisters" is used neutrally. Patriotism is brought up briefly in the final paragraph, but only to explain why it is likely that the subject of American migrants to Canada during the Vietnam War era has not received much attention: "Americans tend to be patriotic, and are apt to be unsympathetic to citizens who leave the country during wartime in acts of legal and political defiance against their government. Perhaps this is why neither Americans in general nor American social scientists in particular have given much attention to the exiles of the Vietnam generation" (final paragraph). There is nothing in this discussion of patriotism that suggests that the author wishes to convey that the migrants are "unpatriotic." "Resister" is used neutrally to denote migrants via the motivation behind their migration-their resistance to participation in the Vietnam War. There is no discussion in the passage of whether or not the author believes the migrations were justified.
Consider the fully protonated amino acid shown. As the pH of a solution of this amino acid is raised, which group deprotonates first? A. I B. II C. III D. IV
Solution: The correct answer is A. As the pH is raised, the most acidic group deprotonates first. Of the choices, two are carboxylic acids, one is a hydroxyl group, and one is a protonated amine. The hydroxyl group is least acidic, while the protonated amine is less acidic (pKa ~ 10) than a carboxylic acid group (pKa ~ 5). Of the two carboxylic acid groups, the one next to the chlorine atoms will be more acidic since its conjugate base (an anion) will be stabilized by an inductive effect which dissipates negative charge building up. A proton on alcohol is much less acidic than one in a carboxylic acid because the negative charge cannot be delocalized to several electronegative atoms. This carboxylic acid proton is less acidic than the one on the left because it is not near two electronegative chlorine atoms. Protonated amines are less acidic than carboxylic acids (pKa of 10 versus 5). The most acidic groups here woudl be the carboxylate hydrogens, so eliminate II(B) and IV(D). Between I and III, I would be more acidic, because the negative charge would be stabilized through inductive effect of the neighboring electronegative chlorines. In the case of III, the negative charge would be relatively unstable due to close charge separation from the positive amino group.
Mice that overexpress PGC-1α specifically in their skeletal muscles are most likely to exhibit which phenotype relative to wild-type mice? A. Lower body weight B. Lower body temperature C. Higher physical activity D. Higher nonfasting blood glucose levels
Solution: The correct answer is A. Based on the passage, overexpression of PGC-1α in skeletal muscle leads to increased subcutaneous fat UCP1 expression, and this UCP1 expression leads to more energy being dissipated as heat. Therefore, mice that overexpress PGC-1α specifically in their skeletal muscles most likely weigh less than do wild-type mice. Based on the passage, increased subcutaneous fat UCP1 expression leads to increased thermogenesis and higher, not lower, body temperature. According to the pathway suggested by the information in the passage, physical activity and exercise are upstream, not downstream of skeletal muscle PGC-1α expression. PGC-1α overexpression in skeletal muscle leads to increased FNDC5 expression. Based on the passage, mice overexpressing FNDC5 exhibit lower, not higher, nonfasting glucose levels. Higher PGC-1a results in more UCP1, which would increase body temp as protons flow down their gradient freely (eliminate B). PGC1a also leads to greater FNCD5 shown in table 1, which in Figure 1 indicates lower blood glucose (eliminate D). Though physical activity might also be affected by PGC1a, the passage only discusses the one-way increase of PGC1a due to physical activity (eliminate C). Answer A makes sense because greater UCP1 would mean greater consumption of nutrients to meet the metabolic demand due to less efficient oxidative phosphorylation.
Which chromatographic technique would most likely separate a mixture of native carbonic anhydrase from carbonic anhydrase photochemically modified by CCl3CO2H? A. Anion-exchange chromatography B. Cation-exchange chromatography C. Gas-liquid chromatography D. Size-exclusion chromatography
Solution: The correct answer is A. Because the passage states that native carbonic anhydrase has a net charge of -2.9 and the modified enzyme would have greater negative charge, anion-exchange chromatography can separate them as this technique separates proteins with different negative charges. The native and modified Enzymes both have net negative charge. In cation-exchange chromatography, both Enzymes would elute together in the void volume. Proteins degrade before they would vaporize. The molecular weights of the native and modified proteins are too close (< 1 kDa difference) to allow separation by size-exclusion chromatography. he photochemical modification adds carboxylic acids which are negative, so an anion exchange chromatography step is ideal for isolating the modified enzymes.
The second purification step is which type of chromatographic separation? A. Affinity B. Size exclusion C. Cation exchange D. Anion exchange
Solution: The correct answer is A. Displacement of the protein from the column in this step involved disrupting the binding of the (His)6 tag to the column. This is a classic example of affinity chromatography. Size-exclusion chromatography separates proteins by molecular weight, not selective column binding. Cation-exchange chromatography separates proteins with different positive charges (or positive versus negative/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction. Anion-exchange chromatography separates proteins with different negative charges (or negative versus positive/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction. he second purification step is described to be through a Ni+2 column - though it is tempting to choose one of the ion exchange options, the polyhistidine tag is selected for due to its unique affinity with Nickel
Which type of cells have a plasma membrane, lack a nucleus, lack most organelles, and are shaped like a flat disc with a concave center? Jack Westin Advanced Solution will be added here. A. Erythrocytes B. Fibroblasts C. Monocytes D. Neurons
Solution: The correct answer is A. Erythrocytes have all of the characteristics listed. Fibroblasts are not flat with a concave center, and they have a nucleus. Monocytes have a nucleus, and they are not flat with a concave center. Neurons are nucleated cells, contain all cellular organelles, and they are not shaped as a flat disc.
The radiation of wavelength 605 nm CANNOT be used to produce the fluorescence radiations depicted in Figure 3 because: A. the energy of the absorbed radiation must be larger than the energy of the fluorescence radiation. B. the energy of the absorbed radiation must be smaller than the energy of the fluorescence radiation. C. the 605-nm radiation has more energy than the 407-nm radiation. D. the 605-nm radiation is not visible.
Solution: The correct answer is A. Fluorescence can occur when the absorbed radiation has a photon energy larger than the photon energy of the radiation emitted through fluorescence. The photon energy is inversely proportional to the radiation wavelength, thus the 605-nm wavelength radiation cannot produce the entire fluorescence radiation spectrum shown in Figure 3 as its photon energy is below that of the fluorescence radiation of wavelength 604 nm. The photon energy of the absorbed radiation must exceed the photon energy of the fluorescence radiation because the energy difference cannot be created from nothing, per the energy conservation principles. The photon energy is inversely proportional to the radiation wavelength according to the formula Energy = Planck's constant multiplied by the speed of light divided by the wavelength, thus the 605-nm wavelength radiation has less photon energy than the 407 nm wavelength radiation. The 605 nm wavelength radiation is visible, but this feature is unrelated to fluorescence. Best approach to these problem types is process of elimination - D is untrue and does not answer the question. C is untrue - higher wavelength means less energy. It comes down to B and A. Absorbed radiation has to be higher energy than fluoresced or emitted radiation due to conservation of energy - all the energy from fluorescence has to come from energy that is absorbed; choose A.
Based on information presented in Table 1, which relationship between pH and charged functional groups is accurate? A. At pH 8.50, the ratio of the cationic to anionic functional groups in the NqrD subunit is equal to 1. B. At pH 7.00, half of the functional groups in the NqrE subunit are protonated. C. At pH 6.30, the net charge of the cationic and anionic functional groups in the NqrA subunit is equal to 1. D. At pH 6.00, the majority of the functional groups in the NqrF subunit are protonated.
Solution: The correct answer is A. Table 1 shows that the isoelectric point of NqrD is 8.50. The ratio of the cationic (+) to anionic (-) functional groups in a protein at its pI is always equal to one. Table 1 shows that the isoelectric point of the NqrE subunit is equal 5.35. At a higher pH, like at pH 7.0, the majority of functional groups in the NqrE subunit will lose their protons and will be deprotonated. Table 1 shows that the isoelectric point of NqrA is 6.30. The net charge of functional groups in a protein at its pI is equal to zero, not 1. Table 1 shows that the isoelectric point of NqrF is 5.25. At any pH higher than 5.25, the majority of the functional groups in the NqrF subunit will be deprotonated, not protonated. The definition of pI is the pH at which the overall charge is neutral - A matches this definition for NqrD, since equal cationic and anionic functional groups would cancel out to 0 net charge. At a pH greater than the pI, you can think of it as there being relatively low proton concentrations in the environment, so the molecule is deprotonated (more negative). Conversely, at a pH lower than the pI, you can think of the environment as being relatively highly concentrated in protons, so the molecule is more protonated (more positive).
Which classification of amino acids applies to the Trp residues after photochemical modification by CCl3CO2H? A. Acidic B. Basic C. Hydrophobic D. Polar neutral
Solution: The correct answer is A. The Trp residues after being photochemically modified by CCl3CO2H have a carboxylic acid group attached to the benzene ring. The group attached to the Trp residues is acidic, not basic. The group attached to the Trp residues is polar and hydrophilic. At pH 7, the group attached to the Trp residues would ionize and develop a negative charge.
According to social psychologists, people tend to respond to words and symbols as if they were the things to which they refer. What would the passage author be most likely to say about this tendency? A. It must be repeatedly overcome by engineers as a project progresses toward its physical realization. B. It accords with the distinction between the hypotheses of scientists and the proposals of engineers. C. It explains the failure to honor great engineers who have turned projects into objects. D. It implies an overemphasis on plans and working models in technological projects.
Solution: The correct answer is A. The author emphasizes that the beauty and appeal of a project before it results in the creation of a physical object is the project's quality as an imaginative fiction under the control of the engineer. The author also associates this conceptual stage with language and symbols: "In this stage of signs, language, and text, people influence the object" (paragraph 4). This suggests that, in the conceptual stage, it is best not to respond to the project as if it were a fixed object that already exists. For instance, the author compares the project to "A fiction with a variable structure" and states that "it is this flexibility to which we must aspire in tracking a technological project" (final paragraph). The tendency described in the question prompt conflicts with the distinction the author makes between the hypotheses of scientists and the proposals of engineers. The author suggests that the physical sciences describe an already existing reality; in contrast, engineering creates objects that do not yet exist and thus have an unknown potential. The author refers to "unhonored" engineers only figuratively, to suggest that engineers have a far greater role in contributing to culture and history than is often acknowledged: "Only a fiction [i.e., a project] that is capable of gaining or losing reality can do justice to engineers, those great unhonored figures of culture and history" (final paragraph). The author does not suggest that people's tendency to respond to words as if they were the actual physical objects contributes to engineers being less honored or respected than they might be. There is no indication in the passage that engineers place too much emphasis on planning, drafting, or creating working models.
One can infer from the passage that the author considers nature NOT to be: A. carefully arranged. B. aesthetically pleasing. C. ecologically complex. D. infinitely mutable.
Solution: The correct answer is A. The author's contention that nature is NOT "perfect" or enormously tidy supports this option: "...one cannot resist the feeling that each of these works is as carefully arranged, composed, and synthetically created as the gardens Monet cultivated at his home in Giverny. The overall compositions are often very simple, yet they seem carefully selected and perfectly balanced—too perfect to be an accident of nature" (paragraph 4). Throughout the passage, the author describes the work of Monet, who paints nature, as aesthetically satisfying. For instance, in paragraph 4, the author writes: "Yet there is much more than sensual pleasure here," certainly indicating the presence of aesthetic pleasure in the work. To the contrary, the author describes natural or organic life as containing "completeness, complexity, and mutability" (final paragraph). In the final paragraph, the author explicitly describes "living beings" (clearly natural) as "mutab[le]." So this option is not supported.
