NSTest2/MCAT

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 gases or liquids.

interphase

(essentially indefinite period of time, resting phase 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.

Pentane is a straight-chain hydrocarbon with the molecular formula C5H12. How many additional structural isomers can be constructed using this molecular formula? A. 0 B. 1 C. 2 D. 3

2 a tertiary and quartenary carbon

Robertsonian translocation

45 chromosomes instead of 46 Translocation in which the long arms of two acrocentric chromosomes become joined to a common centromere, resulting in a chromosome with two long arms and usually another chromosome with two short arms.

Assuming a 95% yield for each coupling step, what would be the final yield for synthesizing a 10 amino-acid length peptide? A. 30% B. 60% C. 80% D. 95%

5% loss each time, so 9 x 5 = 45% loss answer: 60% yield more or less Each time we add an amino acid to the chain, we get a 95% pure yield for that step. Here, imagine that each coupling step has a 95% yield and that we are running the cycle 9 times (since no coupling step is needed for the first of the 10 amino acids), with each cycle being independent from the preceding cycle. Thus, a 10-amino acid peptide would be synthesized in a (0.95)9 = 0.63 ≈ 60% final yield. This can be assured because the passage allows us to assume a 100% yield in each unprotection step. There is no need to calculate it all out. A crude estimate of the yield would be that 9 steps, in each of which 5% is lost (corresponding to a 95% yield), would result in 45% being lost. However, the actual yield will be somewhat higher, because each step does not involve exactly a 5% reduction from the starting percentage. 60% is the best approximation.

The Cannon-Bard theory of emotion would suggest that any aggressive emotions that the children experience as a result of listening to aggressive music would: A. entail simultaneous physiological arousal and a subjective feeling of aggression, which are separate and independent. B. start with a physiological arousal, and this arousal would cause the aggressive emotions. C. stimulate the parasympathetic autonomic response leading to an increased secretion of fight-or-flight hormones. D. start with physiological arousal, and the children would then interpret that arousal given the environmental context.

A is correct. The Cannon-Bard theory of emotion asserts that the physiological arousal and the subjective feeling of an emotion arise from different parts of the brain and are separate and independent of one another.

Which of the following best explains why a Robertsonian carrier may have no health problems due to his or her chromosome rearrangement? A. Genes on the lost p arms can be found elsewhere in the genome. B. There are no genes found on any of the p arms of acrocentric chromosomes. C. All acrocentric p arms contain the same base sequence. D. The long q arm chromosome copies the genetic information from the p arms before they are lost.

A. Genes on the lost p arms can be found elsewhere in the genome. 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.

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? A. Histone deacetylation generally decreases gene expression. B. Histone acetylation tends to promote a more condensed chromatin structure. C. Histone deacetylation reduces the extent to which the gene in question will be replicated. D. Histone acetylation favors the formation of heterochromatin rather than euchromatin.

A. Histone deacetylation generally decreases gene expression. positive plus negative gives tighter packaging, no transcription, less charge, loosens and can get transcription 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 (choice A).

respiration

Alveoli are sacs coated with surfactant, a film that reduces surface tension, allowing the alveoli to remain inflated when the lung is compressed during exhalation. Each alveolus is surrounded by tiny capillaries. An average pair of lungs contains something on the order of 600 million alveoli, resulting in a collectively tremendous surface area for gas exchange. The main mechanism of respiration is known as negative-pressure respiration. When the diaphragm (the muscle at the bottom of the thoracic cavity that separates it from the abdominal cavity below) contracts, the thoracic cavity expands. This causes the parietal pleura to expand, causing a pressure gradient that in turn causes the pulmonary pleura and the lungs to expand. When the lungs expand, the pressure within them decreases. The decreased pressure compared to the external environment causes air to rush into the respiratory tract. Exhalation can be either passive or active. In passive exhalation, the simple relaxation of the diaphragm is enough to cause the lungs to contract, increasing the pressure and expelling air. However, the muscles between the ribs (internal intercostal muscles) and abdominal muscles can be used to force air out more intensely and quickly. This frequently occurs during exercise, but increased reliance on active exhalation even at rest can be a sign of respiratory disease. The final part of breathing is gas exchange. Blood runs through the alveolar capillaries and is separated by a wall only one cell thick from the air that is being breathed in. The deoxygenated blood being returned to the lungs is rich in carbon dioxide and poor in oxygen, while the air being breathed in is rich in oxygen and relatively poor in carbon dioxide. Therefore, oxygen and carbon dioxide can simply diffuse down their respective concentration gradients, although it is important to keep in mind that oxygen is carried by hemoglobin.

Assuming that the children studied are 6-7 years old and are in the normal stage of Piaget's stages of cognitive development, which of the following would most likely be observed among these subjects? A. An ability to mentally manipulate information B. A high degree of symbolic play C. Empathy based on understanding the viewpoint of another child D. An ability to solve puzzle games using simple deductive logic

B is correct. In Piaget's stages of development, children from ages 2 to 7 are in the pre-operational stage of development. At that level, children learn to operate symbolically and engage in a lot of symbolic play (e.g. a stick is a sword, a plastic toy is a cake, a doll is a person, etc.).

egalitarian, patriarchy, matriarchy

egalitarian family, both parents have equal power. In a patriarchy, men have more authority, and in a matriarchy, women have more authority.

