FL 2 biochme/bio

Which of the following biomolecules is an example of a nucleoside?

Adenosine A nucleoside is composed of a nitrogenous base and a five-carbon sugar (ribose or deoxyribose). Adenosine, shown below, is such a molecule. For adenosine, the nitrogenous base is adenine, while the five-carbon sugar is ribose (note the presence of both a 2' and a 3' hydroxyl group). Hypoxanthine Hypoxanthine is a nitrogenous base.

Chemoreceptors, including those of the taste buds in the tongue, olfactory receptors in the nose, carotid and aortic bodies, and chemoreceptor trigger zone of the medulla, are neurons involved in sensing molecules dissolved in gasses or liquids. According to the passage, binding of endocannabinoids to CB1 produces "pain-diminishing analgesic" effects. Activation of sensory neurons known as nociceptors by noxious stimuli mediates the perception of pain.

Baroreceptors are a subclass of mechanoreceptors located in the walls of blood vessels, as well as the atrial and ventricular walls of the right side of the heart. They detect stretch, signaling changes in luminal pressure. Mechanoreceptors are neurons that respond to mechanical pressure or distortion. Cutaneous mechanoreceptors are responsible for somatosensation, while those embedded in muscles and ligaments are responsible for sensing muscular stretch and load, including in the afferent arm of a number of reflex arcs.

Of what significance to the experimental goals is the finding that AAEL004181 has no introns?

For the most part, bacteria lack introns, whereas large eukaryotic genes usually contain several introns. A large, eukaryotic gene without introns suggests a bacterial origin.

What is the difference between the maximum and minimum volume of air in the lungs of an 80-kg man?

From Figure 1, we see the maximum volume held in the lungs is 80 ml/kg, while the minimum is 15 ml/kg. This is a difference of 65 ml/kg, which, for an 80-kg man, amounts to (65 ml/kg)(80 kg) = 5200 ml, or 5.2 L.

Based on the information presented in the passage, which of the following is most likely true about infections of the CNS?

Bacterial infections of the brain are more serious because antibiotics and other treatments may have a hard time passing the blood-brain barrier. This question asks us to determine why bacterial infections of the brain are more serious than normal infections. The passage states that large molecules have a hard time getting past the blood-brain barrier, which would reduce the availability of many potential treatments in the brain.

Which of the following correctly describes the conversion of AMP to IMP by AMP deaminase?

ΔG < 0 and ATP is neither required nor produced. According to the passage, the Keq > 100. This Keq (>> 1), indicates that the free energy change for the reaction is negative, which is highly favorable. Also, if the reaction is spontaneous, ATP is not required to drive it forward. Although ATP could be generated by coupling an energetically favorable reaction to a reaction that generates ATP, there is no ATP production indicated by the passage for this reaction.

What is the probability that the male offspring of an uninfected male and an infected female will be infected with Wolbachia?

100% Paragraph 1 states that the bacterium infects the germ cells of the mosquito and is transmitted from an infected mother via her eggs to her offspring. Thus, the probability of infection of male or female offspring is most likely near 100%.

Meiosis I results in:

2 haploid cells with 23 chromosomes, each chromosome consisting of 2 sister chromatids. Meiosis I results in 2 haploid cells, each with 23 chromosomes consisting of 2 sister chromatids per chromosome. In the male, the sister chromatids are split into 4 gametes during meiosis II. For females, meiosis I results in a secondary oocyte (a gamete) and a polar body. Penetration of the secondary oocyte by a sperm brings on anaphase II. Telophase II produces a zygote and a second polar body. Remember for the MCAT: mitosis results in diploid daughter cells, while meiosis results in haploid cells that go on to produce gametes.

The pKa values for cysteine are shown below. At what pH will a solution of 2.2 M cysteine be isoelectric?

5.07 We need to determine the pH at which Cys will be isoelectric, which means having no net electric charge. There are two ways to solve this question. First, we can use the given pKa values to determine the charge that Cys will carry in each solution. Alternatively, we can recall that an amino acid will be in its zwitterion (i.e. isoelectric) form when the pH of the environment is equal to the pKa of the amino acid. This means we need to take the average of the two most relevant pKa values on cysteine. While Cys is not commonly listed as an acidic amino acid, its side chain includes a thiol (-SH) group, which can be deprotonated/made negative. A way to figure this out is to note that the structure of cysteine at pH = 7 shows that the side group is protonated. So it must be that even though the pKa is 8.33, the thiol is able to act as an acid (reminder, a pKa > 7 does NOT necessarily mean the group is basic). For amino acids with acidic side chains, the isoelectric point is the average of the two most acidic (lowest) pKa values. Alternatively, we can evaluate each choice and calculate the overall charge on the Cys residue. At the pH of choice B (the correct answer), the COOH group will be deprotonated (-1 charge), the amino group will be protonated (+1 charge), and the side chain will be protonated (neutral charge). As such, the overall net charge on the molecule will be 0.

A gene found in the upper left corner of Figure 1 is most likely involved in:

A careful examination of Figure 1 reveals that a gene found in the upper left corner would be expressed at much higher levels in young adult mosquitos than in old adult mosquitos. Thus, the gene likely has something to do with reproduction, as that is a function that would likely be upregulated in younger as opposed to older mosquitos. The best way to tackle this question is by process of elimination, as the passage information contradicts the rest of the answer choices.

The division between prokaryotes and eukaryotes (which include both single-celled organisms such as yeast and multicellular organisms such as ourselves) is arguably the most basic fork in the tree of life. Prokaryotes (including bacteria and Archaea) are defined by the absence of a nucleus and membrane-bound organelles. Bacteria are ubiquitous and play a major role in the maintenance of human health. Most bacteria in the human body are harmless or beneficial, but some are pathogens.

Although bacteria do have specific genus and species names, they are commonly described in terms of their shape. Spherical bacteria are known as cocci, rod-shaped bacteria are called bacilli, and spiral-shaped bacteria are known as spirilla. Bacteria are also classified in terms of how they use oxygen in metabolism. Bacteria that do not require oxygen for metabolism are known as anaerobes. For obligate anaerobes, oxygen is toxic. Aerotolerant anaerobes are similar to obligate anaerobes in that they cannot engage in aerobic metabolism, but oxygen is not toxic for them. Facultative anaerobes can engage in either aerobic or anaerobic metabolism, depending on the circumstances. Bacteria that require oxygen for metabolism are known as obligate aerobes.

Which of the following molecules is/are most likely to have selective protein channels in the BBB to facilitate passage into the brain?

Amino acids A correct answer to this question should be essential to brain function and small enough to be transported via a protein channel. Amino acids are necessary for the production of proteins, which are essential for the function of any cell.

Ploidy refers to how many copies of each chromosome a cell has. In humans, the vast majority of cells are diploid (2n), meaning that they contain two copies of each chromosome (except for the sex chromosomes; females have two X chromosomes and males have an X and a Y chromosome). Such cells are known as somatic cells—that is, the cells of the body. In contrast, germ cells (i.e., ova and spermatozoa) are haploid (n), meaning that they only have a single copy of each chromosome.

