Biology

Question 2

Ask Landon

Question 11

Content -> The sequence of the mRNA transcript is identical to the sequence of the coding DNA strand (again, with uracil rather than thymine).

question 13

Figure 1 analysis Concept: A survival curve graph illustrates the proportion of healthy individuals remaining in a population (y-axis) as a function of time (x-axis). Comparing survival curves of different populations allows researchers to identify which factors lead to an increased or decreased probability of survival.

Question 40

Graph interpretation content: To analyze whether one gene regulates another, researchers can assess variables of interest (eg, cell growth, apoptosis) using engineered cell lines in which one of the target genes is expressed and the other gene of interest is silenced and then reintroduced

Question 43

Just had to find the sequence on the gel Content: The Sanger (dideoxy) method of DNA sequencing analysis uses PCR along with the incorporation of ddNTPs and gel electrophoresis to determine the nucleotide order of target DNA molecules.

question 18

Within the gustatory-salivary reflex, the sensory neuron, which transmits impulses toward the spinal cord (a component of the CNS), makes up the afferent component of the reflex arc. The pre- and post ganglionic fibers, which transmit impulses away from the spinal cord, make up the efferent component of the reflex arc. concept: Afferent signals approach the central nervous system. Efferent signals exit the central nervous system.

question 3

the question says that the mutation occurs in hepatocytes so the chance of passing it down to the child would be 0 content -> mutations that occur in the repro cells are germline mutations and these are ones passed on to offsprings not the somatic mutations

Q4

Asthma is defined in passage as air "trapped" in lungs so that basically means hypoventilation so you're not exhaling which means there will be more CO2 retained in body which will shift bicarb buffer system to the right and increase H+ ions = more acididc so respiratory acidosis. The homeostatic response is increased respiratory rate to get rid of CO2 Educational objective: Respiratory gas exchange functions to remove CO2 from the blood, and CO2 levels affect the pH of the blood by shifting the equilibrium of the bicarbonate buffer system. A decrease in gas exchange will cause respiratory acidosis, and the body will attempt to restore normal blood pH by increasing the respiratory rate.

q23

Had to know that liver produces bile and that teh gallbladder STORES bile. Bile breaks down lipids Educational objective: Bile is synthesized in the liver, stored in the gallbladder, and released into the duodenum to aid lipid digestion. Bile salts, a key component of bile, mechanically digest lipid globules by physically breaking them down into smaller droplets in a process known as emulsification. Subsequently, pancreatic lipase chemically digests emulsified lipids.

q16

Q asking a decrease in which of the folllowing will result in lowered BP. So RAS increases BP and blood volume, aldosterone does the same but through absorption of Na+ and ADH also increases water reabsoprtion the collecting duct so filtrate becomes more concentrated so it increases blood volume and pressure. So a decrease in all three would cause lowered BP Educational objective: Angiotensin II, antidiuretic hormone, and aldosterone all regulate blood pressure. The effect of these hormones on blood pressure can be explained wholly or in part by the modulation of water and salt reabsorption in the kidneys.

question 16

Q asking what is the same between sperm and oocytes? They both contribute the same number of chromosomes to a zygote because one set of chromosomes from mom and one set of chromosomes from the dad to form a diploid zygote because sperm and eggs are haploid. Educational objective: Egg and sperm cells are haploid cells that contribute an equal number of chromosomes to a zygote during fertilization.

q9

data interpretation In the graph you can see that as time increases the HIF-1alpha levels decrease and 0 mins is defined as hypoxia levels while 30, 60 and 120 is defined as normal oxygen level. That would make II and III correct that at low levels of o2, HIF-1 is functional and at normal levels its decreased concept: A heterodimer is composed of two different protein subunits, and its assembly and activity are dependent on the expression and proper function of these subunits. Oxygen concentration, among other factors (eg, temperature, pH), has been shown to influence protein expression.

