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Question 1

Differential Association Theory

Answer 1

Differential association theory suggests deviance is learned through interaction with others engaging in deviance. For example, if gang members carry illegal firearms, new recruits quickly learn that this is “normal” and expected. The passage does not suggest that the initial PD individuals learned their behaviors through interaction.

Question 2

Labelling Theory

Answer 2

Labeling theory suggests that when individuals are labeled as deviant, they will confirm the label by acting deviant. Initial acts (primary deviance) are usually mild but lead to the “deviant” label and social stigma (disapproval by others). Internalization of the deviant label leads to more serious transgressions (secondary deviance). Large-scale deviance (eg, murder) is precipitated by smaller deviant acts (eg, hurting animals).

Question 3

Conflict Theory

Answer 3

Conflict theory suggests that there is a constant struggle for limited resources between social classes (ie, poor, middle class, wealthy). It is a macro-level (large-scale) theory defining deviance as behavior that does not conform to what is defined as acceptable by social institutions (eg, government, law). The first paragraph describes micro-level (small-scale) deviance and does not suggest conflict between social classes.

Question 4

Strain Theory

Answer 4

Strain theory predicts that individuals experience tension (strain) when there is a disconnect between goals and the available means for achieving those goals. For example, a parent who is unable to feed her child (goal) because there is not enough food (lack of means) experiences strain. This strain causes individuals to seek deviant means of achieving the goal.

Strain theory suggests that innovation occurs when individuals come up with new strategies for obtaining goals, which is most applicable to the first paragraph’s description of the initial acts of PD by the Vietnamese villagers.

Question 5

Mental Set

Answer 5

Mental set describes when a problem solver gets stuck on a method that worked in the past but is not right for the current problem. Most villagers were feeding their children as they had always done (eg, certain foods, twice a day), even though this method resulted in malnutrition. Innovative PD strategies involve new methods (eg, feeding children more often), which overcomes a mental set.

Question 6

Functional Fixedness

Answer 6

Functional fixedness prevents a problem solver from conceiving different uses or functions for an object. Using the edge of a coin to tighten a screw is an example of overcoming functional fixedness to solve a problem. Similarly, eating sweet potato plant leaves, not considered “food” by the rest of the villagers, suggests an ability to overcome functional fixedness.

Question 7

Economic Capital

Answer 7

Economic capital describes an individual’s tangible financial assets, such as property and money/income. Money confers power and status in society, so economic capital confers advantage on those who possess it.

Question 8

Social Capital

Answer 8

Social capital includes an individual’s social networks. In other words, it is the people an individual knows who can help that individual advance in society. For example, knowing the president of Harvard could help someone get accepted into Harvard. In this way, social capital can confer advantage, depending on who is part of the social network and how they can help the individual advance.

Question 9

Human Capital

Answer 9

Human capital describes the collective skill and experience possessed by the individuals who are part of an organization. It is not one of the major types of capital that can be possessed by an individual.

Question 10

Cultural Capital

Answer 10

Cultural capital describes all of the nonfinancial and nonsocial network assets that confer advantage in society. For example, a degree from a well-respected university such as Harvard confers prestige. Hard work, talent, intelligence, and physical attractiveness are also examples of cultural capital because these are all valued in society. PD, which includes behaviors that allow individuals to get ahead in society, is a form of cultural capital.

Question 11

Kohlberg Moral Development

Answer 11

Check out Flashcard on UWorld

Question 12

Identity Diffusion

Answer 12

Identity diffusion (low commitment, low exploration): People at this level lack direction, have not explored options, and have not committed to a particular career path or future.

Question 13

Identity Foreclosure

Answer 13

Identity foreclosure (high commitment, low exploration): People at this level have accepted an identity that they have been assigned (typically by a parent or authority figure) without contemplation or exploration.

Question 14

Identity moratorium

Answer 14

Identity moratorium (low commitment, high exploration): People at this level are still trying new activities and thinking about a career path, but have not yet arrived at a decision.

Question 15

Identity achievement

Answer 15

Identity achievement (high commitment, high exploration): People at this level have explored their options and typically feel confident about who they are and what they want to do in the future.

Question 16

Class consciousness

Answer 16

Class consciousness is an awareness of one’s social status in society (which is necessary for social classes to unite in revolution).

Question 17

False consciousness

Answer 17

False consciousness refers to an inaccurate assessment of one’s own status.

Question 18

Heuristics

Answer 18

Heuristics are mental shortcuts that help individuals come to conclusions or make decisions more quickly and without having to consider every single option. Although time-saving, heuristics are not always accurate.

Question 19

Representativeness Heuristics

Answer 19

Representativeness heuristic proposes that individuals tend to compare things to an existing mental prototype when trying to decide the likeliness of something. In other words, when encountering a new situation, individuals tend to rely on mental representations (of people, events) when drawing conclusions.

Question 20

Nonverbal communication

Answer 20

Nonverbal communication involves all of the wordless cues that convey meaning when exchanged between individuals. Some examples of nonverbal communication cues include distance between speakers, body language and other physical movements, facial expressions, and vocal modifications (tone, volume, emphasis, inflection). Research suggests that social interaction is largely influenced by nonverbal communication.

Question 21

Autocommunication

Answer 21

Autocommunication occurs when a message sender is also the receiver. For example, dolphins echolocate by perceiving how the click sounds they have emitted echo back to them. Predator warning calls, communication that alerts other group members (not the individual making the call), is not relevant to auto-communication.

Question 22

DNA Exonuclease activity

Answer 22

Normally, DNA polymerases are equipped with both 5′-3′ and 3′-5′ exonuclease activity that allows them to remove and replace incorrect nucleotides at either end of a DNA strand. However, the passage states that the Klenow fragment (KF) enzyme described in the experiment does not have 5′-3′ exonuclease activity. KF can only proofread DNA in the 3′-5′ direction on the template strand, so only errors at the 3′ end of the growing strand can be repaired.

Question 23

Base excision and Nucleotide Excision enzymes

Answer 23

Base excision repair and nucleotide excision enzymes have endonuclease activity to remove damaged bases and mismatched nucleotides from the middle of a DNA strand, respectively.

Question 24

Fatty acids Fluidity and Permeability

Answer 24

Fatty acids are nonpolar molecules composed of straight hydrocarbon chains with carboxyl groups at one end. Humans synthesize fatty acids with an even number of carbon atoms, and chains usually range from 14 to 18 carbon atoms long. Fatty acids with no carbon-carbon double bonds are described as saturated because each carbon atom has the maximum number of hydrogen atoms possible. In contrast, an unsaturated fatty acid contains one (monounsaturated) or multiple (polyunsaturated) double bonds that may be in either the cis (Z) or the trans (E) configuration.