Aneuploidy results from having too many or too few copies of a given chromosome. This results from nondisjunction in anaphase during cell division. Having only one copy of a chromosome is known as monosomy, and having three copies is known as trisomy. Aneuploidy is commonly discussed as occurring in meiosis, and indeed, this is the only way for aneuploidy to be inheritable. For this reason, nondisjunction during meiosis is the cause of aneuploidies such as Down syndrome (trisomy 21) or Turner syndrome (monosomy X). However, nondisjunction during mitosis can also occur, and this is extremely common in cancer cells.

Diastereomers Enantiomers are mirror images, different in orientation at every chiral carbon, which these two figures are not. Conformers can freely convert between forms with no bond breaking (such as the chair and boat forms of cyclohexane). Here, the hydroxyl group cannot simply rotate into the equatorial "up" position without breaking bonds. Constitutional or structural isomers have fundamentally different connectivity between the atoms. These molecules are stereoisomers, not constitutional isomers.

Anomers are a form of epimer. Epimers are sugars that differ only at one stereocenter, with all other stereocenters being exactly the same. Thus, epimers are a type of diastereomer, and choice C is correct. The stereocenter at which these structures differ is circled below. Note that the structure on the left is the alpha anomer, since the -OH group on its anomeric carbon points in the opposite direction (down from the ring) from the -CH2OH group (which points above the ring). In contrast, the structure on the right, which has its -OH group pointing upward, is the beta anomer.

How many chromosomes are found in a liver cell of a Robertsonian translocation carrier?

As stated by the passage, in ROB, two acrocentric chromosomes have broken at the beginning of the short arm near the point where it meets the long arm. The long arms then fuse together. The resulting chromosome consists of two long arms but no short arms. The short arms are lost, but as all the genes on a short arm are available on the short arms of other acrocentric chromosomes, a Robertsonian translocation carrier will have no health problems due to his or her chromosome rearrangement. However, while other people usually have 46 chromosomes, Robertsonian translocation carriers have 45. We can also see that in Figure 1, in order to create ROB two chromosomes are combined and create two more, one of which is lost. Thus we have a net loss of 1 chromosome. The normal human cell has 46 chromosomes, so 46 - 1 = 45.

The table below gives pKa data for selected amino acids. Which of the amino acids listed will exist predominantly as a molecule with both positive and negative charges at pH of 8.0? I. Arginine II. Tyrosine III. Glutamic acid IV. Glycine

At a pH of 8, all of the molecules as free amino acids will have at least one positive and one negative charge, as the carboxyl terminus will have a -1 charge and the amino terminus will have a +1 charge. According to the given table, arginine and glutamic acid will also have charged side chains.

Two known inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then, lysozyme was also added into the reaction mixture. Assuming Figure 1 represents the reaction conditions, where would the resulting enzyme kinetics curve for this experiment fall?

Be sure to read the question stem carefully! Inhibitor A is a strong acid (H3O+) and inhibitor B is a strong base (OH-). Since these inhibitors are mixed together before being added to the reaction mixture, they would simply neutralize each other and become water before having any effect at all. Thus, the enzyme kinetics would not be affected, and the curve would fall along line 1.

There are four main classes of biomolecules: amino acids/proteins, lipids, carbohydrates, and nucleic acids. All of these classes except for lipids can occur as polymers, as proteins can be seen as polymers of amino acids, DNA and RNA are polymers of nucleic acids, and compounds such as starch, cellulose, and glycogen are polymers of carbohydrates. Amino acids contain a central carbon to which -NH2, -COOH, -H, and -R groups are added, and they combine by peptide bonds to form dipeptides, tripeptides, oligopeptides, and proteins. The key functionality of an amino acid is determined by its side chain (-R). Proteins are the building blocks of the body, and are involved in structural and signaling roles. Lipids are nonpolar (or amphipathic, with both polar and nonpolar areas) molecules that play roles in signaling, structural functions, and energy storage. The main categories of lipids that you must know for the MCAT are fatty acids and the derivatives thereof (triacylglycerols, phospholipids, and sphingolipids), cholesterol and its derivatives (steroid hormones and vitamin D), prostaglandins, and terpenes and terpenoids.

Carbohydrates are important for energy storage and are used in metabolism. They are made up of carbon, hydrogen, and oxygen, often in the ratio Cx(H2O)y. Important monosaccharides include glucose, fructose, and galactose. Key disaccharides are sucrose (glucose + fructose), lactose (glucose + galactose), and maltose (glucose + glucose). Polysaccharides (starch and glycogen) are polymers of glucose that are used for energy storage in plants and animals, respectively. Nucleic acids are involved in the storage and transmission of biological information. They are made up of nucleotides, which have three components: a nitrogenous base, a five-carbon sugar (ribose in RNA and deoxyribose in DNA), and at least one phosphate group. RNA contains uracil (U) instead of thymine (T) and is generally single-stranded, whereas DNA is generally double-stranded. DNA is used for the long-term storage of genetic information, while RNA is used for gene expression.

Which of the following best explains why chemical synapses are slower than electrical synapses? This answer relates to the differences between these modes of transfer between cells. At gap junctions, the cells approach within about 2-4 nm of each other, a much shorter distance than the 20-to-40 nm intracellular space that separates cells at a chemical synapse. In chemical synapses, the chemicals must diffuse over these larger distances, which takes more time. A chemical synapse is shown below; note that it is what we typically think of when we think of a synapse. Indeed, chemical synapses are more common than electrical synapses.

Chemical synapses require the movement of agents through intercellular space. Synapses are small structures at the end of the axon of a neuron that allow the neuron to communicate with another nerve, a muscle cell, or a gland. Neurotransmitters are stored in vesicles in the axon terminal. When the action potential arrives, voltage-gated calcium channels are triggered, allowing Ca2+ to rush into the axon terminal. These calcium ions serve as the signal for the cell to use exocytosis to push the neurotransmitters into the synaptic cleft—the space that exists between the axon and the post-synaptic membrane. The space of the cleft is exceptionally small, such that simple diffusion is enough to very quickly carry the neurotransmitters across the cleft to the post-synaptic membrane. There, the neurotransmitters can act as ligands binding to their receptors. Neurotransmitters must be cleared out of the synaptic cleft quickly. This allows the body to tightly regulate the strength and timing of the signals sent by nerves. Neurotransmitters can either be broken down by enzymes in the cleft (the classic example being acetylcholinesterase, which breaks down acetylcholine) or taken back up by the presynaptic axon for re-use later.

If a person has a mutation in the GLUT-1 gene that results in diminished transport capacity, which of the following is most likely to result?

Cognitive difficulties This question asks us to determine the effects of a change in the GLUT-1 gene. We learn from the passage that GLUT-1 acts mainly in the blood-brain barrier to move glucose into the brain, where it is used more than any other place in the body. The rest of the body will not be directly affected as the brain will be. In fact, De Vivo disease (also known as GLUT-1 deficiency syndrome) is a rare condition caused by inadequate transportation of glucose across the blood-brain barrier, resulting in developmental delays and other neurological problems.

If the DNA of a representative species from each of the major kingdoms was examined, the sequences coding for which of the following would be expected to be most similar?