If the defibrillator described in the passage were fully charged and the entire charge were discharged through a patient in 10 ms, which of the following is closest to the average electrical current that would flow through the paddles? A. 7.5 A B. 15 A C. 22.5 A D. 30 A
Solution: The correct answer is A. The full charge of the capacitor is 25 μF × 3000 V = 75 mC. The average discharge current is (75 mC)/(10 ms) = 7.5 A. If the average discharge current is 15 A, it implies the full charge of the capacitor should have been 10 ms × 15 A = 150 mC. In fact, the full charge of the capacitor is 25 μF × 3000 V = 75 mC. If the average discharge current is 22.5 A, it implies the full charge of the capacitor should have been 10 ms × 22.5 A = 225 mC. In fact, the full charge of the capacitor is 25 μF × 3000 V = 75 mC. If the average discharge current is 30 A, it implies the full charge of the capacitor should have been 10 ms × 30 A = 300 mC. In fact, the full charge of the capacitor is 25 μF × 3000 V = 75 mC. Typically when you look at a circuit and see current - you think of resistors. However, the internal resistor shown is the only resistor and has no otherwise relevant information given. The charge of the capacitor is the hint here. We know that current is charge (Coulombs) per time (s). The time is 10ms, and we need charge Q. For capacitors, Q = CV = 25uF * 3000V, which is 25 x10^-6F *3x10^-3V, or 75 x10^-3 C, or 75mC. I = Q/t = 75mC/10ms = 7.5 Amperes.
Suppose that studies have shown that niche tourists are often very knowledgeable about the subjects of their tourism. How would this information most likely affect passage arguments? A. It would complicate the claim about what tourists expect to find when they visit the Delta region. B. It would support the claim about the effect of museum displays on the image of authentic blues culture. C. It would have no impact on the arguments presented in the passage. D. It would weaken the claim about the effort of communities to accentuate their cultural heritage.
Solution: The correct answer is A. The passage clearly suggests that the "truth" of blues history is more complex and multi-faceted than the dominant images that feature the economic hardship of musicians and communities would have it. The passage says that tourists are "ironically seeking the 'authentic' and 'genuine' blues experience," even as this is somewhat cynically constructed for them (final paragraph). But, if tourists were actually very knowledgeable about the blues, they would be less likely to be seeking this dominant image—so the scenario in the question would complicate passage claims about what tourists are seeking. If visitors were very well informed, as this question indicates, then they would be less likely to take the exhibits simply at face value. The statement in the question is relevant to passage arguments about what tourists expect to discover when they visit the Delta region. The statement in the question is irrelevant to the claim about community efforts to accentuate heritage.
Two open flasks I and II contain different volumes of the same liquid. Suppose that the pressure is measured at a point 10 cm below the surface of the liquid in each container. How will the pressures compare? A. The pressures will be equal. B. Pressure in flask I will be less. C. Pressure in flask II will be less. D. The pressures cannot be compared from the information given.
Solution: The correct answer is A. The pressure at a point 10 cm below the surface of the liquid is the same in both flasks because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. The pressure at a point 10 cm below the surface of the liquid in flask I is the same as the pressure in flask II at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. The pressure at a point 10 cm below the surface of the liquid in flask II is the same as the pressure in flask I at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. The pressures can be compared because both pressures are calculated according to the hydrostatic pressure formula p = ρgd, where ρ is the liquid density, g is the gravitational acceleration, and d is the depth where pressure is measured.
Based on the reported Hill coefficient, in what way do the MCS oligomers affect inhibition? A. As one MCS oligomer binds to the ATPase, it makes it easier for the others to bind, leading to inhibition. B. As one MCS oligomer binds to the ATPase, it makes it more difficult for the others to bind, leading to inhibition. C. A single MCS oligomer binds to the ATPase, leading to inhibition. D. MCS oligomers randomly bind to the ATPase, leading to inhibition.
Solution: The correct answer is A. The reported Hill coefficient of 2.56 indicates positive cooperativity because the value is greater than 1.0. Therefore, when one MCS oligomer binds to the ATPase, it makes it easier for others to bind. When there are sufficient MCS oligomers bound, they lead to inhibition of the ATPase. This would be negative cooperativity, requiring a fractional Hill coefficient. This would indicate no cooperativity, and the Hill coefficient would be 1.0. The Hill coefficient does not indicate whether binding is random. The last paragraph indicates that the hill coefficient is 2.56. Remember that this is the hill coefficient for inhibition of ATPase. A Hill coefficient >1 indicates positive cooperativity, and A describes the mechanism for positive cooperative binding.
What functional group transformation occurs in the product of the reaction catalyzed by Na+-NQR? A. RC(=O)R → RCH(OH)R B. ROPO32- → ROH + Pi C. RC(=O)NHR'→ RCOOH + R'NH2 D. RC(=O)OR'→ RCOOH + R'OH
Solution: The correct answer is A. This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR. This is the reaction catalyzed by a phosphatase. This is the reaction catalyzed by a protease or amidase. This is the reaction catalyzed by an esterase.
Which statement is implied, but not stated, in the passage? A. MIYO could address some of the inefficiencies experienced by local organizations. B. A population health approach focused on population subgroups is not intended to supplant a broad-based approach. C. Internet-based approaches to population health interventions are unlikely to achieve the success of initiatives backed by large national organizations. D. It is very difficult to create targeted health messages for some of the population subgroups described in the passage.
Solution: The correct answer is A. This point is implied here, when inefficiencies experienced by local organizations are described: "Furthermore, because there are few mechanisms for efficient sharing among niche-serving organizations, duplication is bound to occur. One study identified 116 different types of printed cancer education materials that targeted African Americans" (paragraph 4). The author implies that a targeted program such as MIYO could help to minimize such efficiencies—but does not state this explicitly. This is stated, not simply implied, in the final sentence of paragraph 3. This point is neither implied nor stated. The passage focuses on the potential advantages of such initiatives and it does not make the comparison suggested in this option. The passage does not focus on the difficulty of creating targeted health messaging for any particular groups. The passage does explicitly say: "Creating effective, tailored health resources for this near-infinite number of combinations of subgroups and health issues is a daunting task" (paragraph 4), but it never states or implies that it is particularly or "very difficult" to create targeted public health messages for some of the "subgroups described in the passage.
What is the most likely effect of adding a sodium ionophore to a culture of V. cholerae? A. Decreased activity of Na+-NQR B. Decreased production of ATP C. Decreased pH of the periplasm D. Decreased consumption of O2
Solution: The correct answer is B. A sodium ionophore does not interfere with the function of Na+-NQR, but it degrades the sodium gradient (sodium motive force) that is established by the action of Na+-NQR. Ionophores are compounds that bind to ions and facilitate their movements across membranes. A sodium ionophore would collapse the sodium gradient (sodium motive force) that is established by the action of Na+-NQR, resulting in decreased production of ATP. V. cholerae uses an electrochemical gradient of Na+, not of protons, to generate ATP. Therefore the pH of the periplasm is not affected. Since sodium ionophores degrade the electrochemical gradient required for ATP synthesis, they cause a decrease in ATP levels. To compensate for this event, V. cholerae is likely to increase the flux of electrons to the respiratory chain, which results in increased, not decreased, oxygen consumption. The passage states that the bacteria uses a sodium motive force to create energy. This is similar to eukaryotes ATP synthase that uses the motive force of protons to create energy. If we add a sodium ionophore, the gradient would not be able to be established as well (since Na+ will leak through the ionophore), and we won't be able to create as much ATP with a weaker sodium motive force. 3 ReplyShare level 1geh17263·2 mo. ago(1/14, goal 518) - Fl1/Fl2/FL3/FL4/FL5 --> 516/516/517/518/519 You needed to know that an ionophore is something that transports ions to even really approach this question imo. By adding another way for sodium ions to travel across the membrane, you are decreasing the sodium ion gradient. The passage said that vibrio uses that gradient for energy (energy = atp). So if you are taking away that gradient, you are making it harder to vibrio to make ATP 2 ReplyShare level 1PsychologicalCap4961·2 mo. ago(retaker) S/1/2/3/4/5: 509/513/511/518/519/518 Like the other commenter said, you had to know what an ionophore was to answer the question. I didn't so I got it wrong but an ionophore essentially binds to an ion and allows it to freely move across the membrane. This would dissipate the concentration gradient created by the enzyme and essentially decrease the motive force that the bacteria use for energy. I'm not sure if my reasoning is 100% correct but I thought of this question as synonymous to the ETC. If we introduce a sodium ionophore and dissipate the sodium gradient, electrons would nonetheless flow through the chain and reduce O2 at the end. On the other hand, ATP production would decrease because the lack of a sodium gradient. 2 ReplyShare level 2PoopslayertouchOP·2 mo. ago oh that actually makes sense. The protons are being pumped regardless of the gradient but because of the reduction potential right? So O2 is still decreasing BUT ATP synthase is no longer spinning since the protons are not flowing through it. Wow I did not wrap my head around that
In which phase(s) will the MCS precursor be predominantly found after the extraction step? The MCS precursor will: A. be found in the aqueous layer. B. be found in the tert-butyl methyl ether layer. C. be distributed equally between the aqueous layer and the tert-butyl methyl ether layer. D. form a precipitate between the aqueous and tert-butyl methyl ether layers.
Solution: The correct answer is B. According to the passage, the MCS precursor is lipophilic, so it is not water soluble. The passage described the MCS precursor as being lipophilic, which means that it would not dissolve as readily in the aqueous layer. Therefore, it should be found in the tert-butyl methyl ether layer, which is hydrophobic. According to the passage, the MCS precursor is lipophilic, so it prefers to dissolve in organic solvents. The MCS precursor should dissolve in the organic layer.
Based on the passage, what does Robert Lucas most likely mean by the "usual economic forces" (paragraph 3)? A. The tendency of businesses to locate near their competitors B. The tendency to consider expense as a primary driver of choice C. The factors that lead cost of living to be high in many urban areas D. The preference people show for the newest and most valuable items
Solution: The correct answer is B. After quoting Lucas, the passage author goes on to illustrate the point with a question about cost; the location of businesses is not discussed. Lucas says that the "usual economic forces" would suggest that people should not be drawn to cities. The author goes on immediately to ask, "why would young designers live in New York when they could live more comfortably in other cities with much lower costs of living?" The implication is that this question describes the usual economic forces at work, which, in this case, would be the forces that prioritize the tangible items or circumstances (leading to comfort) that one can get for one's money. Lucas is talking about the effects cost of living may have on people's choices about where to live, not about the factors that themselves influence cost of living. There is no discussion in relation to Lucas's claim about any preference people might display for the "newest" or the "most valuable" items, so this option is simply not supported.