The disruption of which membrane component is most likely to result in cellular traffic complications similar to those seen in gap junction disorders? A. Cholesterol B. Glycoproteins C. Glycolipids D. Phospholipids

B is correct. 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 component listed that pass all the way through the cell membrane and facilitate membrane transport. A: Cholesterol does not facilitate membrane traffic, nor does it extend across the entire membrane. C, D: Phospholipids and glycolipids are located on the surface of cell membranes and typically do not extend though the entire bilayer. Glycolipids act to provide energy and also serve as markers for cellular recognition. Phospholipids are a structural component of the membrane and are not involved in traffic/transport.

Based on information in the passage, what type of inhibition best describes the action of NADPH on G6PD? I. Competitive II. Allosteric III. Irreversible A. I only B. II only C. II and III only D. I, II, and III

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

Which of the following is an accurate statement regarding mitosis and meiosis? A. Mitosis results in two haploid daughter cells. B. During mitosis and meiosis II, sister chromatids are separated. C. During meiosis I and meiosis II, sister chromatids are separated. D. Meiosis I results in two diploid daughter cells.

B. 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. A: Mitosis results in two diploid daughter cells. C: During meiosis I, sister chromatids are not separated. D: Meiosis I results in two haploid daughter cells.

If pharmacologists wished to convert morphine into a form available to brain tissue, which of the following changes could be made to its molecular structure to allow for the best chance to use the drug as a direct brain treatment? A. Attach a glucose molecule to the nitrogen atom B. Replace the alcoholic protons with acetyl groups C. Lyse the molecule to make it smaller D. Remove all double bonds from the molecule

B is correct. This question asks us to consider the structure of morphine given in the passage. What needs to change for morphine to enter the brain while maintaining its function? 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. So, replacing them with acetyl groups would allow the drug to become more lipophilic; although the C=O bond in the acetylated molecule is polar, it cannot undergo hydrogen bonding, so the acetylated structure is effectively much more lipophilic than the original structure with -COOH groups. Interestingly, diacetylated morphine is heroin. The effects and pharmacology of heroin are somewhat distinct from those of morphine, although they belong to the same general class of opiates, and heroin does indeed more readily cross the BBB. A: This would only make the molecule bigger and do little for its polarity. Additionally, while glucose can pass through the blood-brain barrier, adding glucose to another molecule would not necessarily allow that molecule to utilize glucose transport proteins. C, D: These would destroy the molecule's structure, making it unlikely to maintain its therapeutic function.

which of the following relationships must hold? (Note: 1 m3 = 1000 L.) A. 100 kPa is equivalent to 100 kJ/L. B. 100 kPa is equivalent to 100 J/L. C. 100 Pa is equivalent to 100 J. D. 100 kPa is equivalent to 100 J.

B. 100 kPa is equivalent to 100 J/L. unit conversion problem: 100 kPa = 100,000 Pa = 100,000 N/m^2 = 105 N/m2 100 J/L x (1000 L/m^3) = 105 J/m^3 = 105 Nm/m^3 = 105 N/m^2 table for conversions to study... #53

Arachidonic acid, released during AEA hydrolysis, is NOT a precursor for the synthesis of what class of molecules? A. Prostaglandins B. Catecholamines C. Thromboxanes D. Phospholipids

B. Catecholamines From first paragraph of the passage- AEA is degraded to ethanolamine and arachidonic acid. Ethanolamine - polar head group in the phospholipid phosphatidylethanolamine arachidonic acid - important precursor for 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.

Of what significance to the experimental goals is the finding that AAEL004181 has no introns? A. It is expected, since introns are excised when heterogeneous nuclear RNA is processed into messenger RNA. B. It suggests that AAEL004181 originated from bacteria. C. It is irrelevant, since both bacterial genes and eukaryotic genes contain introns. D. It challenges the hypothesis that AAEL004181 has a bacterial origin.

B. It suggests that AAEL004181 originated from bacteria. 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. A: The researchers found that not only does the mRNA lack introns (expected for a eukaryote), but the gene itself lacks introns. C: Most bacterial genes do not contain introns.

Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmospheric pressure changes? A. At very high temperatures, air behaves less ideally. B. The volume of a mass of air is not constant. C. The majority of molecules in air are nonpolar. D. Air is not of a uniform composition.

B. The volume of a mass of air is not constant. gases are compressible Hydrostatic pressure for liquids is linear because as depth changes, the density of the liquid remains constant. Gases, however, have densities that change according to the forces applied to them. Gases are compressible, while liquids and solids are not.

oxidative phophorylation

Begins with ETC... ends with ADP to ATP The end purpose of oxidative phosphorylation is to phosphorylate ADP into ATP. We should know for the MCAT that the very final step is ATP synthase allowing protons to move down the electrochemical gradient while forming ATP

According to the data in Figure 1, what is the probability that a male Robertsonian translocation carrier who mates with a normal female will produce a viable offspring?