Unsaturation contributes significantly to membrane fluidity. The carbon-carbon double bonds decrease the melting temperature of fatty acid chains and increase the average space between lipids. As a result, lipids with unsaturated fatty acid chains remain liquid (fluid) at room temperature. The cis configuration is particularly important in cell membranes as it introduces a bend or “kink” in the fatty acid that prevents phospholipids from stacking together and solidifying.

The passage introduces various molecules that have different effects on membrane fluidity. Of the choices listed, the cis bond of the polyunsaturated fatty acid EPA contributes the most to membrane fluidity.

Question 25

Phospholipids and Experimental separation

Answer 25

Phospholipids (phosphatides) are the main structural lipids of the cell membrane. They are composed of a hydrophilic polar head group, which contains a phosphate, and a hydrophobic tail with one or two fatty acid chains attached to a carbon backbone. Each phospholipid has a unique mass, charge, and solubility due to distinct features in the backbone, polar head groups, and fatty acid chains.

Charge
mass
Solubility

Question 26

Membrane fluidity and permeability

Answer 26

Membrane fluidity is determined partially by the concentration of cholesterol and the tail length of fatty acids. Cholesterol is a steroid alcohol that maintains membrane fluidity at an optimal level. At high temperatures, it provides rigidity and stabilizes the membrane; at low temperatures, it increases fluidity and prevents the membrane from solidifying. Short unsaturated fatty acid tails increase membrane fluidity by preventing phospholipids from clustering together, but longer saturated tails such as those typically found in sphingolipids induce lipid clustering and decrease fluidity. As described in the passage, changes in cholesterol levels and sphingolipid concentration can lead to diseases such as atherosclerosis and psoriasis, respectively.

Atherosclerosis results from the accumulation of cholesterol into CCDs in the membranes of endothelial cells. Free fatty acids such as O3FA can inhibit CCD formation by separating cholesterol molecules. Therefore, an increase in free fatty acids would help treat atherosclerosis.

Psoriasis is marked by an increase in membrane permeability due to a lack of ceramides in the cell membranes of the SC. The long fatty acid chains in sphingolipids such as cer-EOS cause lipids to cluster and decrease permeability. Therefore, increased sphingolipid levels can help treat psoriasis.

(Choice A) Decreasing LDL would improve atherosclerosis, but ceramides should increase, not decrease, in psoriasis therapies.

(Choice B) Increasing cholesterol would exacerbate atherosclerosis. In addition, glycolipids are lipids with carbohydrates linked to sphingosine by glycosidic bonds and do not affect membrane permeability.

(Choice D) Terpenes are precursors in cholesterol synthesis, so increased terpene concentration could worsen atherosclerosis by increasing cholesterol production. Saturated lipids such as cer-EOS are necessary for psoriasis treatments.

Question 27

Digestion of Fats

Answer 27

Lipid processing begins in the small intestine (duodenum), where bile salts break down lipid globules into smaller droplets in a process called emulsification. This process results in the formation of spherical structures, known as micelles, composed of a hydrophobic core containing the nonpolar hydrocarbon tails of lipids and an outer shell of polar head groups that make contact with water. The formation of micelles increases the surface area of lipid available for hydrolysis by lipases.

Lipases are enzymes that digest certain emulsified lipids to facilitate their absorption, although some lipids are nonhydrolyzable. Hydrolyzable lipids contain ester bonds that can be cleaved by lipases through the addition of a water molecule (hydrolysis). These lipids include triacylglycerols, phospholipids, sphingolipids, and waxes. Nonhydrolyzable lipids do not contain the ester linkages necessary for lipase digestion. The most common nonhydrolyzable dietary lipids are cholesterol (steroids) and fat-soluble vitamins (A, D, E, and K).

The question states that some dietary lipids are catabolized (broken down) by lipases, whereas others can be absorbed directly. The most accurate description of lipid processing acknowledges that all lipids are emulsified during digestion but only some lipids, including triglycerides and phospholipids, contain hydrolyzable ester bonds.

(Choice A) Waxes are hydrolyzable lipids that consist of fatty acid chains bound to long-chain alcohols by an ester bond.

(Choices B and C) All lipids are emulsified in the aqueous environment of the small intestine. Prostaglandins are nonhydrolyzable lipids that function as signaling molecules in the human body.

Question 28

Emulsification, Micelles, and Lipases

Answer 28

Emulsification increases the surface area of lipids by breaking down large globules into spherical structures called micelles. Micelles have a hydrophobic core, which contains the nonpolar hydrocarbon tails of lipids, and an outer shell of polar head groups, which make contact with water. Lipases can cleave the ester bonds in hydrolyzable lipids such as triglycerides, phospholipids, and waxes by adding a water molecule (hydrolysis).

Question 29

Enzyme Effect

Answer 29

1. Rate constant Arrhenius equation k = Ae(Ea/RT)
2. More transition state
3. Decreasing the activation energy

Question 30

Assumptions of the Michaelis Menten Equation

Answer 30

1. Free ligand approximation - states that substrate concentration [S] is constant during the reaction. This approximation is only true during the initial phase of the reaction, before a significant amount of substrate is converted to product. Substrate can also be depleted when it binds the enzyme to form the enzyme-substrate complex (ES). To ensure that ES formation does not significantly impact [S], the total concentration of enzyme in solution should be much smaller than any substrate concentration tested.
2. Steady-state assumption - states that the concentration of ES remains constant over the course of the reaction, allowing the rate of product formation to remain constant. Once [S] becomes significantly depleted, ES levels decrease and the reaction slows.
3. Irreversibility assumption - states that the reaction proceeds only in the forward direction, and product does not get converted back to substrate. Once enough product accumulates, the reverse reaction occurs at non-negligible levels and further slows the net rate of product formation.

Each assumption holds true only during the initial phase of the reaction, before substrate is depleted on produc accumulates

Question 31

Enzyme affinity

Answer 31

The Michaelis-Menten constant Km is the substrate concentration at which half of the reaction’s maximum velocity is achieved. It depends on both the rate of substrate binding to the enzyme (ES complex formation) and the rate at which the bound substrate is converted to product (product formation). If the rate of product formation is significantly slower than the rate of ES complex formation, as indicated in the question, then Km essentially measures an enzyme’s affinity for a substrate, or the tendency of an enzyme and a substrate to bind and form a complex.

When product formation is rate-limiting, Km can also be thought of as the substrate concentration at which half of the enzymes in solution are bound. A small Km indicates a strong ES complex because only a small amount of substrate is required to achieve a large amount of complex formation. On the other hand, a large Km indicates a weak ES complex that does not readily form. According to Table 1, CSL-174 has a smaller Km for Hip1 than WKLL-ACC does, so CSL-174 forms a stronger ES complex.