Question 3

This question is again referring to the cortisol releasing pathway content: The hypothalamus is a brain structure that regulates numerous endocrine functions through its association with the pituitary gland. Together, the hypothalamus and pituitary gland regulate the synthesis and secretion of hormones that influence metabolism, reproduction, and other important functions.

q1

needed to know the differences btw the types of muscle types concept: Muscle tissue can be classified as cardiac, skeletal, or smooth. Cardiac muscle is striated and under involuntary control. Skeletal muscle is striated and under voluntary control, and smooth muscle is not striated and under involuntary control.

Q7

only eukaryotes have the ability to process the mRNA. Protozoa are single celled eukaryotic organisms concept: Eukaryotic and prokaryotic cells are similar in that both have mechanisms of sexual reproduction and are enclosed by a plasma membrane. In addition, all prokaryotes and some eukaryotes have cell walls. However, eukaryotic cells are unique in that they have membrane-bound organelles as well as a mechanism for intron splicing by the spliceosome.

q28

peritoneal cavity is the space btw the parietal and visceral layers of the peritoneum. The peritoneal cavity is found within the abdomen and contains organs such as the liver, stomach, and intestines. A staple line leak at the junction of the biliopancreatic limb and the alimentary limb would allow gastrointestinal (GI) tract contents to enter the peritoneal cavity. concept: The peritoneum is composed of two membranes that line the abdomen: the parietal layer, which lines the abdominal wall, and the visceral layer, which covers the abdominal organs. The peritoneal cavity is a potential space between the parietal and visceral layers of the peritoneum.

Post-translational modifications

phopshorylation - adding phosphate group proteolysis - breakdown of protein peptide bonds into amino acids (enzymes do this) ubiquitination - adding a marker fro proteasome (complex) to degrade protein glycolysylation - Carbohydrate group on protein, plays a critical role in determining protein structure, function and stability. Structurally, glycosylation is known to affect the three dimensional configuration of proteins.

q12

psg states that partial thickness burn damages the epidermis and dermis, and insulation is through adipose tissue in the hypodermis (subcutaneous layer) and since that isn't damaged by partial thickness burns, the insulation of skin would not be affected Educational objective: Skin has many functions: The subcutaneous layer is composed of adipose cells that insulate the body, epidermal melanocytes prevent UV radiation from damaging the DNA of cells, hair is a keratinized derivative of skin that helps protect the body from external injury, and dermal sweat glands secrete sweat onto the skin surface to regulate body temperature.

q19

somatic = voluntary movements and chewing is the only voluntary movement thats in the asnwer choices concept: The somatic and autonomic nervous systems are the two branches of the peripheral nervous system. Voluntary activities occur through the contraction of skeletal muscles, which is mediated by the somatic nervous system.

q13

A patient is homozygous for a deleterious mutation in a gene coding for a hemoglobin subunit. Expression of this mutant gene would likely produce a portion of the hemoglobin protein with impaired function. Because RBCs are produced from stem cells in the bone marrow, the most effective treatment for this patient would be gene therapy to introduce and activate a functional hemoglobin gene in bone marrow stem cells. This would likely lead to the production of RBCs that contain hemoglobin with normal oxygen-binding abilities. concept: Gene therapy is a technique in which a functional copy of a gene is introduced (eg, via retroviral vectors) into the cells of a patient with a mutation in that gene

Q1

Arteries move away from the body. Thats why pulmonary artery in the heart is deoxygenated because it takes the blood away from the heart to the lungs to get it oxygenated. Deoxygenation blood will have higher CO2 and higher H+ as a result bc of the blood equation. The umbilical vein carries oxygenated, nutrient-rich blood from the placenta to the fetus, and the umbilical arteries carry deoxygenated, nutrient-depleted blood from the fetus to the placenta. CO2 + H2O -> H2CO3 -> H+ + HCO2- concept: Within the pulmonary circuit, the pulmonary arteries carry CO2-rich, O2-poor blood from the heart to the lungs, where CO2 is exchanged for O2. The pulmonary veins then carry CO2-poor, O2-rich blood from the lungs to the heart, which then pumps the oxygenated blood to all body tissues.