If both the capacitor and the power supply in Figure 1 are adjustable, which of the following changes would result in an increase in the charge on the capacitor? A. Decreasing the area of the parallel plates B. Decreasing the separation between the parallel plates C. Removing the dielectric from the capacitor D. Decreasing the voltage of the power supply
Solution: The correct answer is B. Because the capacitance is directly proportional to the area of the parallel plates, a decrease in the area corresponds to a decrease in the capacitance. Given that the charge on the capacitor is directly proportional to the capacitance, a decreased capacitance results in a decrease in the charge as long as the power supply voltage is constant. Capacitance C is inversely proportional to the separation d between the parallel plates according to the formula C =(ϵ0 ϵr A)/d. A decrease in the separation corresponds to an increase in the capacitance. Given that the charge on the capacitor is directly proportional to the capacitance, an increased capacitance results in an increase in the charge as long as the power supply voltage is constant. Capacitance C is directly proportional to the permittivity ϵr > 1 of the dielectric between the parallel plates according to the formula C =(ϵ0 ϵr A)/d. Removing the dielectric essentially means decreasing the permittivity ϵr to 1, which corresponds to a decrease in the capacitance. Given that the charge on the capacitor is directly proportional to the capacitance, a decreased capacitance results in a decrease in the charge as long as the power supply voltage is constant. The charge on the capacitor is directly proportional to the capacitance multiplied by the voltage of the power supply. Decreasing the voltage results in a decrease in the charge unless the capacitance is increased independently.
Suppose that people who live in low-income neighborhoods are less likely than people in affluent neighborhoods to have access to community recreational facilities and lighted sidewalks. How would this affect passage claims? A. It would challenge claims about the "long tail" approach to public health. B. It would support passage claims about access to beneficial resources. C. It would undermine claims about the value of "blockbuster" interventions. D. It would reinforce claims about the funding, capacity, and infrastructure of local organizations.
Solution: The correct answer is B. If anything, the scenario described in the question would support the long tail approach, because the long tail approach targets groups with specific needs, for whom broadly targeted approaches may be ineffective or inappropriate. The passage refers to the fact that some population subgroups have "unequal access to beneficial or health-promoting resources" (paragraph 3), which is exactly what the question refers to. If anything, the scenario described in this question would support those claims, as the point of long-tail approaches is to reach groups who may "bear a disproportionate burden of disease because of heightened exposure to risk or unequal access to beneficial or health-promoting resources" (paragraph 3). The statement in the question is irrelevant to passage claims about the funding, capacity, and infrastructure of local organizations.
Lactase can be classified as which type of enzyme? A. Isomerase B. Hydrolase C. Transferase D. Oxidoreductase
Solution: The correct answer is B. Isomerases are Enzymes that restructure the chemical formula of a compound. They do not break down the chemical bound in a disaccharide such as lactose. The passage notes that lactase breaks down lactose to glucose and galactose. Glycosidic linkages in disaccharides are cleaved via a hydrolysis reaction; therefore, lactase is classified as a hydrolase. Transferases are Enzymes that transfer chemical groups between compounds. They do not break down the chemical bound in a disaccharide such as lactose. Oxidoreductases are Enzymes that catalyze the removal of electrons and hydrogen atoms. They do not break down the chemical bound in a disaccharide such as lactose. Lactase is an enzyme produced by many organisms. It is located in the brush border of the small intestine of humans and other mammals. Lactase is essential to the complete digestion of whole milk; it breaks down lactose, a sugar which gives milk its sweetness. Lacking lactase, a person consuming dairy products may experience the symptoms of lactose intolerance.[1] Lactase can be purchased as a food supplement, and is added to milk to produce "lactose-free" milk products.
The information in the passage best supports which hypothesis? A. Exercise prevents glucose uptake. B. Exercise promotes less effective cellular respiration. C. Exocrine secretions of skeletal muscle act on adipose tissue. D. Endocrine secretions of adipose tissue act on skeletal muscle.
Solution: The correct answer is B. The information in the passage indicates that exercise correlates with increased FNDC5 expression. Additionally, mice overexpressing FNDC5 exhibit lower nonfasting glucose levels, most likely because of higher, not lower, cellular glucose uptake. According to the pathway suggested by the information in the passage, exercise ultimately increases UCP1 levels which in turn degrades the proton gradient that drives oxidative phosphorylation. More energy is dissipated as heat and less is used to synthesize ATP. The passage notes that irisin is secreted into blood and therefore it would be an endocrine, not an exocrine, secretion of skeletal muscle. The passage notes that irisin is a protein formed by cleaving the extracellular domain of FNDC5 which is expressed in skeletal muscles, not adipose tissues.
Which enzyme of the citric acid cycle is NOT directly involved in generation of the dinucleotide required for Na+-NQR activity? A. Malate dehydrogenase B. Succinate dehydrogenase C. Isocitrate dehydrogenase D. Α-Ketoglutarate dehydrogenase
Solution: The correct answer is B. The passage notes that Na+-NQR requires NAD+/NADH for its activity. Malate dehydrogenase does require NAD+/NADH. The passage notes that Na+-NQR requires NAD+/NADH for its activity. Among the options, only the conversion of succinate to fumarate by succinate dehydrogenase which generates FADH2, does not require NAD+/NADH. Malate dehydrogenase, isocitrate dehydrogenase and α-Ketoglutarate dehydrogenase all generate NADH within the citric acid cycle. The passage notes that Na+-NQR requires NAD+/NADH for its activity. Isocitrate dehydrogenase also requires NAD+/NADH. The passage notes that Na+-NQR requires NAD+/NADH for its activity. α-Ketoglutarate dehydrogenase also requires NAD+/NADH.
A 2 kg mass and a 5 kg mass are connected by a massless cord suspended over a massless and frictionless pulley. If the acceleration due to gravity is g, what will be the acceleration of the masses after they are released from rest? A. 2g/7 B. 3g/7 C. 5g/7 D. g
Solution: The correct answer is B. This is smaller than the magnitude of the acceleration of the masses after they are released from rest by a factor of 1.5. It implies that only the 2-kg mass is moving after the release. According to Newton's second law, the net force acting on the 5-kg mass is given by the expression Fnet= 5 kg × a1 = 5 kg × g - T, where a1 is the acceleration after the release and T is the tension in the cord. The net force acting on the 2-kg mass is given by the expression Fnet= 2 kg × a2 = 2 kg × g - T. Because the two masses move simultaneously but in opposite directions after they are released, a1 = -a2 = a. Substituting the expression T = 5 kg × (g - a) into the equation of motion of the 2-kg mass yields -2 kg × a = 2 kg × g - 5 kg × (g - a) = -3 kg × g + 5 kg × a. Then 7 kg × a = 3 kg × g, hence a = 3g/7. This is larger than the magnitude of the acceleration of the masses after they are released from rest by a factor of 1.7. It implies that only the 5-kg mass is moving after the release. This implies the two masses fall freely and separately without being connected by the cord. Draw a pulley system, with 2kg hanging on one end and 5kg hanging on the other end. The net force acting on either mass would be 3kg *g, acting upward on the 2kg mass and downward on the 3kg mass. Newton's second law states sum of all forces = total mass * a of the system, so 3g = (2+5)a, and a = 3g/7.
Based on Figure 2, what is the approximate Ki of the MCS oligomers? A. 12 nM B. 30 nM C. 170 nM D. 800 nM
Solution: The correct answer is B. This is the MCS oligomer concentration at almost the maximum specific activity. In the semi-log plot shown in Figure 2, the Ki is the inflection point of the sigmoidal curve because, by definition, the Ki is the concentration of the inhibitor at which the reaction rate is half of the maximum reaction rate. The inflection point is at approximately 0.030 mM, or 30 nM. This is the MCS oligomer concentration at about 0.05 × specific activity. This is the MCS oligomer concentration at complete enzyme inhibition Ki, like Km, would be inhibitor concentration that results in half of the ATPase's maximum velocity. Note that the x axis is a log scale, and each smaller tick notates a 10th (0.01, 0.02, 0.03, etc). At half of the maximum velocity, the Ki is 2 ticks to the right of 0.01, so Ki = 0.03uM, or 30nM.
If the combined mass of the TPMT substrate and cofactor was determined before the enzymatically catalyzed reaction and then compared to the combined mass of the product and the cofactor after the reaction, the net change in molecular weight will be: A. +15 g/mol. B. 0 g/mol. C. -15 g/mol. D. -16 g/mol.
Solution: The correct answer is B. This is the molar mass gained by the substrate, not the net change in molecular weight of the substrate and cofactor. TPMT is a transferase. It is transferring the methyl group from the cofactor to the substrate. Hence, the total net change in mass would be 0 g/mol. This would be the loss of molar mass by the cofactor, not the net change in molecular weight of the substrate and cofactor. This is the molar mass of CH4.
What is the molecular formula of the heterocyclic aromatic compound pyrrole? A. C2H3N B. C4H5N C. C6H7N D. C8H9N
Solution: The correct answer is B. This is the molecular formula of the non-aromatic heterocycle 2H-azirine. This is the molecular formula of pyrrole, a five-membered aromatic heterocycle containing one nitrogen atom. This is the molecular formula of various forms of azepine, a seven-membered heterocycle, none of which are aromatic. There is a cyclohexene-fused pyrrole with this molecular formula, but not pyrrole itself.
What is the chemical structure of a component found in four of the five cofactors used by Na+-NQR?
Solution: The correct answer is B. This is the structure of adenine. It is only found in FAD. This is the structure of flavin, found in four of the five cofactors used by Na+-NQR. This is the structure of ubiquinone. It is a substrate, but not a cofactor. This is the structure of histidine. It is an amino acid, not a cofactor.
A person pushes on a rolling cart with a force that diminishes with time because the person must walk faster to keep up with the accelerating cart. How much work does the person generate while pushing on the cart? A. 500 J B. 1000 J C. 2000 J D. 4000 J
Solution: The correct answer is B. This quantity corresponds to a calculation error of the area under the force-distance line as (100 N/2) × (20 m/2) = 500 J. The work generated by the person while pushing the cart is equal to the area under the force-distance line, according to the definition of work. The area is equal to 1/2 × (100 N - 0 N) × (20 m - 0 m) = (100 N × 20 m)/2 = 1000 J. This quantity corresponds to a calculation error of the area under the force-distance line as (100 N × 20 m) = 2000 J. This quantity corresponds to a calculation error of the area under the force-distance line as (100 N × 20 m) × 2 = 4000 J. Work is Force times distance for a constant force; for a non-constant force, work is the area under the force-distance curve. Here, we see that this is the area of the triangle, which is (1/2)base*height, or (1/2)100N * 20m = 100N*10m, or 1000J.
Which comparison best determines whether IFNγ is necessary for antidepressant-induced increases in the expression of p11? Expression levels of p11 in: A. wild-type mice versus IFNγ knockout mice, both treated with p11 B. wild-type mice versus IFNγ knockout mice, both treated with an SSRI C. wild-type mice treated with IFNγ versus wild-type mice treated with an SSRI D. wild-type mice treated with IFNγ versus wild-type mice treated with ibuprofen
Solution: The correct answer is B. While it may be possible to measure changes in p11 levels above those that resulted from administration of p11, this is not the best experimental design. Specifically, both groups should be treated with an antidepressant, or SSRI. In order to study antidepressant-induced increases in p11 expression, both test groups must be given an antidepressant, or SSRI. To study the additional effect of IFNγ, one test group should be non-affected (wild-type) and the other should lack IFNγ (IFNγ knockout). Both test groups should be treated with an antidepressant (SSRI) for comparison purposes. Both test groups should be treated with an antidepressant (SSRI) for comparison purposes; ibuprofen is not an antidepressant. The question asks about antidepressant-induced increases, so eliminate A/D since they are not administering antidepressants. Eliminate choice C because only 1 mouse receives an antidepressant. The baseline for the experiment is that antidepressent-induced increase in p11 is a given; the independent variable is the presence of IFNgamma, which is evaluated in choice B.