DNA sequences that are common among different species, phyla, or even kingdoms are called conserved sequences. Conserved sequences tend to remain that way due to the fact that they code for a vital function that is common among disparate species. While the enzymes for DNA synthesis, transcription, and translation do vary between species, they are by far the most conserved of the provided answers with regard to structure and function.

The compounds that neurons use to communicate with each other are known as neurotransmitters. For the MCAT, it is important to be familiar with the general effects of the most important neurotransmitters. Glutamate is an excitatory neurotransmitter and is the most common, as 90% of brain cells are responsive to glutamate. In contrast, GABA is the main inhibitory neurotransmitter of the CNS, and it hyperpolarizes cells to reduce action potential firing. Alcohol binds and activates GABA receptors (in other words, alcohol is a GABA agonist), so the effects of GABA are associated with alcohol intoxication. Glycine is another inhibitory neurotransmitter found in the spinal cord and brainstem that can work in conjunction with GABA.

Dopamine is used in reward and motor pathways. It is particularly associated with Parkinson's disease and the loss of dopaminergic neurons in the substantia nigra. Endorphins suppress pain and can produce euphoria. Serotonin regulates mood, appetite, and sleep in the brain, with low levels associated with depressive mood disorders Neurotransmitters can also be active in the peripheral nervous system. Most notably, acetylcholine activates muscle contraction at the neuromuscular junction. It is used in all autonomic outputs from the brain to autonomic ganglia, and in the parasympathetic nervous system for post-ganglionic connections. Serotonin regulates intestinal movement in the gastrointestinal tract, in addition to its effects in the brain. Epinephrine stimulates the fight-or-flight response, and norepinephrine is used in post-ganglionic connections in the sympathetic division of the autonomic nervous system. It also increases arousal and alertness and focuses attention.

Which of the following is an accurate statement regarding mitosis and meiosis?

During mitosis and meiosis II, sister chromatids are separated. This question is asking us to recall some facts about mitosis and meiosis. Remember that mitosis separates sister chromatids to create two diploid daughter cells. Meiosis I separates homologous chromosomes to create haploid daughter cells, each of which divides again, separating sister chromatids in meiosis II to create two haploid cells.

If a lung is punctured, what effect will this likely have on breathing?

During typical breathing, the diaphragm contracts, flattening it and increasing the volume of the intrapleural space. As a result, the lungs expand and the pressure inside them decreases. The resulting pressure differential between the lungs and the atmosphere causes air to rush in, a process termed inspiration. If the lungs are punctured, air will flow freely between the lung and the intrapleural space, and the lung will not expand normally (since no pressure differential can be maintained). Expansion of the thoracic cavity should still lead to some air flow into the lung, however, although it will also mix freely with the thoracic cavity's air, making it more difficult to absorb oxygen within the damaged lung while also making it harder to maintain a pressure differential with the other lung. The image below depicts control of breathing via contraction and relaxation of the diaphragm.

Promoters are regions of DNA that lie upstream to a given gene and initiate transcription by binding specific transcription factors that contribute to the binding of RNA polymerase. Additionally, expression is upregulated by enhancers, which are DNA sequences that can be located further from the gene of interest, and work by binding transcription factors that twist DNA into a hairpin loop, bringing distant regions into close proximity for transcription to begin. Silencers are the opposite of enhancers in eukaryotic cells; they are regions of DNA to which transcription factors known as repressors bind. Additionally, the methylation of C and A residues can reduce transcription. Methylation is associated with epigenetics, which refers to inheritable phenotypic changes involving mechanisms other than the alteration of the genome itself.

Gene expression can also be regulated on the level of nucleosomes (i.e. chromatin and histones). Acetylation promotes transcription by attaching acetyl groups to lysine residues on histones, making them less positively-charged and causing a looser wrapping pattern that allows transcription factors to access the genome more easily. Finally, non-coding RNA plays a role in gene expression. MicroRNA (miRNA) strands are single-nucleotide strands incorporated into an RNA structure with a characteristic hairpin loop, while small interfering RNA (siRNA) molecules are short and double-stranded. Both tend to be approximately 22 nucleotides in length, and silence genes by interrupting expression between transcription and translation.

Researchers noted that when a lysine residue on a histone is acetylated, its side chain becomes neutral in charge. Combined with the information in the passage, which of the following conclusions will the researchers most likely reach?

Histone deacetylation generally decreases gene expression. Lysine is a basic amino acid and typically has a positively-charged side chain at physiological pH. When lysine is acetylated, this charge becomes neutral. Acetyl group is NH-C=O and methyl group, Since DNA is negatively charged due to its phosphate backbone, the charge on lysine allows for tight histone-DNA interactions thanks to electrostatic attraction between the charged atoms on each molecule. Acetylation of lysine makes the residue neutral, lessening these interactions and promoting a looser structure. Loose chromatin structure is typically associated with euchromatin, the less dense, transcriptionally active chromatin structure that appears light under a microscope. In contrast, histone deacetylation will restore the positive charge to the residue, allowing the electrostatic attractions to return. Therefore, deacetylation of lysine residues on histones should lead to a denser chromatin structure and lowered transcription/gene expression.

Uracil is usually found in: I. tRNA. II. ribosomes. III. single-stranded DNA.

I and II only Uracil is found in any structure made of RNA. Both tRNA and ribosomes are made of RNA (I and II). Single-stranded DNA has thymine rather than uracil (III).

Scientists who wished to prevent cell replication while studying the metabolic function of cells with balanced translocations would be best served by arresting the cells during which phase of the cell cycle?

If the scientists wanted to prevent cellular replication, they would need to halt cell division (mitosis). Interphase is the stage of the cell life cycle including G1, S, and G2 (shown below) that occurs between rounds of division.

Diabetes mellitus results from the dysregulation of insulin. Type 1 diabetes is caused by an autoimmune attack on the pancreatic beta cells, which produce insulin. Type 2 diabetes results from a more gradual breakdown of the degree to which target cells respond to insulin signaling. This loss of responsiveness is termed insulin insensitivity. Individuals with type 1 diabetes are dependent upon the administration of exogenous insulin, because they no longer produce the hormone in sufficient quantity to properly regulate blood glucose levels. Patients with type 2 diabetes are generally initially treated with dietary modifications and/or anti-hyperglycemic medications, but they may eventually require insulin treatment as well. Diabetes is commonly associated with high blood glucose levels (and uncontrolled diabetes can even be associated with excess glucose in the urine). Since insulin signaling promotes the uptake of glucose by cells, impaired insulin functioning will prevent cells from doing so, meaning that high levels of blood glucose will coexist with a state in which the cell has inadequate access to glucose for its own metabolism.

If this occurs, cells may accumulate excess acetyl-CoA molecules that cannot be shunted into the citric acid cycle because the intermediaries of the citric acid cycle, especially oxaloacetate, have been siphoned off to gluconeogenesis. This excess acetyl-CoA can be used to produce acetone, D-β-hydroxybutyrate, and acetoacetate, which are known as ketone bodies. The latter two compounds are both acidic, meaning that when present in the blood at an excessively high level, they can cause the blood pH to drop, resulting in a condition known as ketoacidosis. In patients with underlying diabetes, this condition is known as diabetic ketoacidosis. Ketoacidosis can be smelled on a patient's breath, because acetone accumulates to a noticeable level.