Question 32

Enzyme catalysis process

Answer 32

The question states that the reaction is carried out at saturating substrate concentration. When substrate concentration [S] is significantly higher than Km (ie, saturating), the enzyme essentially operates at maximum velocity Vmax. Vmax is directly proportional to kcat, so the rate of catalysis under these conditions depends directly on kcat. Therefore, the substrate with the highest kcat (CSL-173) will be cleaved the most (and release the most fluorophores), the one with the second highest kcat (CSL-175) will release the second-most fluorophores, and so forth.

Vmax is also directly proportional to the concentration of enzyme [E]. Therefore, the fluorescence generated at 100 pM enzyme should be ten times greater than the fluorescence generated at 10 pM enzyme. Only one graph (Choice B) has both the correct relative intensity of each substrate and a proportional increase in fluorescence when tested at a greater concentration of protein.

(Choices A and D) These graphs do not show a proportional increase in enzymatic activity when enzyme concentration is increased.

(Choice C) This graph depicts a 10-fold increase in activity at 10 pM enzyme relative to 100 pM enzyme. This is the opposite of the expected trend.

Question 33

Cofactors and Coenzymes

Answer 33

Cofactors, including metal ions and coenzymes (small organic compounds), bind to the enzyme’s active site and are required for enzymatic activity. The ACC enzyme is active in the absence of citrate, so citrate is neither a coenzyme nor a cofactor.

Question 34

Gluconeogenesis and Glycolysis notes

Answer 34

The physiological effects of von Gierke disease include low blood glucose and high lactate levels as G6Pase inactivation shuts down the pathways that control them. One of these pathways, gluconeogenesis, consumes lactate to produce glucose, and blocking any enzyme of gluconeogenesis would therefore likely lead to effects similar to those of von Gierke disease. Phosphoenolpyruvate carboxykinase (PEPCK) catalyzes the second step in gluconeogenesis, the conversion of oxaloacetate to phosphoenolpyruvate, and its inactivation can indeed lead to both lactate buildup and glucose depletion in the blood.

(Choice A) Glucose 6-phosphate dehydrogenase (G6PDH) is an enzyme of the pentose phosphate pathway. Although G6PDH uses glucose 6-phosphate as a substrate, it is not involved in gluconeogenesis.

(Choice B) Pyruvate decarboxylase converts pyruvate to acetaldehyde as the first step in the production of ethanol by yeast. Humans do not have this enzyme.

(Choice D) Acyl-CoA dehydrogenase catalyzes the first step in fatty acid catabolism and does not affect blood glucose or lactate levels.

Question 35

Glucose, hormones, and starch

Answer 35

Hypoglycemia can normally be countered with hormones that upregulate gluconeogenesis and glycogenolysis. Normally, glucagon would help counteract hypoglycemia by upregulating gluconeogenesis and glycogenolysis in the liver. However, because von Gierke disease is insensitive to hormones, it inhibits both processes regardless of glucagon levels. Blood glucose levels must therefore be controlled by diet instead.

Starch is a polysaccharide composed of several linked glucose molecules. It is slowly broken down to its constituent glucose molecules in the digestive tract by the enzyme amylase. As glucose molecules are released from starch, they enter the bloodstream and help counteract hypoglycemia.

(Choice A) Insulin induces the uptake of glucose from the blood into cells and upregulates glucose storage (glycogen production) rather than synthesis (glycogenolysis). It would increase the severity of hypoglycemia.

(Choice B) Vitamin A is a lipid-soluble vitamin that is required for vision, immune system maintenance, and growth. It is not involved in glucose homeostasis.

(Choice C) The hormone epinephrine is more commonly released under stress (fight or flight response) than for maintaining glucose homeostasis. In addition, the passage states that von Gierke disease renders patients insensitive to hormones, making these ineffective treatments.

Question 36

Kreb’s Cycle note and Passage notes

Answer 36

Kreb’s cycle is not under hormonal control and check if the passage mentions hormonal sensitivity or not

Question 37

Cori cycle process

Answer 37

During glycolysis, NAD+ is converted to NADH by the enzyme glyceraldehyde-3-phosphate (GAP) dehydrogenase. For glycolysis to continue, NAD+ must be regenerated. Under aerobic conditions, electrons from NADH can be transferred to the electron transport chain (ETC) and ultimately to oxygen. However, in anaerobic conditions, NAD+ cannot be regenerated by the ETC because there is insufficient oxygen to accept electrons. Consequently, NADH donates electrons to pyruvate, which is reduced to lactate in the process. Lactate that builds up from this mechanism must be removed from the system because it can lead to muscle pain and nausea.

The lactate in muscles enters the bloodstream, which carries it to the liver. In the liver, lactate is converted to glucose during gluconeogenesis and is carried back to muscles by the blood. The process of carrying lactate from the muscle to the liver and moving regenerated glucose from the liver back to muscles is called the Cori cycle, which connects gluconeogenesis and glycolysis.

Question 38

Summary of Mitosis steps

Answer 38

Following cell growth and DNA replication during interphase, somatic (non-sex) eukaryotic cells divide via mitosis, producing two genetically identical daughter cells. Mitosis includes the following phases:

1. Prophase: DNA condenses to form chromatids. Each pair of sister (identical) chromatids are joined by a kinetochore to form chromosomes. The nuclear envelope breaks down and centrosomes (microtubule-organizing structures) migrate to opposite poles within the cell. The mitotic spindle is formed as microtubules grow from these centrosomes.
2. Metaphase: Chromosomes attach to spindle fiber microtubules at their kinetochores and align at the metaphase plate, a central plane within the cell.
3. Anaphase: Sister chromatids are pulled apart by the spindle fibers and move toward opposite poles of the cell. This forms two sets of chromosomes within the cell (one set at each cellular pole).
4. Telophase: The nuclear envelope is reformed around each set of chromosomes. Chromosomes decondense and the parental cell undergoes cytokinesis (cytoplasmic division) to produce two identical daughter cells.

In the given scenario, researchers fluorescently labeled various components of actively dividing cancer cells and visualized them under a microscope. Specifically, the nuclear envelope appeared green when viewed by the researchers. In mitosis, the nuclear envelope breaks down during prophase and reforms during telophase. Therefore, green fluorescence would be most intense during telophase as the nuclear envelope reforms.

Question 39

Spermatogenesis and Oogenesis

Answer 39

In humans, the generation of reproductive cells (gametes) occurs in reproductive glands known as the gonads. Before puberty, stem cells called spermatogonia undergo continuous mitotic divisions within the male gonads (ie, testes) to yield identical daughter cells. At puberty, spermatogenesis (sperm production) begins and continues throughout the male’s life.

In spermatogenesis, some of the daughter cells produced from mitotic divisions of spermatogonia become primary spermatocytes. Primary spermatocytes then become mature sperm via meiosis, a process involving two rounds of cell division known as meiosis I and II. A primary spermatocyte that undergoes meiosis I yields two identical haploid cells labeled secondary spermatocytes. Each secondary spermatocyte then undergoes meiosis II to produce four identical haploid spermatids that develop into mature sperm.