When the covalent attachment to alliinase is broken, PLP is still held rigidly in the active site by a salt bridge and a π-stacking interaction. These interactions are most likely provided by the side chains of which amino acids? (Note: The salt bridging amino acid is listed first.) A. Asp and Tyr B. Glu and Ser C. Arg and Tyr D. Lys and Ser
Solution: The correct answer is C. Although Tyr can form a π-stacking interaction, Asp has a negatively charged side chain that will repel, not attract, the phosphate in PLP. Glu has a negatively charged side chain, which will repel, not attract, the negatively charged phosphate in PLP. Lacking an aromatic ring, Ser is not capable of a π-stacking interaction. PLP has a negatively charged phosphate that can make a salt bridge with the positively charged side chain of Arg, and the aromatic ring in the side chain of Tyr can form a π-stacking interaction. Although the positively charged side chain of Lys can form a salt bridge with the phosphate of PLP, Ser does not contain an aromatic ring for a π-stacking interaction. Quick way: The reactant has 3 oxygens, the right compound has 4 oxygens - there is only a total gain of 1 oxygen. The only option that results in an increase of 1 oxygen is option B. .
If the energy of a photon is doubled, which of the following properties of the photon will also double? A. Amplitude B. Wavelength C. Frequency D. Intensity
Solution: The correct answer is C. Amplitude is a characteristic of waves, whereas photons constitute the particle description of light. Therefore, photons cannot be characterized by the property of amplitude. The energy of a photon is given by the relationship E = hc/l. If E is doubled, then wavelength l is halved not doubled. The energy of a photon is given by the relationship E = hf. If E is doubled, then frequency f is doubled, too, as Planck's constant h does not change. Intensity is a characteristic of waves, whereas photons constitute the particle description of light. Therefore, photons cannot be characterized by the property of intensity unless as the number of photons of a certain energy that pass through a surface per unit time.
What structural feature(s) is(are) most important to the functioning of this compound as described in the passage? A. Specific configuration of numerous chirality centers B. Multiple hydrolysable linkages C. Combination of large hydrophobic and hydrophilic regions D. Presence of a reducing sugar
Solution: The correct answer is C. Chirality is important for specific binding to other compounds. This is not a requirement of a detergent. Multiple hydrolysable linkages would facilitate metabolism. Compound 2 was a detergent used to help isolate membrane proteins. Compound 2 was used as a detergent. It liberated a protein from a membrane so that it might be isolated. The combination of large hydrophilic and hydrophobic regions allows Compound 2 to function in this capacity. The presence of a reducing sugar makes the compound susceptible to oxidation and imparts aqueous compatibility. It does not provide all the necessary components of a functioning detergent. The functioning of Compound 2 was used to extract the receptor - which is a membrane protein. A good property of detergents is that they are able to bond to hydrophobic and hydrophilic chemical groups - something that is needed to stabilize a membrane protein which normally is embedded in a similar enviornment (C).
Which of the following is NOT offered as a part of the unconventional representations of blues musicians described in the first paragraph? A. Stylish clothes B. Professional photographs C. Elegant surroundings D. Valuable instruments
Solution: The correct answer is C. Clothes are used as an indicator of success in the image of both Robert Johnson and Muddy Waters. The fact that the photo of Johnson was taken in a professional studio is noted as indicative of success. The surroundings of Johnson and Waters are not discussed. The expensive Gibson guitar held by Waters is noted as an example of success.
Based on Figure 1, which patient is LEAST likely to benefit from therapy with immunosuppressors? A patient whose transplanted kidneys are associated with miRNA expression patterns characterized by: A. high levels of miR-155 B. low levels of let-7c. C. high levels of miR-30a-3p. D. low levels of miR-10b.
Solution: The correct answer is C. Normal kidney transplants have low levels of miR-155, while high levels of miR-155 are associated with acute rejection. Thus, an individual with high levels of miR-155 is likely to benefit from treatment with immunosuppressors. Acute rejection is associated with low levels of let-7c, thus individuals with low let-7c levels are likely to benefit from treatment with immunosuppressors. Based on Figure 1, high levels of miR-30a-3p are associated with normal transplant. These data suggest that such individuals would not benefit from the treatment with immunosuppressors. Low levels of miR-10b are present in individuals experiencing acute rejection. Thus, treatment with immunosuppressors would benefit them. Figure 1 shows that acute rejections show increases in miRNAs for the first 3 miRNAs, and decreases in the last 3 miRNAs. Since these miRNA trends are associated with autoimmune activity leading to rejection, an immunosuppressive agent would be ineffective against a patient with miRNA expression patterns that do NOT follow these trends. Of this, C matches the option that goes against this trend
Which biochemical technique requires a pH gradient? Jack Westin Advanced Solution will be added here. A. Limited proteolysis B. Southern blotting C. Isoelectric focusing D. SDS-PAGE
Solution: The correct answer is C. Proteolysis involves the cleavage of peptide bonds and does not require a pH gradient. Southern blotting is a technique to identify specific DNA fragments and does not require a pH gradient. Isoelectric focusing separates proteins based on their pIs. The technique uses an electric field and a pH gradient which causes proteins to stop moving at a pH equal to their pI. SDS-PAGE is a technique to separate proteins based on their size and does not require a pH gradient.
If only [I] is increased, then [ESI] or [EI] increases. This is an example of: A. the Bose-Einstein Principle. B. the Heisenberg Uncertainty Principle. C. the Le Châtelier's Principle. D. the Pauli Exclusion Principle.
Solution: The correct answer is C. The Bose-Einstein Principle states that a collection of atoms cooled close to absolute zero will coalesce into a single quantum state. The Heisenberg Uncertainty Principle states that one cannot know both the momentum and position of an object with absolute certainty. Le Châtelier's Principle states that in a reversible process, the application of stress to the system will cause the system to respond in a way that will relieve this stress. In this case, the reversible process is E + I forming EI or ES + I forming ESI. In either case, increasing [I] induces stress of the system, and the system relieves that stress by converting I to either more EI or ESI. The Pauli Exclusion Principle states that two or more identical fermions, e.g. electrons, cannot occupy the same quantum state.
Certain types of kidney tumors continuously produce and release EPO. Such tumors most likely have which of the following effects, if any, on erythrocyte production? Jack Westin Advanced Solution will be added here. A. The liver will take over the process of regulating erythrocyte production. B. Erythrocyte production within the bone marrow will cease. C. Constant stimulation of erythrocyte production will occur within the bone marrow. D. There is no effect; erythrocyte production will continue to be regulated by the kidneys based on oxygen levels within the body.
Solution: The correct answer is C. The liver is not a center that directly produces erythrocytes. An increase in circulating EPO will cause an increase, not a decrease, in bone marrow production of erythrocytes. Increased levels of circulating EPO will result in increased bone marrow stimulation to produce more erythrocytes. Circulating EPO will interact with its receptor located on erythrocyte precursors within the bone marrow and will result in increased erythrocyte production.
Based on the passage, which statement is most likely to be true of luftmenschen? A. They tend to value innovation but do not necessarily have innovative ideas. B. They are more likely to be interested in the present than in the future. C. They are likely to be less pragmatic than theoretical. D. They are rarely found outside of urban environments.
Solution: The correct answer is C. There is no passage support for this specific claim about the relationship of luftmenschen to innovation. This option is not supported by the passage; if anything, the luftmenschen, who "lived on air," and are thus portrayed as less than pragmatic, might conceivably be more interested in the abstract future than the concrete present. The passage describes the luftmenschen as those who "lived on air" (first paragraph) and then contrasts the presence of people like this with the reality that "there comes a point when something tangible must be made." These two things together suggest that the luftmenschen are less practical than theoretical. While the passage indicates that "cities are indeed filled" with luftmenschen, this does not necessarily mean that they are rarely found outside of cities.
The graph represents the energy change in an exothermic reaction: A + B → C Which of the following expressions gives the activation energy for the reaction? A. E3 − E1 B. E2 − E1 C. E3 − E2 D. None of the above
Solution: The correct answer is C. This expression gives the activation energy for the reverse (endothermic) reaction. This expression represents the thermodynamic energy of the reaction. This expression gives the energetic barrier that the reactants must overcome before forming product. It is known as the activation energy. Among the responses, E3 - E2 represents the activation energy.
What is the number of neutrons in the nucleus of the atom used to produce laser radiations? A. 48 B. 49 C. 50 D. 51
Solution: The correct answer is C. This is less than the number of neutrons in the 8636Kr atom which contains 36 electrons, 36 protons, and 86 - 36 = 50 neutrons. This is one neutron less than the number of neutrons in the 8636Kr atom which contains 36 electrons, 36 protons, and 86 - 36 = 50 neutrons. The 8636Kr atom contains 36 electrons and 36 protons. Therefore, the number of protons is equal to 86 - 36 = 50 neutrons. This is one more than the number of neutrons in the 8636Kr atom which contains 36 electrons, 36 protons, and 86 - 36 = 50 neutrons.
Which type of reaction has a Keq > 1 and is kinetically fast? Jack Westin Advanced Solution will be added here. A. Endergonic with high activation energy level B. Endergonic with low activation energy level C. Exergonic with high activation energy level D. Exergonic with low activation energy level
Solution: The correct answer is D. A reaction that has a Keq > 1 is exergonic, not endergonic. Additionally, a high activation energy results in a kinetically slow, and not a fast reaction. A reaction that has a Keq > 1 is exergonic, not endergonic. A high activation energy results in slow, not fast, reaction kinetics. A reaction that has a Keq > 1 is exergonic and a low activation energy results in fast reaction kinetics.
What is the name of the ionic compound used to make Buffer B? A. Ammonium formate B. Ammonium carbonate C. Ammonium bicarbonate D. Ammonium acetate
Solution: The correct answer is D. Ammonium formate is NH4HCO2. Ammonium carbonate is (NH4)2CO3. Ammonium bicarbonate is NH4HCO3, a component of Buffer A. Ammonium acetate is NH4CH3CO2, a component of Buffer B.
In which of the following cellular locations does EPO most likely initially bind EPOR in erythrocyte precursor cells? A. Cytosol B. Endoplasmic reticulum C. Nucleus D. Plasma membrane
Solution: The correct answer is D. EPO is a glycoprotein and, therefore, cannot cross the membrane. Consequently, EPOR would not likely be positioned within the cytosol. The endoplasmic reticulum is a site for protein glycosylation, but not the likely site where EPOR localizes. Instead, since EPO is a glycoprotein, it cannot cross the membrane, and must bind to receptors on the cell surface. EPO is not able to cross the membrane to reach the nucleus, as it is a glycoprotein. Consequently, EPOR is not likely positioned within the nucleus. EPO, being a glycoprotein, cannot cross the cell membrane. Consequently, it is likely that EPOR is located on the surface of the plasma membrane.