C is correct. According to Figure 1, there are 6 possible outcomes for the gamete production in a 14/21 translocation carrier. Going from left to right, we have a normal gamete, a balanced translocation (all genetic info is present; this gamete would give rise to another ROB carrier), Trisomy 21, Monosomy 21, Monosomy 14, and Trisomy 14. There are no viable autosomal monosomies, and the only viable autosomal trisomies you should know are Trisomy 21, 18, and 13 (though Trisomy 8, 9, and 22 can also survive to term). Therefore, of the 6 gametes produced by the man, 3 of them contain genetic information that would produce a viable offspring (two normal, one with Down Syndrome). 3/6 = 1/2.

Based on the information in the passage, compared to the T11 residue, the T3 residue is: A. closer to the C-terminus, and that terminus is likely to be positively charged in vivo. B. closer to the C-terminus, and that terminus is likely to be negatively charged in vivo. C. closer to the N-terminus, and that terminus is likely to be positively charged in vivo. D. closer to the N-terminus, and that terminus is likely to be negatively charged in vivo.

C is correct. The T11 (threonine at position 11) residue was introduced in the beginning of the third paragraph with T3 being discussed later in the same paragraph. The amino acid structure of a protein is conventionally written from the N-terminus to the C-terminus, meaning that the T3 residue (which is the third residue in the protein) must be closer to the N-terminus than T11. In vivo means "in the living" which indicates we are talking about the conditions found inside the living human body. Physiological pH ranges from 7.2 to 7.4, and at this pH, the N-terminus tends to exist in the form of a positively charged -NH3+ group.

In severe diabetic hyperglycemia (high blood sugar), insulin cannot effectively induce the uptake of glucose by cells. Chronic hyperglycemia directly leads to the presence of which of these molecules in the urine? I. Proteins II. Glucose III. Ketone bodies A. I only B. I and II only C. II and III only

C is correct. 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. 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. Therefore, Roman numerals II and III are correct. I: Proteins in the urine are not the result of hyperglycemia, but rather damage to the glomerulus. Thus, although individuals with diabetes can have protein in their urine, it is not the direct result of elevated glucose levels, but instead a later complication.

Uracil is usually found in: I. tRNA. II. ribosomes. III. single-stranded DNA. A. I only B. II only C. I and II only D. I, II, and III

C is correct. Uracil is found in any structure made of RNA. Both tRNA and ribosomes are made of RNA. Single-stranded DNA has thymine rather than uracil.

Which of the following molecules is/are most likely to have selective proteins in the BBB to facilitate its passage into the brain? A. Antibodies B. Starch C. Amino acids D. Urea

C. Amino acids C is correct. The correct answer will be a molecule or substance that is essential to brain function. Amino acids are necessary for the production of proteins, which are essential for the function of any cell. A: In general, antibodies are too large to cross the BBB. B: While select monosaccharides (e.g. glucose, fructose) can cross the BBB, large polysaccharides like starch are unable to cross. D: Urea is a waste product that is unlikely to have specialized proteins for entry into the brain.

The concentration of intracellular signaling molecules fluctuates rapidly in dividing cells during the cell cycle. Which of the following experimental techniques would be best to elucidate the mechanism of regulation for these proteins? A. RT-PCR and Southern blot B. Southern blot and northern blot C. Western blot and RT-PCR D. Western blot and Southern blot

C. Western blot and RT-PCR C is correct. 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. A, B, D: Southern blots are used to probe DNA for specific sequences; however, changes of gene expression at a cellular level are due to changes in the transcription and translation rates for the genes, not the number of genes themselves.

A student ran an enzyme inhibition experiment on the enzyme from Figure 1 using known competitive inhibitors. Which of the following could NOT have been the result of his experiment? A. Before the enzyme was saturated, the student finds that the rate of the reaction was 0.75 M/s. B. When the enzyme was saturated, the student finds that the rate of the reaction was 0.85 M/s. C. When the enzyme was saturated, the student finds that the rate of the reaction was 0.95 M/s. D. Before the enzyme was saturated, the student finds that the rate of the reaction was 0 M/s.

C. When the enzyme was saturated, the student finds 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, choice C is correct in that the rate of the reaction from Figure 1 cannot be 0.95 M/s.

Centrosomes, Centrioles, centromere , kinetochore

Centrosomes are portions of the cell that help nucleate microtubules and form the mitotic spindle. Centrioles are composed of tubulin and are portions of the centrosome. The centromere is the portion of the chromosome where the two sister chromatids are linked. The kinetochore is a protein structure that helps associate the mitotic spindle to the sister chromatids. The outer portion of the kinetochore interacts with the microtubules, while the inner portion associates with the centromeric DNA.

meiosis and mitosis/MCAT

https://quizlet.com/349181304/mcat-biology-the-cell-cycle-mitosis-meiosis-flash-cards/

Two known competitive inhibitors were studied to analyze their effects on the reaction rate catalyzed by the enzyme lysozyme. Equimolar amounts of inhibitor A (hydronium ion) and inhibitor B (hydroxide ion) were mixed in a beaker before being added into the reaction mixture containing the substrates. Then, lysozyme was also added into the reaction mixture. Based on Figure 1, where would the resulting enzyme kinetics curve for this experiment fall? A. Above line 1 B. Below line 1 but above line 3 C. Below line 3 D. The same as line 1

D is correct. 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. Be careful, read!!