Which of the following best explains why a Robertsonian carrier may have no health problems due to his or her chromosome rearrangement?

In the figures, we can see that the short chromosomal arms are lost. However, if all of the genes on a short arm are available on the short arms of other acrocentric chromosomes, a Robertsonian translocation carrier will have no health problems due to his or her chromosome rearrangement.

In eukaryotes, the process of asexual cell division is known as mitosis. Mitosis takes place in four phases: prophase, metaphase, anaphase, and telophase. Prophase prepares the cell for mitosis: the DNA condenses such that distinct chromosomes become visible, as sister chromatids (or copies of a given chromosome) join at a region known as the centromere. The kinetochore assembles on the centromere, and is the site where microtubule fibers that extend from the centrosome and form the mitotic spindle attach to pull the sister chromatids apart in later stages of mitosis. Other microtubules known as asters extend from the centrosome to anchor it to the cell membrane. Additionally, the nuclear envelope and the nucleolus disappear, and the mitotic spindle forms. In metaphase, the chromosomes line up at the middle of the cell along an imaginary line that is known as the metaphase plate. In anaphase, the sister chromatids are separated and pulled to opposite sides of the cell by shortening of the microtubules attached to the kinetochores. Telophase can be thought of as the opposite of prophase, as a new nuclear envelope appears around each set of chromosomes and a nucleolus reappears within each of those nuclei. The process of mitosis is completed by cytokinesis.

In contrast, meiosis is a form of cell division that is essential for sexual reproduction. It takes place in germ cells (also known as sex cells). Meiosis differs from mitosis in that it has two stages and results in the formation of four daughter cells, each of which has only one copy of each chromosome (haploid, n), in contrast to mitosis, which generates cells with two copies of each chromosome (diploid, 2n) that are essentially identical to their parent cell.

Which of the following molecules does NOT have an atom with sp2 hybridization?

In each case, you should draw the Lewis dot structure. In methanol (CH3OH), four atoms form single bonds with the carbon atom, which is sp3 hybridized. The oxygen atom has two single bonds, one to carbon and one to hydrogen, as well as two lone pairs of electrons; it therefore has sp3 hybridization as well. Neither atom has sp2 hybridization. Since hydrogen can only ever form one bond, we do not need to consider it here. The structure of methanol is shown below.

Which of the following represents the profile of chromosomes 14 and 21 for a Robertsonian translocation carrier?

In order to be a carrier, a person must have all the requisite genetic information but have the possibility of creating gametes with unbalanced genetic information. This person has copies of both 14 and 21, but there is a translocation between two of the chromosomes which could lead to unbalanced rearrangement in their offspring (i.e. they are a carrier).

Arachidonic acid, released during AEA hydrolysis, is NOT a precursor for the synthesis of what class of molecules? Eicosanoids are a large family of lipids derived from arachidonic acid, a 20-carbon omega-6 polyunsaturated fatty acid with four cis double bonds. Like their parent compound arachidonic acid, eicosanoids have 20 carbons, and they have the additional characteristic feature of a five-carbon ring. The most important eicosanoids are a large family of signaling molecules known as prostaglandins, which have a diverse range of effects, including the modulation of inflammation. Additionally, thromboxanes are involved in the clotting cascade. The enzymes cyclooxygenase-1 (COX-1) and cyclooxygenase-2 (COX-2) are involved in early steps of this pathway, and are targeted therapeutically by non-steroidal anti-inflammatory drugs (NSAIDs), such as aspirin.

In paragraph 1 of the passage, it's written that AEA is degraded to ethanolamine and arachidonic acid. Ethanolamine is included as a polar head group in the phospholipid phosphatidylethanolamine, while arachidonic acid is important as a precursor for the biosynthesis of the eicosanoid signaling molecules: prostaglandins, thromboxanes, and leukotrienes. Neither molecule contributes to the synthesis of catecholamines, which are a class of molecules derived from tyrosine that include dopamine and norepinephrine.

The plasma membrane of eukaryotic cells is primarily composed of a lipid bilayer of amphipathic phospholipids with hydrophilic heads and hydrophobic tails. It also contains cholesterol and membrane proteins. Transmembrane proteins are membrane-spanning proteins with hydrophilic cytosolic and extracellular domains and a hydrophobic membrane-spanning domain. Additionally, peripheral proteins are only transiently attached to integral proteins or peripheral regions of the lipid bilayer, and lipid-anchored proteins are covalently bound to membrane lipids without actually contacting the membrane directly. Membrane transport is accomplished through several mechanisms. Some molecules, like small gasses, can directly diffuse through the membrane. This is known as simple diffusion, and is an example of passive transport because no energy is necessary. Osmosis is a type of simple diffusion in which water moves in or out of the cell to attempt to equalize concentrations of solute. Facilitated diffusion is another form of passive transport where no energy is necessary because molecules diffuse down their concentration gradient, but a transmembrane channel is necessary because the molecule may be too large or polar for simple diffusion. Ions are often transported through facilitated diffusion, and aquaporins are facilitated diffusion channels for water that augment osmosis.

In primary active transport, energy is used directly to move a solute against its gradient through a transmembrane channel. Secondary active transport is a more complicated system in which the energy stored in an electrochemical gradient established via primary active transport is used to facilitate the movement of a solute. An example is the sodium-calcium exchanger, which allows three Na+ ions to flow down their concentration gradient, which was previously established by a primary active transport mechanism, into the cell, while transporting one Ca2+ ion out. Endocytosis is used to ingest larger materials. It is divided into pinocytosis and phagocytosis. In pinocytosis, cells engulf liquid substances, while in phagocytosis, they engulf solid particles. The basic pathway of endocytosis involves recognition of a target molecule at the plasma membrane, followed by invagination and the formation of a vesicle on the inside of the cell. Exocytosis can be thought of as endocytosis in reverse. It is used to release hormones, neurotransmitters, membrane proteins and lipids, and other materials.

meiosis is a form of cell division that is essential for sexual reproduction. It takes place in germ cells (also known as sex cells). Meiosis differs from mitosis in that it has two stages and results in the formation of four daughter cells, each of which has only one copy of each chromosome (haploid, n), in contrast to mitosis, which generates cells with two copies of each chromosome (diploid, 2n) that are essentially identical to their parent cell.

In prophase I of meiosis, homologous chromosomes (i.e., the maternal and paternal copies of a given chromosome) pair up with each other in a process known as synapsis, forming tetrads. While paired up, homologous chromosomes may exchange genetic information in a process known as crossing over. The crossing-over points are known as chiasmata. This process results in recombinant DNA that is another source of variation in sexual reproduction, in addition to the variability inherent to the process. In metaphase I, homologous pairs, which take the form of tetrads, line up at the metaphase plate. The orientation of the homologous pairs is random in terms of which side of the metaphase plate the maternal or paternal copy of a given chromosome in a homologous pair winds up. In anaphase I, the homologous pairs are separated, and one member of each pair is pulled to each side of the cell. In meiosis II, which operates similarly to mitosis, the sister chromatids are split up into two haploid daughter cells.