Which statement would most weaken the likelihood that the MIYO system would help to create evidence-based public health solutions? A. The targeted users of the MIYO system are public health organizations rather than individuals. B. Incorporating the use of the Internet is not the only way to apply the long tail approach to public health interventions. C. The MIYO system does not necessarily make the goal of "blockbuster" interventions less desirable. D. Many public health challenges require social or political solutions that digital products and information alone cannot provide.
Solution: The correct answer is D. Even if the targeted users of the system were organizations rather than individuals, this would not make the targeted MIYO system less likely to be successful. Even if there were other ways besides incorporating the Internet to apply a "long tail," or "niche" market approach to public health interventions, this would not weaken the chance that MIYO can help to create evidence-based solutions to public health problems. The MIYO system does not consist of broadly targeted approaches. Even if such "blockbuster" interventions may still be desired in some circumstances, this would not weaken the likelihood that MIYO could help create public health solutions rooted in evidence. The MIYO system is intended to address public health challenges, and the MIYO system depends on digital products and information. So if digital products and information cannot necessarily provide responses to "many public health challenges," this would clearly weaken the chance that MIYO could help to create the necessary solutions.
In an enzyme-catalyzed reaction where enzyme concentration is held constant and substrate concentration is relatively low, which kinetic parameter will increase with the addition of more substrate? (Note: Other than substrate concentration, assume no other changes to reaction conditions.) A. KM B. kcat C. Vmax D. V0
Solution: The correct answer is D. KM, the rate constant of a reaction, does not change with changes in substrate concentration. Kcat is the reaction turnover number and does not change with changes in substrate concentration. Vmax is the maximum velocity of a reaction and is a constant property which does not change with the addition of more substrate. V0 is the initial velocity of an enzymatic reaction. At low concentrations of substrate and constant enzyme concentration, adding more substrate will increase V0 until the maximal velocity is reached. These describe Michaelis Menten Kinetics - A/B/C are inherent immutable properties of a substrate-enzyme interaction that do not change with substrate concentration. However, the more substrate there is initially, the more enzymatic reactions that will occur (V0 increase).
If the small bumps seen when half of the membrane is peeled away were chemically shown to consist of the lipid cholesterol, how would the Fluid Mosaic Model have to be modified? A. The proteins would have to be embedded less than halfway through the membrane. B. There could be no proteins in the membrane. C. The lipids would have to be embedded in the proteins. D. It would not necessarily have to be altered, but there would be less evidence supporting it.
Solution: The correct answer is D. Not necessarily, as transmembrane proteins could remain embedded within the membrane leaflet that was removed. Even if the small bumps were identified as cholesterol, proteins could still be positioned within the membrane leaflet that was removed. Because the middle of the bilayer is hydrophobic, the lipids do not need to be embedded in the proteins. The Fluid Mosaic Model indicates that the hydrophilic regions of proteins are found on the membrane surfaces, while the hydrophobic regions are buried among phospholipid tails. This arrangement allows proteins to span the membrane. Consistently, the passage mentions that, upon peeling back the top membrane layer, small bumps are observed. This is consistent with the idea that proteins span the membrane. If these bumps were identified as cholesterol, this does not necessarily disprove the model. Instead, it is possible that membrane-spanning proteins remain adhered to the leaflet that has been peeled back.
In contrast to chemical messengers in classical endocrine signaling systems, chemical messengers in paracrine and autocrine signaling systems are: Jack Westin Advanced Solution will be added here. A. secreted by neurons only. B. not likely to bind cell receptors. C. usually transported by ducts. D. not transported by blood vessels.
Solution: The correct answer is D. Paracrine and autocrine chemical messengers can be secreted by cell types other than neurons. Neurosecreted molecules are normally specifically defined as neurotransmitters. Endocrine, paracrine, and autocrine chemical messengers all exert their effects by binding to receptors on or in cells. Ducts transport exocrine factors. Endocrine, paracrine, and autocrine chemical messengers are not exocrine factors. Endocrine signals are released into the blood in order to impact tissues scattered throughout the body. By contrast, paracrine and autocrine chemical messengers act locally. Paracrine chemical messengers act on cells near the cell that secretes the messenger, and autocrine chemical messengers act on the same cell that secretes the messenger. Therefore, paracrine and autocrine factors do not need to be taken up in the blood to act on distant tissues.
Which physical property does NOT change with the amino acid substitution made in TPMT*5? A. Molecular weight B. Hydrophobicity C. Hydrogen bonding capability D. Net charge
Solution: The correct answer is D. The side chain of the L49S variant (TPMT*5) at position 49 has a different molar mass than the side chain of Leu. The L49S variant (TPMT*5) has a hydrophilic residue at position 49, but wild-type TPMT has a hydrophobic residue at that position. The L49S variant (TPMT*5) can hydrogen bond, but wild-type TPMT cannot. The L49S variant (TPMT*5) has the same charge as wild-type TPMT because the amino acid residues do not have charge.
Based on the passage, the "many blues tourists" referred to in paragraph 4 seek an experience most like: A. that of theater-goers who prefer to see plays performed in the language they were written in. B. that of film-goers who generally attend more popular films before they see more obscure films. C. that of sports fans who attend professional sporting events but rarely watch them on television. D. that of viewers who prefer to see artists' drawings that were never advertised for sale.
Solution: The correct answer is D. There is no implication that the tourists seek something specifically "original," or seek to find music that is closest to the way it was originally done or performed. There is no implication that tourists are seeking something particularly popular. There is no implication that the tourists prefer live to televised or filmed events. The "many tourists" mentioned in the passage seek a blues culture "apparently untainted by the more commercial aspects of marketing and tourism" (paragraph 4). This is the only option that mentions an interest in something that is not for sale.
Based on the passage, the microbiome of CD-affected individuals will result in which physiological change? A. Increased polypeptide digestion B. Slower dietary fiber absorption C. Increased amount of propionate D. Decreased immune tolerance
Solution: The correct answer is D. There is no information indicating that any of the bacteria listed in Table 1 impact polypeptide digestion. Dietary fibers are not absorbed by the human intestine. Based on Table 1, the two bacteria that produce propionate, Phascolactobacteria and Odoribacter, are decreased in CD-affected individuals. Thus, the levels of propionate are most likely decreased, rather than increased, in CD-affected individuals. As explained in the passage, molecules such as butyrate, acetate, and propionate inhibit the inflammatory response against commensal bacteria of the GI tract. CD is an inflammatory condition, in which Table 1 indicates that many bacteria producing these anti-inflammatory molecules are reduced during CD. This is consistent with the interpretation that inflammation observed during CD is caused by decreased production of propionate, butyrate, and acetate, which decreases immune tolerance of commensal bacteria. The table caption indicates CD individuals have higher Clostridia and Enterobacter populations and lower bacterial populations of the first 5 listed bacteria. Since propionate is produced by 2 of the first 5 bacteria, this should be reduced rather than increased in CD individuals (eliminate C). Polypeptide digestion is not discussed in any of the bacteria, nor is absorption (dietary fiber is not absorbed regardless). By process of elimination you can choose D, but can also arrive at this answer by recognizing that the SCFAs "attenuate the inflammatory response" (reduce it). It looks like the majority of SCFAs were produced by bacteria reduced in CD individuals, so these individuals have uninhibited inflammation (hallmark of CD), or reduced immune tolerance.
Based on the passage, which of the following is closest to the pressure exerted on the chest by a 10 × 5 cm rectangular paddle during defibrillation? (Note: 1 Pa = 1 N/m2.) A. 5 kPa B. 10 kPa C. 15 kPa D. 20 kPa
Solution: The correct answer is D. This is 1/4 of the actual pressure. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2)=20 kPa. This is 1/2 of the actual pressure. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa. This is 3/4 of the actual pressure. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa. Pressure is the ratio of the force applied perpendicular to a surface and the area of the surface. Because the normal component of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa.
Which Trp residue of carbonic anhydrase can be selectively modified with CHCl3 at 20°C? A. W4 B. W96 C. W190 D. W243
Solution: The correct answer is D. When W4 is modified by CHCl3 at 20ºC using ultraviolet light, W190 and W243 are also modified. W96 is never modified by CHCl3 at 20ºC using ultraviolet light. When W190 is modified by CHCl3 at 20ºC using ultraviolet light, W4 and W243 are also modified. Based on the data in Table 1, only W243 is modified by CHCl3 at 20ºC using ultraviolet light.
Assuming that protein synthesis was underway when the radioactive amino acids were added, which of the following best describes how the radioactivity was distributed in one of the first molecules of Protein X that was completely translated? A. Radioactive amino acids were randomly located throughout the molecule. B. Radioactive amino acids were located only at one end of the molecule. C. Radioactive amino acids were located at both ends, but not in the middle, of the molecule. D. Radioactive amino acids were located in the middle, but not at either end, of the molecule.
That's the MCAT for you... But i'll explain. Nowhere in the passage does it specifically state that B is the correct option; However, we can make a good inference. Look at the other answers... We know that translation has already begun, therefore we know that any answer that says at the beginning of the protein is out of the question (Eliminate C). Next, we can assume that at least one end of the protein will have radioactive amino acids, therefore, any answer that says "neither end" can be eliminated (eliminate D). Finally, it tells us that radioactive amino acids were only added half way through translation; therefore, there is no way for them to be "evenly" distributed throughout the protein. They would have to be primarily on one end of the amino acid (eliminate A). B is the answer. Assuming that protein synthesis was under way when the radioactive amino acids were added, which of the following best describes how the radioactivity was distributed in one of the first molecules of Protein X that was completely translated? A. Radioactive amino acids were randomly located throughout the molecule. Jack Westin Advanced Solution: This answer is incorrect because it implies that protein synthesis is random and that amino acids can be replaced during translation. Proteins are systematically synthesized from N terminus to C terminus; the random allocation of radioactive amino acids is inconsistent with this systematic synthesis B. Radioactive amino acids were located only at one end of the molecule. Jack Westin Advanced Solution: This is a good answer. The question stem notes that protein synthesis was under way, or already initiated, when the radioactive amino acids were added. This suggests that at least the N terminus had been translated, and possibly the middle portion of the protein as well. We have no way of knowing exactly how much of the protein had already been translated from the question stem alone, but it's fair to say that at least the C terminus would contain radioactive amino acids while the N terminus would not. C. Radioactive amino acids were located at both ends, but not in the middle, of the molecule. Jack Westin Advanced Solution: For only the middle of the protein to be free of radioactive amino acids, the center would have had to have been translated before the addition of the radioactive amino acids and the ends afterwards. However, we know that the middle of an mRNA transcript is translated after the N terminus so this answer choice is incorrect. D. Radioactive amino acids were located in the middle, but not at either end, of the molecule. Jack Westin Advanced Solution: This is the opposite of answer choice C and is also incorrect. For this to be true, proteins would need to be synthesized from the outside in with both the N and C termini synthesized at roughly the same time. We know this is not true and that proteins are synthesized in the following order: N terminus -> middle -> C terminus. In the absence of an answer choice that takes into consideration that possibly both one end and the middle have radioactive amino acids, answer choice B remains the best answer.
glucose isomers are...