What type of control does siRNA exert on G6PD expression? A. Transcriptional control B. Promotion C. Repression D. Post-transcriptional control

D is correct. 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. A, B: These controls occur before transcription. In contrast, siRNA prevents mRNA from being translated. C: Repressors are defined as protein molecules that bind with DNA or RNA to prevent eventual translation of a protein. Therefore, siRNA is not technically a repressor.

biology - cell membranes

https://quizlet.com/7711687/chapter-7-biology-flash-cards/

How do the chemical modifications described in the passage differ from eukaryotic post-transcriptional modifications? A. Post-transcriptional modifications are made to mRNA, whereas post-translational modifications only affect enzymes. B. Post-transcriptional modifications are not catalyzed by enzymes. C. Post-transcriptional modifications do not involve the addition or removal of parts of the substrate. D. Post-transcriptional modifications are carried out entirely within the nucleus.

D is correct. The modifications were first mentioned in the first paragraph ("acetylation of lysine residues, methylation of lysine and arginine..."), and then a detailed example was given in the last paragraph. These are all modifications to amino acids and/or proteins. Protein modifications must be 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, 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.

If the DNA of a representative species from each of the major kingdoms was examined, the sequences coding for which of following would be expected to be most similar? A. Photosynthesis B. Cholesterol synthesis C. Protein modification D. DNA synthesis

D. 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.

What is the probability that the male offspring of an uninfected male and an infected female will be infected with Wolbachia? A. 0% B. 25% C. 50% D. 100%

D. 100% matrilineal inheritance - This means that Wolbachia is inherited from the mother. Since the female in this question is infected, we can expect that 100% of her offspring will also be infected. mother to offspring transmission

Following extraction into chloroform, the spectrophotometric absorbance at 820 nm of copper in aqueous solution will be greatest at which pH? A. 5.45 B. 5.35 C. 5.25 D. 5.15

D. 5.15 absorption of copper into aqueous solution will be greatest when the least amount of it has been extracted into chloroform. (info on graph)

While the composition of oxygen and nitrogen in air does not change with altitude, the decreasing temperature at high altitude does change the percent of air that is composed of H2O. Assuming constant relative humidity, which of the following can be asserted about the total grams of H2O in a given volume of air at 3000 m above sea level versus at sea level? A. Assuming constant relative humidity means that air has roughly the same mass of H2O per unit volume at 3000 m above sea level. B. Whether air at very high altitude has more or less mass of H2O per unit volume than it does at sea level depends on the temperature at high altitude. C. Air has significantly more mass of H2O per unit volume at 3000 m above sea level. D. Air has significantly less mass of H2O per unit volume at 3000 m above sea level.

D. Air has significantly less mass of H2O per unit volume at 3000 m above sea level. With decreasing temperature, air is able to hold less H2O. Since colder air can hold less total H2O, this means the same relative humidity would result in less total water in the air.

Which of the following molecules does NOT have an atom with sp2-type hybridization? A. Carbon dioxide B. Carbonate C. Formaldehyde D. Methanol

D. Methanol draw the Lewis dot structure for each 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. The structure of methanol is shown below. remember lone pairs!!!

Diabetes mellitus

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.

What is the wavelength of the photons emitted by the 145Pm-m? (Note: 1 eV = 1.6 x 10-19 J)

E = hc/λ The 145Pm-m is taken up by blood, decays, and emits a 300-keV gamma ray, which can be visualized with a lead-lens camera to take pictures of the target tissue. The energy of the photon is given as 300 keV. This will need to be converted to joules using the equation 1 eV = 1.6 x 10-19 J. The energy of a photon is related to its wavelength via the equation E = hc/λ, where h is Planck's constant (6.62 x 10-34 J•s) and c is the speed of light (3 x 108 m/s). Thus, λ = hc/E Solving for λ, we get: λ = [(6.62 x 10-34 J*s)(3 x 108 m/s)] / [(300 x 103 eV)(1.6 x 10-19)] λ = 4.1 x 10-12 m

Poiseuille's law

describes laminar flow of incompressible fluids through a long cylindrical tube. the flow rate (Q), the pressure drop between both ends of the tube (ΔP), radius of the tube (r), length of the tube (L), and the viscosity (η). It can be written in two equivalent forms, both of which are given below: Q = r^4 dP = 1/r^4 R = 1/r^4 Flow (Q) = deltaPπr^4/ 8Ln Pouiseuille's law (R) resistance: R = 8 η l/ π r^4

embryology

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.

Leon Festinger's cognitive dissonance theory

Festinger's cognitive dissonance theory suggests that incongruence between beliefs and behaviors guides behavior change.

practice mental math - pKa and pH from H and Ka and vice versa

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.