In contrast to spermatogenesis, oogenesis (oocyte production) begins before birth, not at puberty, in the female gonads (ie, ovaries) (Choice A). In the female embryo, oogonia (stem cells) undergo mitosis to produce primary oocytes. Each primary oocyte is surrounded by a saclike structure called a follicle. Although they must also undergo meiosis to mature, primary oocytes begin meiosis I but become arrested at prophase I until puberty. At puberty, one primary oocyte is selected during each menstrual cycle to complete meiosis I.

Unlike in spermatogenesis, meiosis I in oogenesis yields two haploid cells of unequal size, the larger one being a secondary oocyte and the smaller one being a polar body that eventually degenerates (Choice C). The secondary oocyte then begins meiosis II but is arrested at metaphase II. The secondary oocyte completes meiosis II and fully matures only if fertilization occurs. To be fertilized, the secondary oocyte must be released during the ovulation phase of the menstrual cycle. During this phase, the follicle ruptures from the ovary and the oocyte enters the fallopian tube to be fertilized by mature sperm.

(Choice D) Oogenesis begins before birth but ceases in older women when ovarian production of female sex hormones declines (ie, menopause). In contrast, spermatogenesis begins at puberty and continues throughout the male’s life. Neither process occurs continuously throughout an organism’s entire life-span.

Educational objective:
Both spermatogenesis and oogenesis involve cells that undergo meiosis I and II. However, oogenesis in females begins in the female embryo and ends at menopause, whereas spermatogenesis in males does not begin until puberty and continues throughout a male’s life.

Question 40

The Urethra Function in Women and possibly men

Answer 40

The urethra is the canal through which urine exits the body from the bladder. Endometrial implants affecting the urethra would impair urinary function, not fertility/reproduction.

Question 41

Vagina

Answer 41

The sexual organ through which sperm enters the uterus. Endometrial implants within the vagina may impair the passage of sperm and decrease the likelihood that an ovulated oocyte will be fertilized

Question 42

Ovaries

Answer 42

The female gonads that produce gametes (oocytes) and secrete female sex hormones. Endometrial implants attached to the ovary may cause infertility by impairing follicular maturation or ovulation, the release of an oocyte (egg cell)

Question 43

Fallopian tubes

Answer 43

The duct structures lined with motile cilia that transfer the ovulated oocyte from the abdominal cavity toward the uterus; also the primary site of fertilization. Endometrial implants that block or impair the function of the fallopian tubes would prevent a fertilized oocyte from reaching the uterus for implantation

Question 44

Cervix

Answer 44

The barrier separating the vagina and uterus. Endometrial implants that block the cervical canal may impair the passage of sperm and decrease the likelihood of fertilization.

Question 45

Self-tolerance of the Immune system

Answer 45

Normally, the immune system is said to be self-tolerant because it does not attack itself but serves to protect the body from foreign substances (antigens). During the early stages of immune cell development, which occurs in the thymus (T cells) and bone marrow (B cells), cells undergo genetic recombination to rearrange their DNA. This rearrangement leads to the expression of novel antigen-binding receptors and antibodies on the surface of T cells and B cells, respectively. The production of billions of immune cells with varied receptor sets results in an immune system able to target many different antigens. However, the rearrangement also results in the expression of receptors that target endogenous molecules (self-antigens). To avoid rampant immune responses against self, these self-recognizing cells are normally destroyed.

Question 46

Negative Selection and response

Answer 46

Negative selection is the process by which immature T cells and B cells possessing receptors that bind to self-antigens are destroyed. Elimination of these cells occurs either by programmed cell death (apoptosis) or by becoming unresponsive to antigens (anergic). According to the question, rheumatoid factor (RF) is an antibody that mediates an autoimmune response by targeting a patient’s own IgG antibodies. RF is not usually made in high concentrations because B cells and T cells that mediate self-recognition and antibody production are typically destroyed during negative selection.

Question 47

Complement proteins

Answer 47

Complement proteins are blood proteins that increase the effectiveness of antibodies by helping to recruit and activate phagocytes. The question describes RF as an antibody that targets self-IgG antibodies; it is not a complement protein.

Question 48

Self-presentation of antigens by the B-cells

Answer 48

Rearrangement of DNA leads to the expression of surface antibodies/receptors that can identify self-antigens (B cells and T cells). Identification of self-antigens by B cells enables subsequent presentation of the self-antigens. However, these self-recognizing cells are normally destroyed during immune cell maturation.

Question 49

Kidney affects

Answer 49

1. Blood nitrogen levels
2. Blood pH
3. Erythropoietin production

Question 50

Increased Genetic diversity and survival possibility

Answer 50

1. Natural selection
2. Gene flow
3. Random mating

Question 51

Genetic drift

Answer 51

Genetic drift describes the natural variations in allele frequencies of a population due to random genetic changes that are not related to natural selection (eg, sampling error, chance event). Although all populations are affected by genetic drift, its effects are more significant in smaller populations because they have a reduced gene pool. The smaller gene pool cannot buffer random (good or bad) variations in allele frequencies that occur due to chance events. In other words, the probability of a beneficial allele being removed by genetic drift increases for populations whose genetic diversity is low. Similarly, drift increases the probability of a deleterious allele becoming fixed within the population.

Question 52

Natural selection

Answer 52

Natural selection is an evolutionary mechanism by which only beneficial alleles are selected for while unfavorable traits are selected against. These selected changes alter the population’s allele frequencies, which improves the species’ fitness. Consequently, these populations have an improved probability for survival rather than an increased probability for extinction.

Question 53

Gene flow

Answer 53

Gene flow, or the changes in allele frequency due to migration, increases genetic diversity by introducing new alleles to the gene pool. Therefore, small populations with low genetic diversity are more likely to benefit from gene flow and therefore are less likely to go extinct.

Question 54

Random mating

Answer 54

Random mating increases a population’s genetic diversity because of the inherent variations that arise during meiosis (recombination) and sexual reproduction (fusion of gametes). Therefore, the variations or changes to the allele frequencies due to random mating will primarily be beneficial to a population as a whole.

Question 55

Note about the diversity of immune system

Answer 55

Typically, it will increase chances of handling diseases

Question 56

Note about Low-frequency alleles

Answer 56

In general, low-frequency alleles have an increased probability of being eliminated by random events than do high-frequency alleles.

Question 57

Note about DNA and expression

Answer 57

All cells that comprise a multicellular organism contain the same DNA (ie, genome). Therefore, the SRY gene is present in all cells, and this comparison can be made.