The 16 optical isomers of glucose are allose, altrose, galactose, glucose, gulose, idose, mannose, and tallose
Which amino acid exhibits a beta-branched side chain? Jack Westin Advanced Solution will be added here. A. Alanine B. Leucine C. Isoleucine D. Glycine
The AAMC explanation is bad. The question specifically states BETA-branched, not just branched, meaning branching at the beta carbon. Alpha carbon is one carbon away from the carbonyl carbon, while beta carbon is one farther from that. Leucine is branched at the gamma carbon, which is one carbon farther than the beta carbon. Isoleucine is branched at the beta carbon. This should help. Solution: The correct answer is C. Alanine does not contain a branched side chain. Leucine does not contain a branched side chain. Isoleucine contains a branched side chain. Glycine does not contain a branched side chain.
From Table 1, in which metabolic process are GI tract bacteria directly involved? Jack Westin Advanced Solution will be added here. A. Conversion of PS into short chain fatty acids B. Absorption of amino acids C. Fermentation of dietary fibers into peptides D. Absorption of monosaccharides
The GI tract bacteria discussed show that these bacteria are relevant in catabolism of nutrients in the gut - they are not responsible for absorption of nutrients (rule out B/D). Fibers which are made of carbohydrates cannot be easily converted into nitrogenous peptides. However, they can convert polysaccharides (PS) into the short chain fatty acids (butyrate, propionate, acetate). Solution: The correct answer is A. GI tract bacteria convert monosaccharides and polysaccharides into acetate, propionate, and butyrate, which are short chain fatty acids. The table does not indicate a direct role for amino acids in metabolic functioning of GI tract bacteria, but rather that such bacteria convert polysaccharides into short chain fatty acids. Based on the table, dietary fibers are not converted into peptides, but short chain fatty acids (acetate, propionate, and butyrate). Table 1 indicates that GI tract bacteria use monosaccharides as an energy source, but does not indicate whether these bacteria function in monosaccharide absorption.
What is the net volume of fresh air that enters the alveoli each minute, assuming that the breathing rate is 10 breaths/min, the tidal volume is 800 mL/breath, and the nonalveolar respiratory system volume (dead space) is 150 mL?
The amount of air entering the lungs in a single breath, or tidal volume, is given as 800 mL/breath. Of that 800 mL only 650 mL reaches the alveoli per breath (800 mL of air inhaled minus 150 mL of nonalveolar respiratory volume). Therefore the net volume of air that reaches the alveoli each minute is equal to 650 mL/breath multiplied by 10 breaths/min, or 6500 mL.
What is the empirical formula of the conjugate acid of the anion in strontium acetate, Sr(CH3COO)2? CH2O C2H4O2 SrC2H3O2 NaC2H3O2
The anion is acetate, the conjugate acid of which is acetic acid, formula C2H4O2. However, the question asks for the empirical formula, so this must be reduced to the lowest common denominator, which gives CH2O, or Answer A. Answer choice B cannot be an empirical formula, because it can be reduced to lower terms. Answer C is a salt, not an acid. Answer choice D is also a salt rather than an acid, so neither C or D can be the conjugate acid of acetate.
MS2-GFP
The bacterial genes of interest are cat and lacY. The viral or phage gene of interest is binding sequence (bs). The investigators took 6 copies of binding sequence (hence 6xbs) and inserted it upstream of the bacterial genes cat and lacY. They did this so that when the bacteria transcribe the operon containing cat and lacY, the viral DNA 6xbs will be transcribed too. The viral MS2 protein, which is fused to green fluorescent protein, binds to 6xbs mRNA and fluoresces. This reveals the location of the cat and lacY transcripts because they are connected to 6xbs.
Which adaptive immunity cell type is most effective at removing a cancerous cell from the body without the assistance of other immune cells? Cytotoxic T Lymphocyte Regulatory T Lymphocyte Helper T Lymphocyte Natural Killer Cell
The correct answer is A since the Cytotoxic T Lymphocytes are very specific for the cells that they kill and they are designed to kill virus-infected or cancerous cells. Regulatory (Answer B) and Helper (Answer C) T Lymphocytes are very specific, but they only aid in the removal of cancer cells by preventing wrongful activation (regulatory) or by activating B cells (helper). Further, the clue given in the stem, "without the assistance of other immune cells," negates the possibility of a helper cell such as the helper T Lymphocyte. Natural killer cells (Answer D) also kill cancerous cells, but they are innate immune cells and the stem specifies adaptive immune cells.
Assume that the rate of protein synthesis is greater when the culture is incubated at 45°C than at 37°C. Which of the following graphs shows how temperature will influence the results in Figure 1?
The graphs here show the quantity of radioactivity on the y axis (which is directly proportional to the amount of protein) over time on the x axis. Rate of synthesis would then be shown as slope (change in quantity of radioactivity/protein over change in time). That's why you were right to look for a higher slope to represent the higher rate of synthesis. edit. As for why the maximum level of radioactivity didn't change, it's potentially the result of a equilibrium between protein synthesis and degradation (e.g. faster synthesis causes faster degradation). But without further information we can't know where the maximum radioactivity should plateau at, so the plateau height is irrelevant to choosing an answer; our only solid criteria here is higher slope. The answer to this question is B because in this experiment, the rate of protein synthesis is directly proportional to the level of radioactivity, and option B is the only option that exhibits a faster increase in radioactivity at 45 °C than at 37 °C.
An ice cube at 0°C and 1 atm is heated to form steam at 100°C and 1 atm. Ignoring heat loss to the surroundings, what part of the process uses the most heat? (Note: Specific heat of water = 1 cal/g°C. Heat of fusion = 80 cal/g. Heat of vaporization = 540 cal/g.) A. Melting the ice cube B. Heating all the water from 0°C to 50°C C. Heating all the water from 50°C to 100°C D. Vaporizing all the water
The highest heat consumption will be the one with the most calorie usage. Note that you can eliminate B and C without calculating, because the temperature difference is the same and the specific heat of liquid water is stated to be constant and independent of temperature - if B and C are equivalent, neither of them can be the single highest correct answer. Melting the ice cube is fusion, and would be 80cal/g. Vaporizing the water is - vaporization?! - which is 540cal/g and is the highest. Solution: The correct answer is D. Melting the ice cube requires only 80 cal/g, which is less than the heat of vaporization. Heating all the water from 0°C to 50°C requires 50 cal/g, which is less than the heat of fusion. Heating all the water from 50°C to 100°C requires 50 cal/g, which is less than the heat of fusion and the heat of vaporization. Vaporizing all the water requires 540 cal/g, which is a greater heat requirement than specific heat or heat of fusion.
p38 phosphorylation
The p38 mitogen-activated protein kinase (MAPK) is activated through phosphorylation primarily by MAPK kinase 3 (MKK3) and MKK6 in response to cellular stress and cytokines. The p38 MAPK pathway functions in the control of differentiation, the blockade of proliferation, and in the induction of apoptosis (1).
What is the best experimental method to analyze the effect of tdh2 gene deletion on the rate of histone acetylation? Comparing histone acetylation in wild-type and Δtdh2 cells by: Western blot
The process of Histone Acetylation and deacetylation is part of regulating whether or not a piece of DNA is open or closed for transcription. However, histone "acetylation" is post translational because the acetyl groups are removed from or added to histone proteins and NOT DNA. Therefore it is a post translation modification all by itself. The modified proteins as a result happen to associate with DNA to regulate whether it's condensed or open for transcription. That's what I was thinking too, but I guess I can understand why it's wrong. If they were testing for the affect of acetylation on gene expression, then that would be the way to go. But in molecular biology research, when you want to test for something, you test for it directly. To test for acetylation of a protein, you do a western blot and detect with an antibody that would bind an acetylated protein. To test for changes in gene expression due to acetylation, you would do RT-PCR, but that's not what they were doing. Can't get em all man
Which experimental evidence suggests that the purified hMPRα obtained by the researchers was in its native state? The hMPRα that was obtained: A. retained both the Compound 1 and (His)6 tags. B. was purified by two separate chromatography steps. C. exhibited a binding affinity for progesterone that was similar to that exhibited by native hMPRα. D. had a nearly identical molecular weight to hMPRα obtained elsewhere.
The tags that were used for extraction could still interfere with the protein's native function/binding, so rule out A. Similarly, purification by different steps only accounts for purity of yield - it does not ensure retention of enzymatic function (rule out B). D does not indicate retention of native state, as a completely unfolded protein would still have the same Mw. Choose C by process of elimination. C indicates that the protein retained its native function, therefore its native state. Solution: The correct answer is C. The retention of these tags does not confirm conformational stability. A protein can denature during any chromatography step. The fact that hMPRα required two steps for purification means that it was more likely to have denatured. The only way to tell for certain that a protein is still in its native state is to compare its functioning to the levels observed for protein believed to be in the native state. The researchers compared the binding affinity of progesterone of hMPRα both before (when it is assumed to be in its native state) and after releasing it from the membranes (which could conceivably cause denaturation). The fact that the affinities were the same suggests that the isolated hMPRα is indeed in the native state. Molecular weight is a measure of the stability of the primary structure. Activity requires full conformational stability.
endothelial cells
The thin layer of cells that line the interior surface of all blood vessels. Only one cell is thick in capillaries. connected by tight junctions and have a nucleus
exogenous proteins
Those from dietary sources
Medieval Times
Time period from roughly 500-1500. Also known as "Middle Ages"
Capacitance and permittivity
Two things: Q = CV Dielectrics will ALWAYS increase capacitance (C). For an isolated and charged capacitor, Q is constant because there's no battery source that can add any further charge. Since C increases with dielectric, V must decrease. For a charging capacitor, V is constant and equal to the voltage of the battery source. Since C increases with the dielectric, Q must increase. Why is Capacitance higher with greater Permittivity, but the Electric Field Weaker? Capacitance = charge that can be stored per unit area. So if the Permitivvity is higher, its like the resistance between the two plates is higher, so I reason more charge can be stored before the charge difference breaks and charge jumps from one plate to the other. While for an electric field, does the electrc field weaken because it is harder for a charge to move given that force put on it so the net force is weakened? The electric field weakens because the insulating medium has induced dipole moments caused by the electric field, essentially creating its own electric field, but in the opposite direction. The individual molecules that make up the medium will have a partially positive end and a partially negative end due to the electric field created by the capacitors (induced dipoles). The partially positive ends are attracted to the negative plate, and the partially negative ends are attracted to the positive plate (their fields are traveling in opposite directions). While quite small in magnitude compared to that of the electric field created by the capacitors, the induced dipoles still work against the capacitor's electric field, thereby weakening the overall strength.
Native PAGE SDS-PAGE Isoelectric focusing
Types of electrophoresis *Native PAGE*: maintains the protein's shape, but results are difficult to compare because the mass-to-charge ratio differs for each protein; most useful to compare the molecular size or the charge of proteins known to be similar in size *SDS-PAGE*: separates based on mass alone; denatures the proteins and masks the native charge so that comparison of size is more accurate, but the functional protein cannot be recaptured from the gel *Isoelectric focusing*: separates proteins by their isoelectric point; the protein migrates toward an electrode until it reaches a region of the gel where pH = pI of the protein; positively charged proteins migrate toward (negatively charged) cathode, negatively charged migrate toward (positively charged) anode
Michael-Menten Equation
V= (Vmax x S) (Km+S) The graph has three orders 1. When the substrate concentration is low compared to enzymes = First order so inc substrate would increase the rate of the reaction LINEAR V=kS 2. When the substrate concentration is medium, the order is mixed, still increasing the concentration would increase the rate to an extrent NON-LINEAR 3. When the subtrate concentration is extremely high, increasing the substrate wouldn't increase the rate because the enzymes are working on their maximum velocity zero order V=Vmax plateu
Enzymes alter the rate of chemical reactions by all of the following methods EXCEPT: A. co-localizing substrates. B. altering local pH. C. altering substrate shape. D. altering substrate primary structure.