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. pay attention to units!!!

pH and molarity problem, practice with pH

Hydrochloric acid is a strong acid and completely dissociates in aqueous solution. In this solution, the hydronium ion concentration is 10.6 M, which can be approximated as 10 M to make the math easier. The pH is the -log of the hydronium ion concentration: -log[10] = -log[101] = -1. While the typical pH range is normally thought of as ranging from 0 to 14, if the concentration of hydronium ion is greater than 1 M, negative pH values are possible. It is also possible to have pH values greater than 14, i.e. if the hydroxide concentration is greater than 1 M.

hydration/solvation

If the potassium ions were able to surround the neodymate ions more efficiently, then it should follow that the neodymate ions are more easily separated from the gallium during the wash.

Biochemical Analytical Techniques

Immunoassays operate on the principle that antibodies cultured from antibody-producing cells can bind with great specificity to a protein antigen of interest. A protein in a mixture can be detected using a radioimmunoassay (RIA); in an RIA, the protein concentration can be assessed indirectly by measuring the extent to which the protein competes with a radioactively labeled standard for antibody binding sites. A related technique is known as the enzyme-linked immunosorbent assay (ELISA). In general, ELISA uses a solid-phase enzyme immunoassay that detects the presence of an antigen. First, a sample containing an unknown amount of antigen is applied to a solid-phase supporting structure. After the antigen contained in the sample becomes attached to the solid support, a specific detection antibody is applied to, and binds with, the antigen. The antibody is then covalently linked to an enzyme directly or through a secondary antibody that is conjugated with an enzyme. Between steps, the plate is often washed with a detergent to rinse unbound proteins or antibodies. Following addition of the enzyme's substrate, a reaction occurs that produces a visualizable signal. The intensity of this signal is related to the quantity of protein antigen present in the original sample. A separate category of analytical techniques combines the use of hybridization (binding of complementary nucleic acid strands) and electrophoresis (movement of molecules toward a charged electrode). For the MCAT, you should be aware of Southern blotting, western blotting, and northern blotting. Southern blotting is a technique used to identify specific DNA sequences. Western and northern blots are used to identify protein and RNA sequences, respectively. In these sequences, the molecules of interest undergo gel electrophoresis to separate them by size and then are transferred to a nitrocellulose membrane that can be heated, at which point probe analysis can be performed. (The process is slightly different for western blots, where antibodies are used instead of DNA/RNA probes.) Two final techniques to be aware of are the polymerase chain reaction (PCR) and Edman degradation. PCR is essentially "laboratory DNA replication," and uses a thermostable DNA polymerase and successive cycles of denaturation, annealing of primers, and extension of a new complementary strand to produce many copies of a sequence of interest. In contrast, Edman degradation is a technique used to sequence proteins via successive cleaving of terminal amino acid residues.

acid base nuetralization

In a neutralization reaction, an acid and a base react with each other to produce a salt and water. The actual neutralization that takes place can be thought of as a "canceling out" of each proton from the acid with one hydroxide ion from the base, forming a molecule of H2O. Despite the name, neutralizations do not always result in a solution of neutral pH (pH = 7 under standard conditions), since the product salt may be acidic or basic. A classic example of an acid-base neutralization reaction is the reaction between HCl (a strong acid) and NaOH (a strong base): HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l). The amount of base (or acid) required to fully neutralize an acid (or base) is given by the equation NacidVacid = NbaseVbase, which relates the normalities and volumes of the two solutions. For monoprotic species, where normality and molarity are equal, this simplifies to the fundamental concept of "moles acid = moles base," while for polyprotic species, it can be described as "moles H+ = moles OH−." In a titration, complete neutralization occurs at a position termed the equivalence point. The above equation, then, is only valid at the equivalence point, not at any random point over the course of the titration.

shifts in reactions

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.

John B. Watson's Little Albert experiment

John B. Watson's Little Albert experiment involved the use of classical conditioning and stimulus generalization to cause a healthy young boy to fear furry animals and objects.

gametes and fertilization, meiosis

Meiosis I results in 2 haploid cells, each with 23 chromosomes consisting of 2 sister chromatids per chromosome. male- sister chromatids split into 4 gametes during meiosis II. 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 to produce gametes.

Ploidy, Aneuploidy, and Nondisjunction

Ploidy - how many copies of each chromosome a cell has. ==humans - most are diploid (2n), two copies of each chromosome (except for the sex chromosomes; females have two X chromosomes and males have an X and a Y chromosome). somatic cells - 2n, body cells germ cells (i.e., ova and spermatozoa) - haploid (n), single copy of each chromosome. Aneuploidy - too many or too few copies of a given chromosome. Causes: nondisjunction in anaphase during cell division. ---monosomy, one copy ---trisomy, 3 copiesAneuploidy is common Aneuploidy occurring in meiosis, only way for aneuploidy to be inheritable. Results in: aneuploidies - Down syndrome (trisomy 21) or Turner syndrome (monosomy X). **nondisjunction during mitosis can also occur, extremely common in cancer cells.