Question 58

Speciation

Answer 58

A speciation event leads to the formation of a new species from an existing species. Speciation may occur if two populations of a single species are separated for an extended period and become so genetically distinct that individuals are no longer able to reproduce with one another. The graph shows that black mice were already present at generation 0, so it is impossible for black mice to have been derived from either a speciation event in white mice or from a mating event between white mice.

Question 59

Allopatric speciation

Answer 59

Allopatric speciation occurs when a species separates into two separate groups which are isolated from one another. A physical barrier, such as a mountain range or a waterway, makes it impossible for them to breed with one another. Each species develops differently based on the demands of their unique habitat or the genetic characteristics of the group that are passed on to offspring.

Question 60

Peripatric speciation

Answer 60

When small groups of individuals break off from the larger group and form a new species, this is called peripatric speciation (2). As in allopatric speciation, physical barriers make it impossible for members of the groups to interbreed with one another. The main difference between allopatric speciation and peripatric speciation is that in peripatric speciation, one group is much smaller than the other. Unique characteristics of the smaller groups are passed on to future generations of the group, making those traits more common among that group and distinguishing it from the others.

Question 61

Parapatric speciation

Answer 61

In parapatric speciation (3), a species is spread out over a large geographic area. Although it is possible for any member of the species to mate with another member, individuals only mate with those in their own geographic region. Like allopatric and peripatric speciation, different habitats influence the development of different species in parapatric speciation. Instead of being separated by a physical barrier, the species are separated by differences in the same environment.

Question 62

Sympatric speciation

Answer 62

Sympatric speciation (4) is controversial. Some scientists don’t believe it exists. Sympatric speciation occurs when there are no physical barriers preventing any members of a species from mating with another, and all members are in close proximity to one another. A new species, perhaps based on a different food source or characteristic, seems to develop spontaneously. The theory is that some individuals become dependent on certain aspects of an environment—such as shelter or food sources—while others do not.

A possible example of sympatric speciation is the apple maggot, an insect that lays its eggs inside the fruit of an apple, causing it to rot. As the apple falls from the tree, the maggots dig in the ground before emerging as flies several months later. The apple maggot originally laid its eggs in the fruit of a relative of the apple—a fruit called a hawthorn. After apples were introduced to North America in the 19th century, a type of maggot developed that only lays its eggs in apples. The original hawthorn species still only lays its eggs in hawthorns. The two types of maggots are not different species yet, but many scientists believe they are undergoing the process of sympatric speciation.

Question 63

Dipole Moment

Answer 63

This separation of charges across the bond results in a vector quantity called a dipole moment. Therefore, the magnitude of the dipole moment increases as the difference in electronegativity between the two atoms in the bond increases.

Question 64

Spontaneity and Reaction Rate

Answer 64

Spontaneity of a reaction does not provide a direct correlation to the rate of a reaction

Question 65

Nernst Equation

Answer 65

Usually used when the same ions are available in varying concentrations. In this case, think which has the most ions and as such will become reduced in the process.

Question 66

Standard reduction potentials in ETC and a General relation

Answer 66

Standard reduction potentials are indicative of whether a molecule will spontaneously gain or lose electrons. In a spontaneous process, ΔE0′ is positive. The ETC is a series of spontaneous processes, so all of the standard reduction potentials must fall within the range of NAD+ (E0′ = −0.320 V) and O2 (E0′ = 0.816 V), and each step in the chain must have a larger ΔE0′ than the previous step.

Question 67

ATP production and Energy input

Answer 67

ATP production requires energy input, which is measured by ΔG. The number of ATP molecules that can be produced depends on the energy available.

Question 68

Gibbs Free energy and Reduction Potential

Answer 68

ΔG (Gibbs free energy) depends on ΔE (reduction potential) and is related by the Nernst equation: ΔG = −nF ΔE.

Question 69

Electron Transduction

Answer 69

NADH and FADH2, which are produced by catabolic oxidation of lipids, proteins, and carbohydrates, provide energy for ATP production by passing high-energy electrons through the ETC. This process is also known as electron transduction.

Question 70

Electromotive Force

Answer 70

The electromotive force Eocell for a cell is the difference between the standard reduction potential of the reaction at the cathode and the anode. The standard reduction potential of a particular metal or molecule is the potential (in volts) required to reduce the compound. The more negative the value of Eo for any given compound, the less likely it is to be reduced. The equation below is used to determine the standard potential for an oxidation-reduction pair for a particular electrochemical cell:

E∘cell=E∘cathode−E∘anode

For an electrolytic cell, this value is negative, indicating that the oxidation-reduction reaction is not spontaneous.

During the charging phase of a rechargeable battery, an external potential is applied to force the oxidation-reduction reaction to proceed in a nonspontaneous direction. Because Eocell for this battery is a negative value (−2.0 V), the cell is functioning as an electrolytic cell.

Question 71

Electrolytic and Galvanic Cell Intricacies

Answer 71

When a battery is charging, it is an electrolytic cell, and when it is discharging, it is a galvanic cell. The difference between charging and discharging is determined by which electrode is the cathode and which is the anode. Although electron flow is in the opposite direction when a battery is charging as when it is discharging, electrons always flow from the anode to the cathode. Reduction occurs at the cathode, and oxidation occurs at the anode.

PbSO4 is the solid product that accumulates on both electrodes after the battery has been discharged. When the lead-acid battery is charging, electrons flow toward the lead electrode, making it the cathode. Because reduction occurs at the cathode, PbSO4 is reduced.

(Choice A) The lead electrode is the anode and is oxidized when the battery is discharging and the battery is acting as a galvanic cell, but it is the cathode when the battery is charging.

(Choices B and D) In a lead storage (lead-acid) battery, H2SO4 is the electrolyte. Although the electrolyte serves as a charge carrier, it is neither oxidized nor reduced by the reaction.

Question 72

Charging batteries

Answer 72

Charging a battery involves forcing the oxidation-reduction reaction to run in a nonspontaneous direction by applying an external voltage. Any circuit has some amount of internal resistance that will deplete a portion of the applied voltage. Therefore, a higher voltage than that produced by discharging the battery is required to recharge it.

The passage states that a NiCd battery produces a potential of 1.3 V, so the reverse reaction would require more than 1.3 V of applied potential to drive the reaction in the nonspontaneous direction.

(Choice A) For the nonspontaneous reaction to occur, the applied potential must be greater than (not equal to) the potential that the oxidation-reduction reaction spontaneously produces when discharging.

(Choices C and D) Although the electromotive force for a nonspontaneous reaction in an electrolytic cell is negative, the applied potential must be positive to overcome the negative potential of the reverse reaction.

Educational objective:
When charging a battery, an external potential must be applied to force the oxidation-reduction reaction in the nonspontaneous direction. The reverse reaction requires more than the potential produced by the battery because of internal resistance.