The reactions catalyzed by xanthine oxidase in Figure 2 are reversible. If Keq for the conversion of hypoxanthine, water, and oxygen to xanthine and hydrogen peroxide is approximately 1, which change in an equilibrium mixture of the five compounds will NOT necessarily increase the ratio of xanthine to hypoxanthine once equilibrium is reestablished?

Increasing temperature For the reaction described, Keq is approximately 1. As we do not know whether this is an endothermic or exothermic reaction, changing the temperature may drive the reaction forward, or it may drive it backwards. In that case, a change in temperature might not necessarily lead to a change in concentration of reactants and products. So increasing temperature may increase this ratio, but it may decrease it as well!

Which of the following changes occur immediately following the consumption of a carbohydrate-rich meal?

Insulin secretion increases; G6PD activity increases. From outside knowledge, we know that insulin secretion increases after a meal to help the cells take up glucose from the bloodstream. The passage states that the PPP is a parallel (alternative) path to glycolysis, so it makes sense that it would also be activated when blood sugar levels rise after a meal.

The cell cycle can be divided into a resting phase (interphase) and cell division (mitosis or meiosis). Resting phase is also known as Gap 0 (G0). During this period, the cell just goes about its business; in fact, many fully-differentiated cells in the body remain in G0 for long periods of time. Because it can last for an essentially indefinite period of time, resting phase is often considered not to be a proper part of the cell cycle itself.

Interphase is when a cell prepares for division, and it can take up approximately 90% of the time of the cell cycle. Two major things happen during interphase: growth and DNA replication. However, interphase is broken into three stages: Gap 1 (G1), synthesis (S), and Gap 2 (G2). During G1 and G2, the cell grows, and during S, DNA is replicated. The fact that S is located between G1 and G2 allows checkpoints. The G1/S checkpoint, also known as the restriction point, is when a cell commits to division. The presence of DNA damage or other external factors can cause a cell to fail this checkpoint and not divide. The G2 checkpoint that takes place before cell division similarly checks for DNA damage after DNA replication, and if damage is detected, serves to "pause" cell division until the damage is repaired. Throughout interphase, chromatin is loosely packaged (euchromatin) to allow transcription and replication.

What is the correct molecular formula for isopropyl alcohol?

Isopropyl alcohol is a three-carbon chain with a hydroxyl group on the middle carbon: CH3 - CHOH - CH3 Alternatively to writing out the formula (as shown above), you could draw out the chemical structure of isopropyl alcohol. (Note that the central atom in the diagram below is bound to an "invisible" hydrogen atom.)

Acidic species can be strong (for the exam, assume that strong acids completely dissociate in water) or weak. When an acid dissociates, it releases a proton to make the surrounding solution acidic. However, weak acids only partially dissociate and at equilibrium coexist in a deprotonated state (A−) and a protonated state (HA), according to the equation HA ⇌ H+ + A−. The concentration ratio of products and reactants is constant given fixed conditions and is called the acid dissociation constant (Ka). Ka is defined by the equation below.

Ka expresses how easily an acid releases a proton (i.e. its strength). In addition, this equation shows how the dissociation state of weak acids vary according to the [H+] level in the solution. A commonly tested family of acids on the MCAT are carboxylic acids (those containing -COOH), such as lactic acid and amino acids, which normally have a Ka of approximately 10−3 to 10−6. For example, the Ka values for lactic acid (HC3H5O3) and nitrous acid (HNO2) are 8.3 × 10−4 and 4.1 × 10−4, respectively. The pKa values for these acids are 3.1 and 3.4, respectively, which are simpler expressions that are easier to understand and compare. The smaller the pKa value, the stronger the acid (since as X increases, pX decreases). Therefore, the pKa values above tell us that lactic acid is a stronger acid than nitrous acid.

Reversible chemical reactions eventually reach a state known as equilibrium in which the forward and reverse reactions are proceeding at the same rate. However, the concentrations of products versus reactants at equilibrium are generally different. The relative preponderance of products versus reactants is described by the equilibrium constant (Keq), which is the concentration of products at equilibrium (raised to the power of their stoichiometric coefficients) divided by the concentration of reactants (raised to the power of their stoichiometric coefficients). For a sample reaction X(g) + 3 Y(g) ⇌ 2 Z(g), Keq = [Z]2/[X][Y]3. An important caveat is that solids and pure liquids should not be included in the equilibrium expression, and it is also important to note that Keq varies with temperature. A reaction that favors the products—in other words, that has a numerator greater than its denominator— will have a large Keq, while a reaction that favors the reactants will have a small Keq. Many other constants in chemistry with an uppercase K, such as Ka and Kb in acid-base chemistry and Ksp in solubility chemistry, are in fact specialized versions of equilibrium constants, so your knowledge of equilibrium processes can be applied to those contexts as well.

Le Châtelier's principle is an MCAT-critical concept related to equilibrium. This principle states that if an equilibrium mixture is disrupted, it will shift to favor the direction of the reaction that best facilitates a return to equilibrium. A simpler version is that a reaction will shift in the direction that relieves the stress put on the system, where "stress" can refer to a change in reactant or product concentration, temperature, pressure, or volume. Thus, increasing the reactant concentration or decreasing the product concentration will shift the reaction towards the product side, and vice versa. In reactions with gases, increasing the volume (decreasing the pressure) will shift the equilibrium to the side with more moles of gas (and vice versa). For a reaction where ∆H > 0, increasing the temperature will shift the reaction toward the products, while decreasing it will shift the reaction toward the reactants; the opposite pattern is found if ∆H < 0.

Insulin is a peptide hormone released by the beta cells of the pancreas in response to high blood glucose levels. Its basic function is to reduce blood glucose levels by promoting the transport of glucose into cells via insulin receptors, which activate membrane-bound glucose transporters. The glucose transported into the cell can be used immediately through glycolysis; alternatively, muscle and liver cells can store the glucose as glycogen, and adipocytes (fat cells) can mobilize fatty acids to store downstream byproducts of glucose metabolism in the form of triglycerides. Insulin upregulates all those processes, as well as protein synthesis. Glucagon is a peptide hormone released by the alpha cells of the pancreas, and its mechanism and function are essentially the opposite of insulin. Glucagon is released in response to low glucose levels and has the effect of increasing blood glucose levels by promoting glycogenolysis and gluconeogenesis in liver cells.

Other hormones can affect blood glucose levels as well. Cortisol (the main example of a class of hormones known as glucocorticoids) is released by the adrenal cortex. It is associated with long-term responses to stress and increases blood glucose levels. Epinephrine, which is released by the adrenal medulla and plays a major role in the fight-or-flight response to immediate stress, also raises blood glucose levels. In addition, growth hormone can increase blood glucose levels due to its antagonistic effects on insulin. Nonetheless, on Test Day, if you see a question about blood glucose levels, you should immediately think of the insulin-glucagon pair, unless the passage or question points you specifically in the direction of other hormones that may affect blood glucose levels.