What this is saying is enzymes bring substrates together so that they may react(think ATPase, brings ADP+Pi) B) Enzymes change internal pH creating a favorable enviroment for a rxn to occur that may not occur in the body. So some rxns need an acidic/basic place to do that in. C) Enzymes shape the shape of substates. (Think DHAP to GAP by TIM) D) Primary structure is the sequence of amino acids in a substrate. Also , question is talking about the RATE, what can an enzyme do to speed up the rate of a rxn. D doesn't really make sense here D. The enzyme (a protease to be specific) does not directly alter the substrate's primary structure. Rather, it only helps brings the substrates together so that the reaction occurs more rapidly. If the enzyme did directly cause a change in primary structure (i.e., a chemical reaction), then the enzyme is not considered a catalyst. Thus, Choice D is false, and the correct answer to this question. Enzyme's can change the pH very locally, in the region of the active site. While the entire solution may remain at pH 7.4, in the local area around the reacting substrate, the enzyme may donate or accept protons, changing the pH but only at that specific area
Which statement is best supported by the data using the anti-AcTubK40 antibody?
When βOHB is inhibiting:↑ [ βOHB ] → ↓ deacetylation activity → ↑ Acetylation level According to Figure 1, βOHB does not change AcTubk40 acetylation level, but changes the other two, showing that βOHB is specific to the other two, thus it is not a general deacetylase inhibitor.
Which action(s) could contribute to the positive inotropic effect of digoxin on cardiac myocytes? Decrease transport of Ca2+ to the extracellular environment. Increase availability of intracellular Ca2+ to bind to troponin. Increase overall Ca2+ stores in the sarcoplasmic reticulum. A. II only B. III only C. I and II only D. I, II, and III
Which action(s) could contribute to the positive inotropic effect of digoxin on cardiac myocytes? Decrease transport of Ca2+ to the extracellular environment. Increase availability of intracellular Ca2+ to bind to troponin. Increase overall Ca2+ stores in the sarcoplasmic reticulum. Jack Westin Advanced Solution: A positive inotrope increases the force of contraction and contractility which is mediated by intracellular calcium levels. An increase in intracellular calcium will positively affect contractility. A. II only Jack Westin Advanced Solution: Roman numeral II explicitly mentions an increase in intracellular calcium; it will increase contractility. We would need to evaluate the other roman numerals before deciding if only roman numeral II is true. B. III only Jack Westin Advanced Solution: Roman numeral III, an increase in calcium storage in the sarcoplasmic reticulum, will mean that more calcium is available for release intracellularly. This will increase contractility. Because both roman numerals II and III are true, neither answers A nor B are correct. C. I and II only Jack Westin Advanced Solution: Because roman numerals II and III are both correct, this answer choice must be incorrect. D. I, II, and III Jack Westin Advanced Solution: I, II, and III- This is the only answer that shows that both roman numerals II and III are correct meaning that this is the correct answer and that roman numeral I must also be correct. On test day we would select this answer and move on. We're not yet at test day so let's evaluate roman numeral I as well. Decreasing transport of calcium to the extracellular environment means that there will be more calcium in the intracellular environment which will increase contractility. Roman numeral I is therefore correct as well.
Which component of the electron transport chain is defective in cells from an SDH-linked paraganglioma tumor? A. Complex I B. Complex II C. Complex III D. Complex IV
Which component of the electron transport chain is defective in cells from an SDH-linked paraganglioma tumor? A. Complex I Jack Westin Advanced Solution: Succinate dehydrogenase is not found in Complex I of the electron transport chain. The corresponding Complex I enzyme is NADH reductase. B. Complex II Jack Westin Advanced Solution: This is correct, complex II uses succinate dehydrogenase. It is the only enzyme involved in both the citric acid cycle and the electron transport chain. C. Complex III Jack Westin Advanced Solution: SDH is found in complex II and not complex III. D. Complex IV Jack Westin Advanced Solution: SDH is found in complex II and not complex IV.
Which statement correctly identifies an enzyme involved in DNA replication and describes its primary function? A. Ligase catalyzes the binding of RNA primers to DNA via phosphodiester bonds. B. Primase catalyzes the replacement of RNA primer nucleotides with DNA nucleotides. C. Helicase catalyzes the separation of the parent DNA strands at the origin of replication. D. Topoisomerase catalyzes the joining of adjacent Okazaki fragments into a continuous strand of DNA.
Which statement correctly identifies an enzyme involved in DNA replication and describes its primary function? A. Ligase catalyzes the binding of RNA primers to DNA via phosphodiester bonds. Jack Westin Advanced Solution: A. Primase and not ligase is the enzyme that lays down the RNA primer. Ligase joins the Okazaki fragments formed during replication of the lagging strand. B. Primase catalyzes the replacement of RNA primer nucleotides with DNA nucleotides. Jack Westin Advanced Solution: B. As noted above, primase lays down the RNA primer on the lagging strand but is complementary to the DNA strand. It does not replace any of the original nucleotides. C. Helicase catalyzes the separation of the parent DNA strands at the origin of replication. Jack Westin Advanced Solution: C. Helicase "unzips" the strands of DNA at the beginning of replication. This occurs at the origin of replication. This is the best answer. D. Topoisomerase catalyzes the joining of adjacent Okazaki fragments into a continuous strand of DNA. Jack Westin Advanced Solution: D. Topoisomerase prevents the re-coiling of the DNA strands during replication by addressing the DNA strands' tendency towards supercoiling. As noted in answer choice A, ligase is the enzyme responsible for joining the Okazaki fragments. A Primase B DNA pol2 D Ligase 3 ReplyShare level 2MCAT2019QuestionsOP·4 yr. ago For A, why would a primase do this? Why do RNA primers ever bind to DNA via phosphodiester bonds. I didn't think this really happened??
Which type(s) of restriction enzyme(s) can recognize the HIF binding sequence? A restriction enzyme that has: a four-base recognition sequence a six-base recognition sequence an eight-base recognition sequence A. I only B. II only C. III only D. I and II only
Which type(s) of restriction enzyme(s) can recognize the HIF binding sequence? A restriction enzyme that has: a four-base recognition sequence a six-base recognition sequence an eight-base recognition sequence Jack Westin Advanced Solution: The first paragraph indicates that the HIF binding sequence is CCCCGGGC. Notice that the first and last nucleotides are not complementary and are thus not a part of the palindrome. Recall that a palindromic sequence is a sequence where both DNA strands have the same sequence when read from 5' to 3.' Another way to think about this is that the complement of the sequence is the same as reading the sequence in reverse. The palindromic sequence is six nucleotides long and is CCCGGG. A. I only Jack Westin Advanced Solution: A four base recognition sequence is smaller than the six nucleotides that make up the palindromic sequence so the restriction enzyme should be able to recognize the HIF binding sequence. B. II only Jack Westin Advanced Solution: We already established that roman numeral I is correct so this answer choice is incorrect. C. III only Jack Westin Advanced Solution: An eight base recognition sequence is too long for a six nucleotide palindrome. If you selected this answer, you might have missed that the first and last nucleotides in the sequence are not complementary and therefore are not a part of the palindromic sequence. D. I and II only Jack Westin Advanced Solution: Time to evaluate roman numeral II. A six base recognition sequence would be able to recognize the six nucleotide palindromic sequence so both roman numerals I and II are correct making this the correct answer. Restriction enzymes can be really tricky for a lot of students so we've broken down the topic here: https://jackwestin.com/resources/mcat-content/recombinant-dna-and-biotechnology/restriction-enzymes
Within the intestines, unabsorbed fats are broken down into fatty acids by intestinal bacteria. Given this, excess unabsorbed fats most likely have which of the following effects within the intestines? A. They increase the osmotic pressure within the intestines, leading to diarrhea. B. They decrease the osmotic pressure within the intestines, leading to diarrhea. C. They increase the osmotic pressure within the intestines, leading to constipation. D. They decrease the osmotic pressure within the intestines, leading to constipation.
Within the intestines, unabsorbed fats are broken down into fatty acids by intestinal bacteria. Given this, excess unabsorbed fats most likely have which of the following effects within the intestines? A. They increase the osmotic pressure within the intestines, leading to diarrhea. Jack Westin Advanced Solution: If there is an excess of unabsorbed fats, there will be a disruption of normal electrolyte and water reabsorption. The sodium absorption that is usually stimulated by short chain fatty acids and the bile salt reabsorption that occurs with each meal will decrease. The excess solutes in the intestine that would otherwise have been absorbed increase the osmotic pressure, and there will be increased fluid in the intestines which in turn promotes diarrhea. B. They decrease the osmotic pressure within the intestines, leading to diarrhea. Jack Westin Advanced Solution: The presence of additional unabsorbed solutes in the intestines will increase osmotic pressure, not decrease the osmotic pressure. C. They increase the osmotic pressure within the intestines, leading to constipation. Jack Westin Advanced Solution: An increase in osmotic pressure within the intestines would mean more water is remaining in the intestines, promoting diarrhea and not constipation. If the question were asking about a scenario that would result in less fluid in the intestinal lumen, then constipation would be an appropriate expectation. D. They decrease the osmotic pressure within the intestines, leading to constipation. Jack Westin Advanced Solution: As noted above, the presence of additional unabsorbed solutes in the intestinal lumen will increase the osmotic pressure compared to if the fats had been appropriately absorbed.
Based on the passage, is CatL expression sufficient for VSV-EGP infection of the mouse cell lines presented in Figure 1? (Note: In these experiments, assume that values of <3% of total cells infected are too low to be measured accurately.)