Question 73

Electrolytes

Answer 73

While the electrolyte is neither oxidized or reduced, the concentration can be concentrated or diluted.

When a lead storage battery is discharged, lead oxide (PbO2) gains electrons, or is reduced. Based on the oxidation-reduction reactions for the lead storage battery, water is one of the products from the reduction of PbO2.

This oxidation-reduction reaction occurs in an acidic medium. Before discharging, the concentration of H2SO4 is 4 M, but after the battery is completely discharged, the acid is diluted by the additional water, lowering the concentration of H2SO4. Because the concentration of H2SO4 changes when the battery is discharged, it is often used as an indicator of when the battery needs to be charged.

(Choice B) The concentration of PbSO4 increases as the battery is discharged, which requires SO42− ions to come out of solution. This decreases the concentration of H2SO4.

(Choice C) It is true that the concentration of H2SO4 does not remain constant; however, the SO42− ion is not one of the species being oxidized or reduced.

(Choice D) While it is true that the H+ ions are not oxidized or reduced, the H2SO4 concentration does not remain constant, since H+ ions (from H2SO4) are used in the formation of water during the reduction half-reaction.

Question 74

Diamagnetic and paramagnetic

Answer 74

Electrons fill the shells and subshells of an atom in order of increasing energy as described by the atom’s electron configuration. As stated by the Pauli exclusion principle, each orbital within a subshell can hold a maximum of two electrons, which must have opposite spins. In schematic representations, these electrons are represented by arrows and the spin is indicated by the orientation of the arrow (up or down). The occupied orbitals are often represented by blanks or boxes. Hund’s rule states that electrons fill orbitals in such a way as to maximize the number of unpaired electrons.

As a sublevel is filled, electrons form pairs only after all orbitals contain at least one electron. The magnetic properties of an atom (or ion) depend on whether or not unpaired electrons remain after all electrons are assigned to an orbital. If unpaired electrons remain, then the atom is paramagnetic and the unpaired electrons will interact with a magnetic field. If all electrons are paired, then the electrons will not interact with a magnetic field and the atom is diamagnetic.

The electron configurations of the alkaline-earth metals end with a completely filled s-block. Therefore, the electrons occupying the corresponding s orbital in the respective valence shells are paired. Accordingly, Ca and Sr have no unpaired electrons and are both diamagnetic (Numbers I and III). The Ca+2 ion results from losing both electrons from the valence shell, but because all the shells below the valence shell are fully filled (no unpaired electrons), Ca+2 is also diamagnetic (Number II).

Question 75

Beta Decay

Answer 75

In all forms of beta decay, the mass number remains unchanged, while the atomic number increases (β−-decay) or decreases (β+-decay and electron capture) by 1. β−-decay converts a neutron into a proton and emits an electron. β+-decay and electron capture convert a proton into a neutron (the opposite of a β−-decay).

Question 76

Second Ionization Energy

Answer 76

The second ionization energy is the energy required to remove the second of two electrons from an atom. The second ionization energy tends to increase across a period and to decrease down a group; however, ionizations involving core electrons are higher energy than those involving valence electrons.

Question 77

Disproportionation Reaction and Frost Diagrams

Answer 77

A disproportionation reaction is a redox reaction in which both the oxidation and reduction occur to atoms of the same element. Disproportionation reactions can be easily identified by assigning oxidation numbers and then comparing the oxidation number of the given element in the reactants with the oxidation number of the same element in the products. If a compound undergoes disproportionation, some of the compound molecules supply atoms of a particular element for oxidation, and other molecules of the compound supply atoms of the same element for reduction.

Frost diagrams provide a way of predicting which chemical species are likely to undergo disproportionation. As shown in Figure 2 of the passage, a species is likely to undergo disproportionation if its position on the Frost diagram lies above a line connecting the points of two adjacent species because ΔG° < 0 with respect to the mean of the potential gradient.

Based on the above criteria and an examination of Figure 1, MnO2 is least likely to undergo disproportionation because its point on the Frost diagram lies below the line connecting the points of two neighboring species.

(Choices A and B) On the Frost diagram, H3MnO4 is positioned above a line connecting MnO2 and HMnO4−, and MnO43− is above a line connecting MnO2 and MnO42−. Therefore, in both cases, ΔG° < 0 relative to the mean and these species are likely to undergo a disproportionation reaction.

(Choice C) The points for Mn2+, Mn3+, and MnO2 are not perfectly linear, and Mn3+ lies slightly above the line connecting Mn2+ and MnO2. Although disproportionation is unlikely due to the small difference, Mn3+ is not the least likely to undergo disproportionation.

Question 78

Oxidation and Reduction

Answer 78

In an electrochemical reaction, oxidation and reduction always occur simultaneously. The electrons lost by the atom that is oxidized are gained by another atom that is reduced. As a result, the chemical species containing the atom being oxidized is the reducing agent that causes reduction in another atom by giving up the electrons involved in the reduction in the other atom. Conversely, the species with an atom that is reduced functions as the oxidizing agent that causes oxidation by taking electrons from (oxidizing) another atom.

In the given reaction, the chlorine atom in ClO3− is reduced while the iodine atoms in I2 are oxidized. This can be determined by examining the changes in the bonds to oxygen between the species, by assigning oxidation numbers, or by reading the oxidation numbers for chlorine directly from the Frost diagram in the passage (Figure 1). During the reaction, the chlorine atoms lose bonds to oxygen and have a decrease in oxidation state from +5 to −1. In contrast, the iodine atoms gain bonds to oxygen and have an increase in oxidation state from 0 to +5.

Therefore, I2 acts as the reducing agent by giving up electrons and causing ClO3− to be reduced. In the process, I2 itself gets oxidized because it is losing the electrons.

Question 79

Formal charge and Oxidation number

Answer 79

Formal charge and oxidation state (oxidation number) are two different methods used to account for electrons around an atom. Formal charge and oxidation state are not necessarily equal because these values signify different aspects related to chemical bonding. Formal charge assesses the allocation of charge to the atoms in a Lewis structure based on the bonding configuration of the atoms. Oxidation state assesses the gain or loss of electrons by an atom (relative to the elemental valence configuration) due to bond formation.

FC = Group valence - (nonbonding elections + 1/2bonding electrons)

OS = Group valence - Non bonding electrons - Bonding electrons

Question 80

Reduction Potential and Reduction

Answer 80

On a Frost diagram, NE° (the y-axis) is proportional to the standard Gibbs free energy ΔG° according to the relationship

ΔG° = −FNE° = −nFE°

where n is the number of moles of electrons transferred during the electrochemical process, F is the Faraday constant, and E° is the standard reduction potential. Because reactions with a ΔG° < 0 can occur spontaneously, a redox process with a larger positive E° is more favorable than a redox process with a smaller E°. On a Frost diagram, the slope of a line segment joining two species is equal to E° for the couple. As a result, when comparing line segments, the segment with the greatest positive slope is the more favorable reaction.