How do the chemical modifications described in the passage differ from eukaryotic post-transcriptional modifications?

PTMs were first mentioned in paragraph 1 ("acetylation of lysine residues, methylation of lysine and arginine..."), and then a detailed example was given in the final paragraph. These are all modifications to amino acids and/or proteins. Protein modifications are nearly always post-translational modifications. Such modifications can take place in a variety of locations within the cell, such as the interior of the endoplasmic reticulum or the cytoplasm. In contrast, nearly all post-transcriptional modifications (those performed on mRNA) occur in the nucleus; these include the addition of the poly(A) tail, the addition of the 5' cap, and splicing.

Given that siRNA directly reduces the synthesis of proteins, what type of control does siRNA most likely exert on G6PD expression?

Post-transcriptional control Due to its structure, siRNA is only able to bind to other RNA strands, not to DNA or protein. Therefore, it must interfere with gene expression after transcription has already occurred, but before translation. Specifically, it prevents the translation of mRNA corresponding to the target protein.

Which of the following is NOT a function of the blood-brain barrier?

Protection of the brain from carbon dioxide poisoning The passage states that the blood-brain barrier protects the brain from harmful agents that are large or polar. Carbon dioxide (shown below) is both small and nonpolar, making it highly lipid-soluble. Thus, it will freely pass through the BBB.

A pharmacologist wishes to convert morphine into a form available to brain tissue. Which of the following changes could be made to morphine's molecular structure to improve its ability to cross the BBB while maintaining its therapeutic effects?

Replace the proton of each alcohol with an acetyl group This question asks us to consider the structure of morphine given in the passage. We are told in the passage that lipid-soluble molecules are able to pass through, while hydrophilic molecules are not. Morphine is hydrophilic in part because of its hydroxyl groups. Although the C=O bond in the acetylated molecule is polar, it would not donate hydrogen bonds. As a result, the acetylated compound would be more lipophilic than morphine. Diacetylated morphine is called heroin and, while the effects and pharmacology of heroin are somewhat distinct from those of morphine, heroin does indeed cross the BBB more readily.

A spontaneous reaction is one that can proceed without the help of some additional, external force. Importantly, spontaneity does not relate to the rate of the reaction, or the speed at which it progresses—spontaneous reactions can be very slow! Rate is a kinetic parameter that is determined largely by the activation energy. In contrast, spontaneity is a thermodynamic parameter and relates to ∆G, or the change in Gibbs free energy. The most basic definition of spontaneity is ∆G < 0, meaning that the reaction has a negative change in Gibbs free energy. Spontaneity is associated with exothermic (∆H < 0) reactions and those that increase entropy (∆S > 0) through the equation ∆G = ∆H - T∆S. Note that a reaction can still be spontaneous even if it fulfills only one of these requirements. For example, a spontaneous reaction may be endothermic (∆H > 0) as long as this is balanced by a sufficiently positive ∆S.

Spontaneity can also be defined in terms of other chemical concepts. By definition, a spontaneous reaction will lead to more products being present than reactants. This means that the equilibrium constant (Keq), which is broadly defined as [products]/[reactants], will be greater than 1 for a spontaneous reaction. This relationship is encoded in the equation ∆G°rxn = ∆RT ln(Keq). Additionally, in the context of electrochemical cells, spontaneous reactions are associated with positive cell potentials (E° > 0). On Test Day, it is important to automatically recognize that spontaneity is equivalent to ∆G < 0, Keq > 1, and E° > 0.

The disruption of which membrane component is most likely to result in cellular traffic complications similar to those seen in gap junction disorders?

The cell membrane is composed of several different components, each responsible for different functions. Membrane transport is most likely to be affected if the disruption occurs in components that span the entire membrane. Transmembrane proteins (many of which are glycoproteins) are the only components listed that pass all the way through the cell membrane and facilitate membrane transport. Phospholipids are located on the surface of cell membranes and typically do not extend through the entire bilayer. Phospholipids are a structural component of the membrane and are not involved in traffic/transport.

Isoelectric focusing is an electrophoretic technique that separates proteins according to their isoelectric point (pI). The pI refers to the pH at which the net charge on a protein will be 0; this parameter varies across proteins, such that isoelectric focusing is a good way to separate proteins in a relatively specific manner. Isoelectric focusing requires the use of a gel that has a stable pH gradient. A protein that is in a region of the gel where the pH is below its pI will be positively charged and will migrate towards the negatively-charged plate (the cathode). The protein will cease to migrate when it reaches the region of the gel where the pH of the gel equals the pI of the protein. At this point, the protein will stop migrating because the net charge of the protein will be 0, so it will not experience any electric force. The proteins analyzed using isoelectric focusing will be separated based on the relative number of acidic and basic residues that they contain.

The charge of a peptide at a given pH can be roughly predicted based on how many acidic and basic residues they contain. The side chains of the acidic amino acids, aspartic acid and glutamic acid, will be protonated and uncharged (-COOH) below a pH of roughly 3 and deprotonated and negatively-charged (-COO-) above that pH. Correspondingly, the side chains of the basic amino acids (histidine, lysine, and arginine) will be protonated and positively-charged at pH levels below the pKa values of their side chains (roughly 6.0, 10.5, and 12.5, respectively). The pI can also be estimated for amino acids in isolation. For diprotic amino acids (i.e., those amino acids for which the side chain is neither acidic nor basic), the pI can straightforwardly be captured as the average of the two pKa values. For triprotic amino acids (i.e., those that have an acidic or basic side chain), the pI can be obtained by averaging the two acidic pKa values for acidic amino acids or the two basic pKa values for basic amino acids.

Which components of cells are physically connected by a gap junction?

The cytoskeleton of one to the cytoskeleton of the other The answer is given directly in paragraph 2, which mentions "...two hemichannels (connexons), one embedded in each cell's plasma membrane and anchored to the cytoskeleton of the cells." This means that the cytoskeleton of one cell is connected to the cytoskeleton of another. A schematic of a gap junction is shown below.

Fertilization takes place in the Fallopian tube, when a sperm cell encounters a secondary oocyte. The sperm cell passes through the corona radiata, a layer of follicular cells surrounding the oocyte, and the zona pellucida, a layer of glycoproteins between the corona radiata and the oocyte. This triggers the acrosome reaction, in which digestive enzymes are released that allow the nucleus of the sperm cell to enter the egg. The secondary oocyte completes meiosis II, creating a second polar body and a mature ovum. Then, the haploid nuclei of the sperm cell and the ovum merge, creating a diploid one-cell zygote. As the zygote travels to the uterus, it undergoes a series of mitotic cell divisions known as cleavage. Once the zygote has cleaved into a mass of 16 cells by three to four days after fertilization, it is known as the morula. By three to five days after fertilization, the morula develops some degree of internal structure and becomes a blastocyst, with a fluid-filled cavity in the middle known as the blastocoel. The blastocyst implants in the uterine endometrium and further differentiates into the gastrula. The gastrula has three layers: the ectoderm, the mesoderm, and the endoderm. These layers eventually go on to form specific organs and components in the body.