Question 5

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direction of dynein and kinesin

Dynein towards the negative (nuclues) Kinesis towards the periphery of the cell (positive)

Question 21

Educational objective: Erythrocytes are biconcave, disc-shaped cells containing hemoglobin, the carrier protein that delivers oxygen to body tissues. Erythrocytes contain no mitochondria as they expel their organelles during the maturation process in the bone marrow. As a result, mitochondrial mutations do not affect erythrocytes.

q15

Look at fig 1 and all you have to do is look at x intercepts for both for the threshold and find the difference so 200-85=115 Educational objective: A threshold is a set value that must be surpassed for a particular event to take place.

Question 20

More Angptl4 = less LPL activity = more triglyceride content -> A constitutively active gene is transcribed at a relatively constant rate regardless of current cell conditions. In a knockout model, an existing gene is replaced or disrupted, which leads to an absence of the protein product.

Difference between prokaryotic and eukaryotic reproduction

Prokaryotes: - Prokaryotes reproduce asexually by binary fission; - they can also exchange genetic material by transformation, transduction, and conjugation. Eukaryotes: - Transfection is the process by which genetic material, usually in the form of a plasmid, is introduced into eukaryotic cells - mitosis and meiosis

q11

Psg states that OA causes pain and restricts joint movement so can cancel C and D cuz they talk abt calcium storage for OA. And osteoporosis is decreased bone density and mass so it would correlate to structural support so A Educational objective: The skeletal system facilitates mobility, provides the body with structural support, is the site of hematopoiesis, protects the body's internal organs, and serves to store fat, calcium, and phosphate.

What is the function of smooth ER and rough ER?

Smooth ER: has a lot of metabolic functions depending on the cell type like lipid synthesis (testes/ovaries), drug/poison detoxification (liver), and calcium ion storage (muscle) Rough ER: produce proteins

Q1

There is less efficient gas exchange bc in the psg they say that in emphysema theres a breakdown of alveolar walls. That would decreases the alveoli surface area leading to less gas exchange. concept: The alveoli are highly efficient at gas exchange due to their structure and extensive surface area. The destruction of alveolar walls in emphysematous lungs is expected to decrease the lungs' ability to perform respiratory gas exchange.

question 25

adrenal medulla secretes epinephrine and norepi which work like glucagon so they will inhibit anyway pathways that insulin promotes. Thats why A is the correct answer Concept: Energy metabolism refers to the processes by which the body manages cellular energy stores. The adrenal gland increases the body-wide level of free cellular energy sources by releasing glucocorticoids, norepinephrine, and epinephrine.

Question 45

asking Landon

Question 24

it went from a drastic reduction from one generation to the next and thats why its population bottleneck over natural selection In natural selection, alleles that produce environmentally beneficial phenotypes are more likely to be selected for and passed on to the next generation than alleles that are harmful. In this scenario, only organisms that primarily express the deleterious mutant mtDNA survived and reproduced, a result opposite of what would be expected had natural selection occurred. Educational Objective: Natural selection is the tendency for alleles that make an organism better suited for survival and reproduction to be passed along to the next generation. Bottleneck events reduce genetic diversity and change the allele frequencies of a population in a random way.

q16

q says connecting 2 bones so it has to be a ligament concept: Joints are structures of the musculoskeletal system where bones articulate (ie, interact) and can range in mobility from freely moveable to immoveable. Within moveable joints, strong connective tissue structures called tendons attach muscle to bone whereas ligaments attach bone to bone.

q3

the box whisker plot shows that sCD18 levels are decreased in cRA patients because its lower than eRA and HC and there is overlap of the interquartile range meaning that both groups differ in mean and data spread. Since sCD18 levels can be used to determine disease status, you can use figure 1 to support the claim that eRA and cRA patients can be differentiated by assessing their sCD18 levels. Concept: Box-and-whisker plots are used to visually represent the median and distribution of the data into quartile ranges.