Comparing the slopes of the line segments between ClO4− and ClO3− for both pH 0 and pH 14 on the Frost diagram of chlorine (Figure 1 in the passage), the slope for acidic conditions (pH 0) is greater than the slope for basic conditions (pH 14). Therefore, performing the experiment under acidic conditions will make the reduction of ClO4− more favorable during the reaction.

Use the GRAPHS!!!!

Question 81

Enzymes and Activation

Answer 81

In contrast to the thermodynamics of a reaction, which indicate whether a reaction will be spontaneous or not, the kinetics of a reaction describe how long a reaction will take to reach completion under given conditions. A reaction may be spontaneous, but it may also proceed so slowly that it does not reach completion in a reasonable amount of time. A catalyst increases the rate (kinetics) of the reaction by decreasing the activation energy required for the reaction to progress at a given temperature. However, a catalyst does not affect the amount of products produced or the enthalpy (thermodynamics) of the reaction.

The decomposition of H2O2 is an exothermic reaction. No matter how long the decomposition takes, the reaction will produce a certain amount of heat per mole of reactants. This means that the identity of the catalyst does not affect the heat of the reaction. Therefore, the reaction using Fe(NO3)3 as the catalyst will result in the same change in temperature as the reaction using NaI as a catalyst, assuming the same amount of H2O2 is used.

Question 82

Why the use of UVs for Thin-Layer Chromatography

Answer 82

Thin-layer chromatography (TLC) is a technique that separates components of a mixture based on polarity, and can be used to monitor reactions. The stationary phase (TLC plate) is made of a polar material, usually silica (SiO2). Therefore, polar molecules will have a greater affinity for the TLC plate and travel a smaller distance up the plate than nonpolar molecules. After separation of the reaction mixture, the components are visualized. Molecules that have UV chromophores, which include double and triple bonds, carbonyls and conjugated systems, can be visualized with UV light.

The aromatic rings of the indole group on the tryptophan side chain and the Fmoc group are made up of conjugated systems of double bonds, and therefore can absorb UV light. Because tryptophan, Fmoc-Cl, and compound 2 are visible on a TLC plate under UV light and differ in relative polarity, the addition of the Fmoc protecting group to tryptophan can be monitored by TLC.

Question 83

HPLC vs Reverse HPLC

Answer 83

High-performance liquid chromatography (HPLC) is a purification technique used for small sample sizes. The instrumentation consists of a sample injector, a column (stationary phase), solvent under pressure (mobile phase), a detector, and a computer for data acquisition. Two types of columns—normal-phase (NP) or reverse-phase (RP)—can be used, depending on the polarity of the compounds being separated.

NP-HPLC is used to separate relatively nonpolar compounds and consists of a polar stationary phase (typically silica) and a nonpolar mobile phase. RP-HPLC is used to separate polar compounds and has a nonpolar stationary phase (typically C-18 alkyl hydrocarbon) and a polar mobile phase. Nonpolar compounds in a mixture will interact more with the stationary phase than polar compounds, causing nonpolar compounds to have a longer retention time.

The only structural difference between Compound 2 and tryptophan is the Fmoc protecting group on Compound 2, which contains a polycyclic aromatic hydrocarbon and an ester. The large nonpolar hydrocarbon group decreases the polarity of Compound 2 relative to tryptophan. Because tryptophan has a shorter retention time and is more polar than Compound 2, tryptophan interacted less with the stationary phase than Compound 2. Therefore, the mobile phase is polar and the stationary phase is nonpolar. These conditions indicate that an RP-HPLC column was used.

(Choice B) An RP-HPLC column was used but RP columns have a nonpolar stationary phase and a polar mobile phase.

(Choices C and D) In NP-HPLC, nonpolar compounds have a greater affinity for the mobile phase (nonpolar), and therefore travel through the column faster than polar compounds. Because tryptophan is more polar than Compound 2, it would have had a longer retention time than Compound 2 on a NP-HPLC column.

Educational objective:
In high-performance liquid chromatography (HPLC), two types of columns—normal-phase (NP) or reverse-phase (RP)—can be used, depending on the polarity of the compounds being separated. NP-HPLC is used to separate nonpolar compounds and consists of a polar stationary phase and a nonpolar mobile phase. RP-HPLC is used to separate polar compounds and consists of a nonpolar stationary phase and a polar mobile phase.

Question 84

Imide

Answer 84

An imide is a functional group with a nitrogen atom bound to two acyl groups (two carbonyl carbon atoms). Imides are structurally similar to acid anhydrides but include bonds with the acyl groups instead of oxygen atoms.

Question 85

Carbocation stability

Answer 85

Tertiary > Secondary > Primary > Methyl

Question 86

Carboanion stability

Answer 86

methyl > primary > secondary > tertiary

Question 87

Conjugated system and Resonance forms

Answer 87

A conjugated system can be identified by alternating single bonds and double or triple bonds, or in other words, by alternating π bonds (p orbitals) separated by σ bonds. Conjugation allows for the delocalization of electron density, meaning electrons can be distributed through the alternating system’s π bonds. In general, conjugation and electron delocalization serve to stabilize molecules such as carbanions, carbocations, and radicals by creating a more favorable charge distribution, therefore reducing charge density.

The quinonoid intermediate (Figure 3) delocalizes the negative charge of a carbanion over a large conjugated system of π bonds whereas the enolate (Figure 1) delocalizes the negative charge over a relatively small region via its resonance forms.

Question 88

Functional groups and Priority

Answer 88

For compounds with multiple functional groups, the highest-priority functional group should be used as the basis for assigning a compound name. This group determines the compound name ending (suffix), and all lower-priority functional groups are named as substituent branches using unique prefixes associated with each. Carbon atoms of the longest continuous carbon chain within the structure should be numbered to specify the location of the substituents. The selected carbon chain must incorporate the highest-priority functional group and must be numbered in the direction that assigns it the lowest possible number. Stereocenter configurations are also specified in parentheses using the same numbers at the beginning of the compound name.

In Compound 1, four functional groups are present, with an aldehyde group as the highest-priority group. Therefore, the compound is named as an aldehyde using the –al suffix. The longest continuous carbon chain containing the aldehyde is six carbon atoms long, and the chain is numbered to give the lowest position to the aldehyde carbon. The amine, ketone, and alcohol groups are subsequently located along the position reference numbers and are named as substituents using their respective prefixes. The configurations of the three stereogenic carbon atoms are then specified by position number at the beginning of the name to give (2R, 3S, 5S)-2-amino-5-hydroxy-3-methyl-4-oxohexanal.