The ectoderm primarily gives rise to the nervous system and epidermis (skin), as well as related structures like hair, nails, and sweat glands, and the linings of the mouth, anus, and nostrils. The process through which the nervous system is formed from the ectoderm is known as neurulation. The mesoderm generates many of the structures present within the body, including the musculature, connective tissue (including blood, bone, and cartilage), the gonads, the kidneys, and the adrenal cortex. The endoderm is basically responsible for the interior linings of the body, including the linings of the gastrointestinal system, the pancreas and part of the liver, the urinary bladder and part of the urethra, and the lungs.

The concentration of intracellular signaling molecules fluctuates rapidly in dividing cells during the cell cycle. Which of the following experimental techniques would be best to elucidate the mechanism of regulation for these proteins?

Western blot and RT-PCR Rapidly dividing cells undergo mitosis under the influence of specific signaling molecules. These molecules are expressed when their genes are transcribed, then are translated into proteins. In order to gain the best understanding of how a signaling protein's levels are regulated, both the protein and mRNA levels would need to be studied. Western blotting gives us information about the amount of protein expressed in a cell, while RT-PCR gives us information about the amount of RNA expressed.

The process of going from DNA to RNA—more specifically, messenger RNA (mRNA)—is called transcription. Transcription takes place in the nucleus, and it results in the creation of an mRNA copy of a gene that can then be transported to the cytosol for translation into a protein. The DNA helix must be unzipped for transcription to take place, which means that some of the same machinery used for DNA replication has to be engaged, especially enzymes like helicase and topoisomerase. RNA polymerase is the enzyme responsible for RNA synthesis. In eukaryotes, it binds to a promoter region upstream of the start codon with the assistance of transcription factors, the most important of which is the TATA box. RNA polymerase travels along the template strand in the 3'-5' direction, synthesizing an antiparallel complement in the 5'-3' direction. The template strand is known as the antisense strand, and the opposite strand is known as the sense strand, because it corresponds to the codons on the mRNA that is eventually exported to the cytosol for translation.

The immediate product of transcription in eukaryotes is not mRNA, but heterogeneous nuclear RNA (hnRNA). hnRNA must undergo a set of post-transcriptional modifications to become mRNA. Examples commonly tested on the MCAT include the 3' poly-A tail, the 5' cap, and splicing. The 3' poly-A tail is a string of approximately 250 adenine (A) nucleotides added to the 3' end of an hnRNA transcript to protect the eventual mRNA transcript against rapid degradation in the cytosol. The 5' cap refers to a 7-methylguanylate triphosphate cap placed on the 5' end of an hnRNA transcript. Similarly to the 3' poly-A tail, it helps prevent the transcript from being degraded too quickly in the cytosol, but it also prepares the RNA complex for export from the nucleus.

Which of the following phylogenetic trees is most likely given the data in Table 1?

The key idea here is that of parsimony: we want the tree that requires the fewest "events" (mutation, transfer, etc.) to produce the genetic picture in Table 1. The passage uses the idea that adjacent genes were transferred to support its argument for horizontal gene transfer, so we can assume that AAEL004181 and 004188 transferred together at the same time. That means that albopictus should be the first branch off the tree, as it lacks both genes (eliminate choice A). Since the three remaining species all contain segments a-c of AAEL004181, we can assume that the original gene consisted of these parts, with d, e, and f as later add-ons. Further complicating matters is the fact that simpsoni appears to have lost AAEL004188, so it should be on a different branch than mascarensis and aegypti. Mascarensis is the most closely related species to aegypti, since it is only missing the intervening d and e segments of AAEL004188.

A spontaneous reaction is one that can proceed without the help of some additional, external force. Importantly, spontaneity does not relate to the rate of the reaction, or the speed at which it progresses—spontaneous reactions can be very slow! Rate is a kinetic parameter that is determined largely by the activation energy. In contrast, spontaneity is a thermodynamic parameter and relates to ∆G, or the change in Gibbs free energy.

The most basic definition of spontaneity is ∆G < 0, meaning that the reaction has a negative change in Gibbs free energy. Spontaneity is associated with exothermic (∆H < 0) reactions and those that increase entropy (∆S > 0) through the equation ∆G = ∆H - T∆S. Note that a reaction can still be spontaneous even if it fulfills only one of these requirements. For example, a spontaneous reaction may be endothermic (∆H > 0) as long as this is balanced by a sufficiently positive ∆S. Spontaneity can also be defined in terms of other chemical concepts. By definition, a spontaneous reaction will lead to more products being present than reactants. This means that the equilibrium constant (Keq), which is broadly defined as [products]/[reactants], will be greater than 1 for a spontaneous reaction. This relationship is encoded in the equation ∆G°rxn = −RT ln(Keq). Additionally, in the context of electrochemical cells, spontaneous reactions are associated with positive cell potentials (E° > 0). On Test Day, it is important to automatically recognize that spontaneity is equivalent to ∆G < 0, Keq > 1, and E° > 0.

Which of the following Lineweaver-Burk plots correctly represents the change in enzyme kinetics of the reaction catalyzed by PRPP synthetase in the presence and the absence of purine nucleotides?

The passage indicates that purines act as competitive inhibitors of PRPP synthetase. Competitive inhibitors bind free enzyme, increasing Km for the reaction catalyzed. In a Lineweaver-Burk plot, the x-intercept of the graph equals -1/Km, while the y-intercept equals 1/Vmax. Thus, in the presence of purines, plotted values should change such that the y-intercept remains the same and the x-intercept moves closer to the y-axis (the value of -1/Km increases as Km increases due to the addition of the competitive inhibitor).

Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD?

The passage states that high levels of NADPH inhibit G6PD, and Equation 1 shows that the substrate of G6PD is G6P. Because the structures of G6P and NADPH are very different, it is unlikely that NADPH competes with G6P at the active site; thus, this is not competitive inhibition (I). The passage also never suggests that the inhibition of G6PD is irreversible; rather, it is likely to be dynamic based on the amount of NADPH available (III). Therefore, NADPH most probably binds to a site that is not the active site, which is characteristic of allosteric inhibition (II). A generic model of allosteric inhibition is included below. Because the structures of G6P and NADPH are very different, it is unlikely that NADPH competes with G6P at the active site; thus, this is not competitive inhibition (I).

Which statement best explains why the lung volume in Figure 1 never drops below 15 mL/kg?

The residual volume is such that the lungs are at their minimum volume under maximum intrapleural pressure.

Based on passage information, cardiac disorders involving gap junction dysfunction would most likely manifest clinically through: arrhythmia. The passage discusses the role of gap junctions in conducting nerve impulses through cardiac muscle with the appropriate timing needed for prompt, regular contractions. Disrupting this flow would mean that some parts of the heart would contract before others, resulting in an abnormal heart rhythm or arrhythmia.