Down's Syndrome (Trisomy 21)

Risk increases w/ maternal age. Caused from nondisjunction during meiosis. ClinMan: small square head, upward slant of the eyes, small low set ears, fat pad on the back of the neck, open mouth with protruding tongue, Simian crease, and MR. Also assoc w/ congenital heart defects, ocular issues, leukemia, respiratory complications. Diag: parental screening/amniocentesis, hormone levels, 4D ultrasound. Treatment: symptomatic and supportive.

socialization (primary/secondary, agents of socialization)

Socialization is the lifelong process through which people inherit, develop, and disseminate social norms, customs, and belief systems. It is through socialization that we develop the habits and skills necessary for successfully living in society. Primary socialization refers to the learning of acceptable actions and attitudes during childhood, mostly from observation of our parents, siblings, friends, teachers, and other authority figures. Secondary socialization refers to the process of learning what is acceptable and appropriate in a smaller, more focused section of society. Learning how to behave at school or in the workplace are examples of secondary socialization. Anticipatory socialization refers to the process by which we prepare for future changes that we anticipate. For example, a security officer who will be switching to the night shift in a few weeks may prepare by shifting his or her sleep cycle, so as to anticipate the demands of the shift change. Resocialization is the process through which we get rid of old behaviors in order to take on new ones. The training of soldiers to obey orders and behave within the rigorous confines of military life is an example of resocialization. The factors that drive and have the most influence over our socialization, known as agents of socialization, change throughout our lives. When we are children, the primary drivers of socialization are our parents and family life. When we are teenagers, our social circles have a strong influence on our socialization. By the time we are adults, the workplace, as well as media, can have a strong effect on how we develop our habits.

strong acids and bases to also include

Strong acids:: HClO4 (Perchloric acid) Strong bases: LiOH CsOH (cesium hydroxide)

structural isomer

Structural isomers describe the different ways that the atoms in a compound can be connected. 3 main sub-catergories of structural isomers ----Chain isomers- different arrangements of the carbon 'skeleton.' ---Functional isomers- isomers where the molecular formula remains the same, but the type of functional group in the atom is changed. (ie. a compound with an oxygen atom in addition to several carbon atoms and the corresponding number of hydrogens could be an alcohol with an -OH group, or an ether with a C-O-C group. ---Positional isomers have a given functional group in different locations (e.g., 1-pentanol vs. 2-pentanol).

synapses

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.

electron transport chain

The ETC is composed of four large protein complexes (Complexes I-IV) embedded in the inner mitochondrial membrane and two small electron carriers shuttling electrons between them. Complex I is known as NADH dehydrogenase, II is known as succinate dehydrogenase, III is known as cytochrome bc or c, and IV is known as cytochrome c oxidase. Electrons are released from NADH and FADH2 through a series of reactions. In the ETC, the energy released from the series of electron transfers is used to pump H+ across the membrane. The unequal concentrations of H+ ions across the membrane establishes an electrochemical gradient, leading to chemiosmosis, or the passive diffusion of the protons down their concentration gradient, which is coupled to ATP synthase. This proton movement generates 90% of the ATP synthesized during oxidative phosphorylation. The electrons passing through the electron transport chain gradually lose energy until eventually they are donated to O2, which accepts two H+ ions and is transformed into water. If the proton gradient is disrupted or destroyed, chemiosmosis can become uncoupled from the ETC, resulting in little to no ATP generation despite the transfer of electrons carrying on. Many poisons and toxins act by uncoupling the proton gradient from ATP synthase.

Ka and Kd

The association constant (Ka) can also be defined, using the mathematical formalism of equilibrium constants, as [ES]/[E][S], where [ES] is the concentration of the enzyme-substrate complex, [E] is the concentration of the enzyme, and [S] is the concentration of the substrate. The dissociation constant (Kd) is then the inverse of Ka, and can be defined as [E][S]/[ES].

Gene Expression

The central dogma of molecular biology states that information is passed from DNA to RNA to protein. This means that when a cell needs more of a certain protein, it can increase the degree to which the gene corresponding to that protein is transcribed. Transcribing more or less of a gene in response to the cell's needs is known as gene expression. It plays a major role in the differentiation of organs in multicellular organisms, and can also vary on shorter time scales in response to changing environmental conditions. The details of how gene expression is regulated in eukaryotes are quite intricate, but you should be aware of some key concepts for the MCAT. 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.

In most cases, the solubility of ionic substances in water increases with temperature, while the opposite pattern is observed for gases.

This is because higher temperatures provide gases with more kinetic energy that they can use to escape the solution. Additionally, pressure favors the solubility of gases.

DNA in Hershey-Chase Experiment

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 (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.

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

The cytoskeleton of one to the cytoskeleton of the other

Prokaryotes and Eukaryotes

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 spirilli. Bacteria are also classified in terms of how they use oxygen in metabolism. Since bacteria by definition do not have membrane-bound organelles, their structure is simpler than that of eukaryotes. However, they have some key structures, including a cell wall containing peptidoglycan that encloses the cell membrane, ribosomes that are differently sized from those of eukaryotes (30S and 50S subunits, instead of the 40S and 60S components in eukaryotes), rotating flagella, and a single circular chromosome. Structural differences between eukaryotes and prokaryotes are often specifically targeted by antibiotics.

hydrogen sulfide

The formula for hydrogen sulfide is H2S. Sulfur is in Group 16, directly under oxygen, so the structure of hydrogen sulfide would be similar to that of water: bent. The Lewis dot structure of H2S also predicts a bent structure due to the presence of the two lone electron pairs on the sulfur atom. Water has a H-O-H angle that is slightly less than the idealized tetrahedral angle of 109° due to the fact that the lone pairs occupy more space around the oxygen atom than do the bonding pairs. The sulfur lone pairs occupy even more space, and the H-S-H angle is around 92°, as shown below.