Question 89

BH3 and NaBH4 reduction and oxidation

Answer 89

Reduction of a functional group results in a gain of electrons from the transfer of a hydride, a decreased oxidation state of carbon, and a decreased number of C–O bonds. Several reagents can be used to reduce a functional group, including borane (BH3) and NaBH4. These reducing agents transfer one or more hydrides to the carbon atom being reduced. BH3 reacts most readily with carboxylic acid carbonyl groups and therefore will selectively reduce carboxylic acids. As a result, the intermediate aldehyde will be further reduced to a primary alcohol. NaBH4 reduces the reactive carbonyls—ketones and aldehydes—and will not reduce the less reactive esters and carboxylic acids.

If a compound that contains a ketone, an ester, and a carboxylic acid is reacted with BH3, the carboxylic acid will be selectively reduced to a primary alcohol, leaving the ketone and ester intact. When the same compound is reacted with NaBH4, only the ketone will be reduced, generating a secondary alcohol. Therefore, the products of the two reactions will be different.

Question 90

Retro-aldol

Answer 90

A retro-aldol reaction is the reverse of an aldol condensation. Either the aldol product (β-hydroxy ketone) or the dehydration product (enone or enal) is heated and treated with base, causing the carbon-carbon bond between the α- and β-carbons to break. The retro-aldol reaction yields two products: either two ketones, two aldehydes, or one of each.

The products depend on the substituents on the carbonyl carbon and the β-carbon. If the carbonyl carbon is bonded to a hydrogen atom before the reaction (ie, it was already an aldehyde), it will still be an aldehyde after the reaction. If the β-carbon is bonded to a hydrogen, it will become an aldehyde. If the carbonyl or β-carbon are not bonded to hydrogen, they will become ketones.

Compound 4 is an enone in which the β-carbon is bonded to hydrogen and the carbonyl carbon is not. Therefore, a ketone will form from the carbonyl and an aldehyde will form from the β-carbon when Compound 4 is heated and treated with a base.

Question 91

Aldol condensations

Answer 91

Aldol condensations are important carbon-carbon bond-forming reactions where two carbonyl compounds (ketones and/or aldehydes) are joined together by nucleophilic addition to form the aldol product (a β-hydroxy ketone). The carbonyls can come from separate compounds (intermolecular aldol condensation) or from a dicarbonyl compound (intramolecular aldol condensation). The aldol product can then undergo dehydration to form an α,β-unsaturated carbonyl compound.

An aldol condensation as described in the passage requires two carbonyl substrates (Compounds 1 and 2) (Choice A). The α-carbon (carbon adjacent to the carbonyl) on Compound 1 is deprotonated when base is added. The intermediate formed is stabilized through resonance as the negative charge on the α-carbon is delocalized to the carbonyl oxygen, forming an enolate (Choice C). The nucleophilic enolate (from Compound 1) then attacks the carbonyl of Compound 2. A new carbon-carbon bond is formed (the aldol product), in which the carbonyl adjacent to the nucleophile remains intact. Finally, –OH elimination takes place to give a conjugated product, an α,β-unsaturated enone (Choice D).

The dehydration step requires the α-carbon to be deprotonated rather than protonated. The resulting carbanion is an enolate that can eliminate the –OH on the β-carbon and form a double bond between the α- and β-carbons, giving a conjugated product.

Question 92

State and Process functions

Answer 92

A system is in a state of thermodynamic equilibrium if the temperature of the system is constant and uniform throughout its volume and there is no flow of energy. A system is in thermodynamic equilibrium with another system or its surroundings if both have the same temperature.

State functions (or state quantities) describe the equilibrium state of a system as a relationship between various thermodynamic variables and are independent of the path taken by the system to arrive at its present state. State functions include a system’s pressure, volume, and temperature.

Process functions (or path functions) describe the path taken by a system to transition from one equilibrium state to another. A system transitions from one state to another due to a net flow of energy in the form of heat transfer or work. For example, the loss or gain of heat is a process function because it describes the path taken by a system from its current pressure, volume, and temperature to a different set of values.

In the Hampson-Linde cycle described in the passage, work is done by the compressor on the nitrogen gas to decrease its volume. Therefore, work is a process function because it describes what was done to the system to change its state (Choice C). A system’s entropy is a measure of its current state of disorder and does not depend on how the system arrived at that state (Choice B). Therefore, the entropy of nitrogen is a state function.

Question 93

State Functions

Answer 93

State functions (or state quantities) describe the equilibrium state of a system as a relationship between various thermodynamic variables and are independent of the path taken by the system to arrive at its present state. State functions include a system’s pressure, volume, and temperature. Such as entropy

Question 94

Process Functions

Answer 94

Process functions (or path functions) describe the path taken by a system to transition from one equilibrium state to another. A system transitions from one state to another due to a net flow of energy in the form of heat transfer or work. For example, the loss or gain of heat is a process function because it describes the path taken by a system from its current pressure, volume, and temperature to a different set of values. Such as work

Question 95

Conduction

Answer 95

Conduction is the transfer of heat through direct physical contact.

Question 96

Convection

Answer 96

Convection is the transfer of heat through the flow of fluids. Fluids absorb heat from hotter regions and deliver it to colder regions.

Question 97

Radiation

Answer 97

Radiation is the transfer of heat through electromagnetic radiation, such as infrared light. Radiation heat transfer is significant only for high temperatures.

Question 98

Heat Energy

Answer 98

Heat is energy that flows from high temperatures to low temperatures. Heat can be transferred through direct physical contact (conduction), through the flow of fluids (convection), or through emission of electromagnetic energy (radiation)

Question 99

Centration

Answer 99

In psychology, centration is the tendency to focus on one salient aspect of a situation and neglect other, possibly relevant aspects. Introduced by the Swiss psychologist Jean Piaget through his cognitive-developmental stage theory, centration is a behaviour often demonstrated in the preoperational stage.

Question 100

Venturi Effect

Answer 100

The Venturi effect describes the decrease in pressure observed concurrently with an increase in fluid velocity. The venturi effect is modeled by Bernoulli’s equation, which describes the conservation of energy in fluids.

Question 101

Bernoulli’s equation

Answer 101

One application of Bernoulli’s equation relates to the travel of fluids between two points (A and B) within a fixed conduit (a pipe or channel through which a fluid passes). Bernoulli’s equation is useful in modeling the flow of ideal fluids because ideal fluids are non-compressible (ie, unchanging density). Fluid incompressibility requires that changes in conduit geometry (ie, cross-sectional area) cause changes in fluid velocity because the volumetric flow rate through interconnected segments of a conduit must be equal. Consequently, changes in conduit geometry are accompanied by changes in fluid pressure or potential energy, maintaining the conservation of energy.

Question 102

Schiff Base

Answer 102

Imine

Question 103

Enzymes DO NOT AFFECT

Answer 103

1. Thermodynamics (enthalpy) of a reaction

2. The amount of products produced