The rhythm of heart contractions is controlled by the sinoatrial (SA) node, which is found at the roof of the right atrium. The cells in the SA node periodically send out action potentials, much like nerve cells. Gap junctions between the cardiac muscle cells allow the action potential to propagate throughout the tissue, causing contraction. The action potential flows from the SA node into the atria, but not into the ventricles because of a layer of insulating tissue. This causes both atria to contract, pushing the blood forward into the ventricles. The atrioventricular (AV) node allows the action potential to pass through to the ventricles after the atria have contracted. At this point, the ventricles must contract to push the blood out of the heart. Since the ventricles are larger than the atria, it is a bit more difficult to get them to contract together, so the signal is sped through the bundle of His and Purkinje fibers to all the muscle cells of the ventricles. Deoxygenated blood returns to the right atrium via the superior and inferior venae cavae and the coronary sinus, which drains the coronary veins. From there, it is pumped into the right ventricle through the tricuspid valve. From the right ventricle, it goes to the pulmonary arteries through the pulmonary semilunar valves. After becoming oxygenated, it is returned to the heart via the pulmonary veins, and enters the left atrium. It is pumped through the bicuspid valve from the left atrium to the left ventricle, and then the left ventricle pushes the blood into circulation (more specifically, through the aortic semilunar valves into the ascending aorta).

Which of the following is NOT a part of the respiratory system?

This is a straightforward anatomy question with no information from the passage needed. The lacteals (shown below) are structures in the intestines associated with absorbing fat into the lymphatic system. The epiglottis is part of the respiratory system. It is a flap of cartilage that covers the trachea to ensure that food properly remains in the digestive tract.

In severe diabetic hyperglycemia (high blood sugar), insulin cannot effectively induce the uptake of glucose by cells. Assuming otherwise healthy nephrons, chronic hyperglycemia directly leads to the presence of which of these molecules in the urine?

This question requires outside knowledge about glucose metabolism. If cells cannot take up glucose, it will remain in the blood and eventually be excreted in the urine when it builds up to the point that it cannot be reabsorbed by the nephron (II). In a state of extended hyperglycemia, the body relies on fat metabolism to generate energy, which produces ketone bodies that are also excreted in the urine (III).

Which phase of mitosis is likely to be interrupted first if microtubule polymerization is inhibited in a cell?

This question requires us to remember the workings of mitosis. In mitosis, the spindle fibers that move chromosomes are made up of microtubules. They are attached to chromosomes in metaphase and pull them apart in anaphase. Therefore, the first phase to be interrupted by lack of microtubules will be metaphase.

Given that ΔG° is negative for a particular reaction, what can be said about the equilibrium constant K? K < 1 K = 1 K > 1

This would correspond to a positive ∆G°. This would correspond to ∆G° = 0. Given that ∆G° is negative, and using the equation ∆G° = -RT ln(K), ln(K) must be positive for ∆G° to be negative. K > 1.

Histones are proteins that act as spools for DNA to wind around. They are composed of various subunits known as H1, H2A, H2B, H3, and H4. The core of a histone contains two dimers of H2A and H2B and a tetramer of H3 and H4, while H1 serves as a linking unit. Approximately 200 base pairs of DNA can be wound around a histone, and the complex formed by DNA and a histone is known as a nucleosome.The phrase "beads on a string" is often associated with the appearance of nucleosomes under electron microscopy, and chromatin refers to the structure formed by many nucleosomes. On a biochemical level, charge plays a major role in the interactions between histones and DNA. Histones are highly alkaline and are positively charged at physiological pH, which facilitates their interaction with the highly negatively charged phosphate groups on the backbone of DNA. Modifications like acetylation of histones reduce that positive charge, making histones interact with DNA less closely, which in turn facilitates transcriptional activity.

Two distinct forms of chromatin exist: euchromatin and heterochromatin. Euchromatin is a loose configuration that is difficult to see under light microscopy and allows DNA to be readily transcribed. Throughout interphase (i.e., most of the cell cycle), DNA generally exists as euchromatin. Heterochromatin is the tightly coiled, dense form of chromatin that is visible during cell division and is present to a lesser extent even during interphase.

The central dogma of molecular genetics states that information flows from deoxyribonucleic acid (DNA) to messenger ribonucleic acid (mRNA) to protein. The Hershey-Chase experiment, conducted in 1952, helped demonstrate DNA's key role as genetic material. The researchers used radiolabeled sulfur and phosphorus to distinguish between proteins (which contain sulfur atoms present in cysteine and methionine residues, but not phosphate groups) and nucleic acids (which contain phosphate groups but no sulfur). Radiolabeled bacteriophages, which inject genetic material into bacterial cells, were added to bacterial cell cultures. It was determined that the bacterial cells post-transduction contained radiolabeled phosphate, not sulfur, indicating that the genetic material in question is DNA. DNA is organized in a double helix of antiparallel strands, with a sugar-phosphate backbone connected by phosphodiester bonds on the outside and nitrogenous bases on the inside. DNA contains four such bases: adenine (A), cytosine (C), guanine (G), and thymine (T); RNA contains A, C, and G, but has uracil (U) in place of thymine (T). Of these, adenine and guanine are purines, which indicates that they contain two fused rings in their structures. In contrast, cytosine, thymine, and uracil are pyrimidines and contain only a single ring. Complementary base pairing dictates that adenine pairs with thymine in DNA, adenine pairs with uracil in mRNA, and cytosine pairs with guanine in both molecules. For the DNA double helix, the interior of the structure is stabilized by hydrogen bonds between base pairs (two hydrogen bonds for AT pairs and three for CG pairs), as well as hydrophobic interactions between stacked nitrogenous bases.

Unlike DNA, RNA tends to exist in single-stranded form, rather than as a double-stranded double helix. Additionally, the sugar in RNA is ribose, which contains a hydroxyl (-OH) group on its 2' carbon. (DNA contains deoxyribose, which lacks this 2' hydroxyl group.) In addition to mRNA, which codes for protein production, several forms of non-coding RNA exist. These include transfer RNA (tRNA), which assists in translation, and small interfering RNA (siRNA) and microRNA (miRNA). siRNA and miRNA differ in their structure: miRNA strands are single-nucleotide strands incorporated into an RNA structure with a characteristic hairpin loop, while siRNA molecules are short and double-stranded. Both tend to be approximately 22 nucleotides in length, and both silence genes by interrupting expression between transcription and translation.

A student ran an enzyme inhibition experiment on the enzyme from Figure 1, under the same conditions but using known competitive inhibitors. Which of the following could NOT have been the result of a trial from their experiment?

When the enzyme was saturated, the student found that the rate of the reaction was 0.95 M/s. The rate of the reaction cannot exceed Vmax, which is 0.85 M/s. Vmax represents the rate that is reached when the enzyme is saturated with substrates. When the enzyme is not yet saturated, the rate can reasonably fall along a range of values that are all less than Vmax, even zero. Thus, the rate of the reaction from Figure 1 cannot be 0.95 M/s unless the conditions were changed in some way, such as increasing the enzyme concentration.

Folate is a common supplement given to pregnant women to prevent neural tube defects. The most likely reason for this is:

folate plays a role in ectodermal induction. Neural tube defects are defects in the central nervous system. The nervous system is derived from the ectoderm. It can be concluded from the given information that folate is important for neurulation, or the induction of the ectoderm to differentiate into the nervous system. Remember, the ectoderm (shown below) is the outer layer of the gastrula. It gives rise to the nervous system, epidermis, hair, nails, teeth, and sweat glands.