Histones and DNA Packaging

The human genome contains approximately 3 billion base pairs. To fit into human nuclei, which are generally about 6 μm (6 × 10−6 m) in size, DNA must be compressed. Subdividing the genome into linear chromosomes accomplishes some of this task, but the rest of the job is done by histones and chromatin. 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. 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. 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.

Viable trisomies

Trisomy 21, 18, and 13 (though Trisomy 8, 9, and 22 can also survive to term).

isolectric point from titration curve

The pH corresponding with the middle of the first sharp rise in pH, at 15 mL of added NaOH solution, is the isoelectric point.

membrane transport

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 gases, 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.

Pyruvate dehydrogenase (PDH)

The pyruvate dehydrogenase complex converts pyruvate (a three-carbon molecule that is produced by glycolysis) into acetyl-CoA, a two-carbon molecule that is fed into the citric acid cycle for further metabolism. As the name implies, the pyruvate dehydrogenase complex is indeed a complicated structure composed of multiple molecules. It contains three distinct enzymes that are physically linked with each other (pyruvate dehydrogenase being the most important for the MCAT, but pyruvate dehydrogenase is linked to dihydrolipoyl transacetylase and dihydrolipoyl dehydrogenase, which assist in some of the maneuvering involved with the coenzymes). The complex also requires the action of no fewer than five coenzymes: thiamine pyrophosphate (TPP), FAD, NAD, CoA, and lipoate. It is useful to note that several of them (thiamine, FAD, NAD, and CoA) have components derived from B vitamins. The conversion of pyruvate to acetyl-CoA is accompanied by the production of 1 molecule of NADH, which can eventually enter the electron transport chain to generate ATP, and 1 molecule of carbon dioxide. In eukaryotes, the pyruvate dehydrogenase complex is located within the mitochondrial matrix.

biomolecules

There are four main classes of biomolecules: amino acids/proteins, lipids, carbohydrates, and nucleic acids. All, 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.

Mechanical advantage (MA) inclined plane What is the mechanical advantage of using the ramp, as opposed to lifting the gurney straight up? (Ignore any effects of friction.)

answer: 2 One form of mechanical advantage represents an ability to move a load (do work) using less force than simply lifting the load directly. inclined plane: MA = hypotenuse / height From trigonometry, we know this as sin-1. We're told the ramp makes a 30º angle with the ground, so think: 30-60-90 triangle hypotenuse is 2x longer than side, so 2x/x = 2

anaerobes

bacteria that do not require oxygen for metabolism

experiments in authority

Zimbardo's Stanford prison experiment focused on the effects of power and authority on individuals. Participants designated as "guards" were given power over participants designated as "prisoners," and over time, the guards began to exhibit progressively more abusive and problematic behavior. Milgram's electric shock experiment also relates to authority. This experiment indicated that participants were willing to administer painful stimuli to others if instructed to do so by an authority figure. In reality, the "others" in the study were actors who were simply pretending to be shocked.

In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinning. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp?

both equally likely to slide ma = mgsinƟ - μsmgcosƟ, masses cancel out a = gsinƟ - μsgcosƟ In order for the gurney to slide down the ramp, the force pulling it downward (mgsinƟ) must be greater than the static frictional force (μsFN = μsmgcosƟ). The net force on each gurney is thus Fnet = (mgsinƟ) - (μsmgcosƟ). Since net force is equal to the product of mass and acceleration, we can rewrite this equation as ma = (mgsinƟ) - (μsmgcosƟ), where an acceleration greater than 0 means the gurney will slide down the ramp. The mass of the gurney is present in all terms and can be canceled, meaning that it is not a factor in whether the gurney slides.

facultative anaerobes

can engage in either aerobic or anaerobic metabolism, depending on the circumstances.

A 50-kg child is riding on a playground merry-go-round. If the radius of the circular path of the merry-go-round is 5.0 m and the frequency is 0.1 hertz, what is the force required to keep the child on the ride? A. 25 N B. 100 N C. 1000 N D. 2500 N

centripetal force frequency = 0.1 Hz, so period = 10 s. speed = V = 2πr/T = 2(3.14)(5.0 m)/(10 s) = 3.14 m/s. Using this in the centripetal force equation: Fc = mv^2/r = (50 kg)(3.14 m/s)^2/(5.0 m) Approximate the square of the speed at 10 m2/s2. Fc = (50)(10)/(5) = 100 N

H or OH when given molarity

concentration equal to the molarity of the solution Because strong acids and bases completely ionize, the [H+] and pH (or [OH−] and pOH) are easily calculated, since the concentration of protons or hydroxyl groups is equal to the molarity of the solution. Thus, a 0.4 M solution of KOH has a [OH−] of 0.4, and a 0.007 M solution of HI has a [H+] of 0.007. For solutions so dilute that the H+ or OH− normally present in pure water (1 × 10−7 M) is comparable or larger to the amount produced by acid/base ionization, the ions present due to water must be considered.