Yes, because VSV-EGP infects cells expressing CatL better than it infects cells not expressing CatL Yes, because VSV-EGP infects cells expressing CatB better than it infects cells not expressing CatB No, because VSV-EGP does not infect CatB-/- cells expressing CatL better than it infects CatB-/- cells not expressing CatL No, because VSV-EGP infects cells expressing both CatB and CatL better than it infects cells expressing CatB but not CatL C
Which experimental approach(es) can be used to analyze the effect of ROS on the lifespan of yeast? Comparing the lifespans of: wild-type yeast versus yeast lacking antioxidant enzymes wild-type yeast versus yeast overexpressing antioxidant enzymes yeasts growing in the presence or absence of hydrogen peroxid
You can think of it as a positive/negative experimental controls. I is the presence without antioxidant enzyme (meaning they can't process ROS to form something safer) --> negative control. II is the presence of many antioxidant enzymes (meaning they can process ROS more quickly) --> positive control. If ROS has a direct causal effect on yeast lifespan, than you would expect yeast without antioxidant enzymes (I) to have shorter lifespan and yeast with more antioxidant enzymes (II) to have a longer lifespan
Microfilaments
a small rodlike structure, about 4-7 nanometers in diameter, present in numbers in the cytoplasm of many eukaryotic cells. involve in gross movements within the cell cell division when a cell divides into two cells amoeba picking up food made of actin dynamic so actin polymerization and depolymerization helix
protein glycosylation
addition of carbohydrates to proteins -each step is strictly dependent on the previous step I think some low yield stuff that I kind of had in the back of my mind as a backup was for glycosylation, which might be a good way to justify educated guesses in BB. Asparargine or arginine side chains can be glycosylated by N-linked glycans, and O-linked glycans are usually attached to Ser, Tyr, and Thr. Let me know if I'm wrong. It's been a few months since I took the MCAT and this is pretty low yield information, since usually the question will give you other information to get to these answers. Glycosylation refers to the systematic addition of one or more carbohydrates to proteins or lipids. This process requires specific glycosyltransferases, usually located in the smooth endoplasmic reticulum and the Golgi apparatus.
ostensibly (adv)
apparently or purportedly, but perhaps not actually
mixed inhibitors
bind at other site than active site, does not prevent substrate from binding. can bind to either enzyme alone or enzyme-substrate complex; has either competitive or uncompetitive effects; all lower Vmax to some extent I was a bit confused on this at first as well. So essentially with mixed, uncompetitive, and noncompetitive inhibition, the inhibitor is going to bind on an allosteric site on either the lone (i.e. "empty") enzyme or the enzyme-substrate complex (i.e. "filled" enzyme). In noncompetitive inhibition, it is equally likely to bind on either the empty enzyme or filled enzyme, and thus there is no change in Km as the two possibilities cancel out. In mixed, it can prefer the empty enzyme (thus increasing Km cuz the empty enzyme is less likely to be filled) or filled enzyme (thus decreasing Km cuz it's more likely to be filled). In uncompetitive inhibition, it only binds to filled enzymes. If you have an uncompetitive inhibitor, the inhibitor can only bind to the Enzyme-Substrate complex (and that prevents the complex from making the product). For an uncompetitive inhibitor, both Km and Vmax decreases. However, a mixed inhibitor binds to EITHER the free enzyme OR the enzyme-substrate complex. However, it does so with unequal affinities (unlike noncompetitve inhibition). Therefore depending on if the mixed inhibitor has a stronger affinity for the free enzyme or the complex, Km will increase/decrease! Hope this helps (: edited 4y ago Similarities: Both noncompetitive mixed inhibitors bind at allosteric sites, and bind to both enzyme and the enzyme substrate complex. Vmax decreases in both cases. Differences: Noncompetitive: equal binding affinity to both enzyme and enzyme-substrate complex (km no change) Mixed: differing binding affinity to enzyme and enzyme-substrate complex If enzyme more affinity, km increases If enzyme-substrate complex more affinity, km decreases. Hope this helps! 2 ReplyShare level 1djp219· 4y ago513 (129/126/128/130) Mixed inhibition has equal affinity for substrate and E-S complex, so therefore affects both Km and Vmax while NC inhibition lowers Vmax but doesn't affect Km
endoderm
bronchi, bladder, stomach, urethra
When viewing an X ray of the bones of a leg, a doctor can tell if the patient is a growing child, because the X ray shows:
cartilaginous areas in the long bone The epiphysis is normally referred to as the growth plate. The physis itself is the cartilage between the two ossification centers. Note however that the epiphysis in very young children is not ossified yet, so then there's only an empty area at the end of the bone (of course not really empty, but fully made up by cartilage so not visible on x-ray).
In humans, eggs and sperm are most similar with respect to:
cell size genome size time required for development the numbers produced by a single individual. b both egg and sperm contain haploid number of chromosomes so most similar with regards to genome size
Topoismerase
corrects "overwinding" ahead of replication forks by breaking, swiveling, and rejoining DNA strands I'm reviewing DNA replication currently, and I'm trying to conceptualize the positive supercoiling that occurs with helicase and the negative supercoiling that occurs with DNA Topoisomerase II/Gyrase to counteract that strain. Can anyone recommend an easier way to visualize this? Any youtube videos perhaps? Thanks in advance! Its neither. Topoisomerase is a gyrase enzyme. It relieves positive supercoiling upstream from helicase. Endo/exo nucleases are involved in DNA repair. Endonucleases remove pieces of DNA on the interior of the strand and exonucleases remove DNA either at the 5' or 3' ends (typically as the DNA pol is synthesizing it).
Paramecium is
eukaryote
precipitious
extremely steep
mast cells
found in the connective tissue of the dermis; respond to injury, infection, or allergy by producing and releasing substances, including heparin and histamine
Oratories
groups of clergy who banded together for the purpose of prayer, meditation, and mutual support as they participated in discussions about how they might reform the church
Work done by longer and shorter limbs
he answer to this question is C because work is equal to the product of the force applied to an object and the distance over which the distance is moved. Individual N has shorter limbs (and likely a shorter ground to shoulder distance), thus this individual will perform less work because the muscles contracted a shorter distance, and the object was moved a shorter distance. A is incorrect because the stem indicates that Individual M has longer limbs than does Individual N. Given this, Individual M will have done more work than Individual N because the distance the muscles contracted when moving the load from ground to shoulder (also a greater distance for Individual M) is greater. B is incorrect because work is equal to the product of the force applied to an object and the distance over which the distance is moved. Thus, moving an object a longer distance would result in more, not less, work being done. D is incorrect because while it is true that Individual M did more work, work is not dependent on speed, and thus, the rate of muscle contraction is not relevant to the amount of work done by the individuals Assuming two muscles have the same CSA, the longer muscle would perform more work because work is the product of force times the distance shortened.
Based on Table 1, what is most likely associated with a reduction in gram-positive bacteria? A. Increase in acetate production B. Increase in use of acetate C. Decrease in pH D. Decrease in PS production
he gram-positive bacteria are Roseburia, Ruminococcaceae, and Clostridia. If these were to be reduced, acetate production would be reduced; note that even though Roseburia consumes acetate and you could argue for less acetate levels, note that the answer talks about acetate production specifically and not acetate levels or concentration. Similarly, for B, reduction in Roseburia would result in less acetate consumed. This specific wording is how acetate consumption AND acetate production could both be reduced with reduction of g+ bacteria. Ruminococcaceae convert protons into acetate (consumption of protons would lead to fewer H+, or higher pH); reduction of this process results in a more acidic (lower pH) environment (choose C). PS production is irrelevant for g+ bacteria. Solution: The correct answer is C. Based on Table 1, a decrease in gram-positive bacteria (Ruminococcaceae and Clostridia) will be linked to a reduction, not an increase, in acetate production. According to Table 1 only Roseburia use acetate as a source of energy. Roseburia are gram-positive bacteria. Thus, a reduction of gram positive bacteria is most likely to be associated to a reduction, not an increase, in the use of acetate. As indicated by the table, Ruminococcaceae consume H+. Reduction in gram-positive Ruminococcaceae results into higher levels of H+ and, as a consequence, an acidic pH. None of the bacteria assessed, regardless of gram status, produce polysaccharides (PS). Consequently, reducing the number of gram-positive bacteria will not likely affect PS production.
Micro tubules
hollow tubule tublin protein help cell transport materials within itself resist shape changes largest element of cytoskeleton provide a track along which the vesicles move through the cell pull chromosomes to the opposite ends forms mitotic spindle
Ionophores
hydrophobic molecules that dissolve in lipid bilayers and increase permeability to specific inorganic ions
Scientists are hopeful that NPY can be used in combination with GnRH to treat certain cases of female infertility. Individuals with a deficiency in what receptor system would be most likely to benefit from such a treatment? LH NPY GnRH LH+NPY
in which receptor". Anyway, I had the same logic as you. Thought if they had defective LH receptors, maybe a large enough increase in the amount of circulating LH could overcome the defect. I think, however, this is too nuanced and it's one of those "best answer" questions. If the LH receptors are broken, even in increase in circulating LH won't help. On the other hand, if the GnRH receptors are defect and not releasing enough LH, then presumably another pathway to release LH (via NPY) would make up for this
mito damage not making ATP but increase in heat https://youtu.be/X9bO9CVx_xs
it really depends on the type of disruption, this here scenario would happen when something known as an uncoupler (bc it uncouples the gradient from the atp production) is inserted into the inner membrane. essentially this opens another "door" for protons to reenter the matrix and bc flowing down a gradient releases energy but theres no "use" for it, its just given off as heat. aside from toxins, there is actually a physiological uncoupler known as thermogenin (aptly named) which is present in brown fat in babies and helps them keep warm, awww. for the above scenario they should say something along the lines of the inner membrane has become more permeable, use the word uncoupler, etc. aside from decreased atp, lower matrix and higher intermembrane space ph, this will also lead to increased oxygen consumption (counterintuitive for some students) as the body tries to compensate for the leaking gradient by putting etc into overdrive. now we can also have a toxin directly disrupt one of the complexes in which case either mitochondria are completely ****ed -> death orrr very niche but true that disrupting only complex i or complex ii alone would still allow some atp to be made, since they both drop electrons off w/ coq and that can go on to complex iii even if theres much less of it
enzyme activity and free energy of ion transport
kinetic and thermodynamic factors respectively
After a section of a DNA strand containing a UVR-induced lesion is removed and resynthesized, the newly synthesized strand is rejoined to the remainder of the DNA strand by what type of bond?
re synthesis is done by DNA polymerase 1 which does hydrogen bonding too so now it would be DNA ligase and Phosphodiester bonds rejoin key
goblet cells function and location
secrete mucus lining that protects the stomach from acid and self-digestion located in the inner lining of many organs (stomach, large intestine, and small intestine) a column-shaped epithelial cell found in the respiratory and intestinal tracts, which secretes the main component of mucus.
RNA dependent RNA polymerase
the enzyme that uses RNA as a template to produce RNA. By in large absent from the host cells, it has to be either carried in by the virions or encoded by the viral genome usually have DNA dependent RNA polymerase +ssRNA can be directly translated into proteins so they don't need RdRp -ssRNA NEED RdRp because the final RNA cannot be directly translated, it must be synthesized via the enzyme RdRp. -ssRNA have RdRp because of the fact that the genome that they have (RNA) is not the final template/final product like mRNA. The genome they have serves as a template for the final RNA which is synthesized via RdRp. No eukaryotic cell has this enzyme-which is why they must make one and bring it with them. Again, I do not think RNA viruses have to enter the nucleus. They are either transcribed directly in the cytoplasm or replicated via RdRp into the final product in the CYTOPLASM. The only virus that enters the nucleus is DNA viruses because they need access to host DNA polymerases, which all eukaryotic cells have to replicate their DNA. They then use host RNA polymerases to transcribe their DNA and etc. From a medical standpoint, this is why RNA viruses have so many mutations, their enzymes have very little error reading involved, compared to DNA viruses. Why does COVID have so many variants? (it's a +ssRNA virus). A RT is a type of transcriptase used in retroviruses. Makes the viral RNA into dsDNA which incorporates into cell nucleus. A -ssRNA carries a RDRP that is able to make RNA from RNA. Our cells make more RNA from making more DNA, so the virus needs to carry this. A +ssRNA can just use host machinery but will still need to code for an RDRP in order to replicate its genome.
potency
the strength of something such as a drug or alcoholic beverage.
Km
the substrate concentration at which the velocity is half its maximal value at this Km = S