Stereoisomers

involve different ways that substituents can be positioned. Cis-trans isomerism on the MCAT most frequently involves alkene bonds. If the highest-priority groups for each carbon are on the same side of the molecule, that molecule is denoted as the cis or Z isomer. If they are on opposite sites, the isomer is trans or the E isomer. Cis-trans isomers generally have similar chemical but different physical properties.

Homotropic regulation

is when a molecule serves as a substrate for its target enzyme, as well as a regulatory molecule of the enzyme's activity. O2 is a homotropic allosteric modulator of hemoglobin.

work measured in

joules!! it is = to energy (KE or PE)

The __________ is the site of attachment of spindle fibers to sister chromatids during mitosis. Possible Answers: centriole centromere centrosome kinetochore

kinetochore kinetochore assembles on centromere, 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. (and in meiosis II)

Eicosanoids

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. most important eicosanoids - large family of signaling molecules, 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.

At what pH will a solution of 2.2 M cysteine be isoelectric? A. 1.96 B. 5.07 C. 8.18 D. 9.15

look to see when will be 0 and use the two pKa values where it would change, then divide by 2 B is correct. 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's 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 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's. pKa = (8.18 + 1.96) / 2 = 10/2 or approximately 5 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.

six phase changes:

melting or fusion (solid to liquid), evaporation (liquid to gas), sublimation (solid to gas), condensation (gas to liquid), freezing (liquid to solid), and deposition (gas to solid). --endothermic phase changes—melting, evaporation, and sublimation—require the input of heat --exothermic phase changes—condensation, freezing, and deposition—release heat into the environment. Endothermic processes involve the breaking of bonds or intermolecular interactions, which require a source of heat. In contrast, exothermic processes typically involve bond formation or an increase in intermolecular force strength.

Mechanoreceptors

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.

kinship of affinity

one in which individuals are related by choice, such as through marriage, rather than through blood, such as the other choices.

obligate anaerobes and obligate aerobes

oxygen is toxic. bacteria that require oxygen for metabolism

half equivalence point and equivalence point

pay attention on graph!! pH=pKa at half equivalence point molarity can be figured from equivalence point

primary and secondary transport

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.

primary kinship, secondary kinship, tertiary kin

primary kinship involves a direct relationship, such as that between a brother and sister or between a father and daughter. secondary kinship exists between an individual and "the primary kin of that person's primary kin." For example, the relationship between a grandson and his paternal grandmother would typically be secondary kinship, since the grandson is primary kin with his own father, who is primary kin with his mother (the grandmother). tertiary kin is one step further removed, and can refer to primary kin of one's primary kin's primary kin (three levels!) or to the secondary kin of one's primary kin, such as one's husband's grandmother.

Which of the following represents the chromosome 14, 21 profile for a Robertsonian translocation carrier?

question #8 reference question on test

When 2 moles of hydrofluoric acid are added to 100 mL of water, the resulting solution has a pH equal to 4. What is the percent dissociation of HF? A. 0.0005% B. 0.05% C. 0.5% D. 5%

remember: pH to 4 = -log[H+] === 10^-4 Original [HF] = 2 moles / 0.100 L = 20 M [H+]/[HF] = (10-4 M H+) / (20 M HF) = 5 x 10-6 % dissociation = 5 x 10-6 x 100% = 5 x 10-4 % dissociation = 5 x 10-4% = 0.0005% Remember that HF is a weak acid, meaning that it does not fully dissociate in water. In many weak acid problems, you must use an ICE table and the Ka of the acid to calculate the pH of an acidic solution.

aerotolerant anaerobes

similar to obligate anaerobes b/c cannot engage in aerobic metabolism, but oxygen is not toxic for them.

parallel circuit, one bulb out

still a circuit, so other bulbs will light Even after unscrewing one of the parallel light bulbs, there is still a circuit for current to flow from the positive terminal to the negative terminal of the battery through the two remaining light bulbs.

structural strain theory of deviance and anomie

structural strain theory traces the origins of deviance to the tensions that are caused by the gap between societal goals and the means people have available to achieve said goals. The related concept of anomie describes social instability caused by the breakdown of social bonds, such as social norms, between individuals and communities.

Hans Eysenck

studied personality with a strong focus on the biological perspective, which considers personality differences to be the result of biological factors.

Baroreceptors

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.

RNA Transcription

transcription is DNA to mRNA—in the nucleus, results in creation of an mRNA copy of a gene that can then be transported to the cytosol for translation into a protein. 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. template strand - antisense strand sense strand - opposite 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. In splicing, noncoding sequences (introns) are removed and coding sequences (exons) are ligated together. (Remember that exons are expressed). Each gene normally has multiple distinct exons that can be ligated in different combinations; that is, if a gene had a set of four exons named A, B, C, and D, possible alternate splicing combinations could include ABCD, ABC, ACD, ABD, BCD, and so on. This dramatically increases the amount of different, but related proteins that can be expressed from a single gene. Splicing explains why there are over 200,000 proteins in the human body, but only approximately 20,000 genes. Splicing is carried out by the spliceosome, a combination of small nuclear RNAs (snRNAs) and protein complexes.