question 16

the drug in the question is a Ach agonist meaning it will activate Ach. In the passage they say that salivary glands use ach. So, administering this drug will increase salivation due to stimulation of ach receptors (mAChRs) concept: Agonists are ligands that activate the receptors they interact with and induce downstream effects. Antagonists are ligands that inhibit the receptors they interact with and block downstream effects.

question 18

the homodimer from the passage is 450 AA 1AA = 110dal so the weight of 450 AA is about 50kDa (multiply them together 450 AA in LPL homodimer × 3 nucleotides = 1,350 nucleotides content -> The total number of nucleotides in an mRNA molecule can be calculated by multiplying the number of amino acids in the protein by the number of nucleotides (ie, three) in a codon. The approximate molecular weight of a protein can be obtained by multiplying the number of amino acids in the protein by the average molecular weight of an amino acid (110 Da).

question 19

the question asking what can you use to measure the gene count in VEGF. You are basically measuring the DNA so southern blot and DNA sequencing. SN(O)P DR(O)P concept: DNA sequencing and Southern blotting are DNA assays that may be used to assess the relative quantity of genes between tissue types. Northern blotting is an RNA assay used to assess gene expression in different tissues.

q2

The dissociation of ADP and Pi causes the power stroke. And the binding of ATP to myosin causes actin to be released from myosin head. concept: When ATP bound to a myosin head hydrolyzes, the myosin head shifts back and attaches to the actin filament. Dissociation of ADP and Pi leads to a power stroke, which shortens the sarcomere. A new ATP molecule binds to the myosin head, causing it to dissociate from the actin filament. The action of many myosin heads cycling through this process leads to muscle contraction.

Question 22

if the large ribosomal subunit is impaired that means the whole ribosome is impaired which means that ti will not bind to the rough ER content -> Ribosomes translate mRNA sequences into proteins in eukaryotes (80S ribosomes: 60S large subunit and 40S small subunit) and prokaryotes (70S ribosomes: 50S large subunit and 30S small subunit). The ribosome-mRNA complex translocates to the rough endoplasmic reticulum to synthesize secretory, lysosomal, or integral membrane proteins.

question 8

An electric signal in the pre-synaptic neuron travels down the axon which conducts the signal to the axon terminals. From there, chemical messengers called neurotransmitters are released into the synaptic cleft (synapse), the region between the axon terminals and the dendrites of the next neuron. Neurotransmitters bind to receptors on the dendrites of the post-synaptic neuron, altering the electric potential of the cell. Lastly, the change in electric potential spreads to the cell body (soma). concept: The electric signal from the axon of the pre-synaptic neuron is converted to the release of neurotransmitters into the synaptic cleft between neurons. Neurotransmitters then bind to receptors on post-synaptic dendrites, changing the electric potential of the cell.

question 29

Apoptosis is important for normal patterning and development. Ex. cells in between fingers (webbing) are removed through apoptosis to sculpt the hand. If you abnormally activate apoptotic pathways (through oxidative stress) then you can result in embryonic anomalies (ex NTDs). Oxidative stress is due to when the body can't detoxify ROS (reactive oxygen species) like peroxides and free radicals. ROS are naturally generated from cellular reactions involving reactions that take place in mitochondria (via oxidative phosphorylation) and peroxisomes (via β-oxidation of long-chain fatty acids). The passage states that NTDs can be caused by increased embryonic oxidative stress during crucial developmental periods; therefore, a child born with myelomeningocele, a severe NTD, most likely experienced oxidative stress in utero. Compared to a healthy newborn, a newborn with myelomeningocele likely had a higher incidence of apoptosis induced by oxidative stress during gestation, as well as: Higher frequency of cell damage at critical developmental periods during gestation (Choice A) Higher cellular concentrations of free radicals during gestation (Choice B) Lower levels of enzymes with antioxidant capabilities during gestation (Choice C) Educational objective:Apoptosis (programmed cell death) is crucial for normal embryonic patterning and development. Oxidative stress occurs when the body is unable to detoxify naturally generated reactive oxygen species, which then cause considerable cell damage upon accumulation. This damage can result in abnormal apoptosis and congenital malformations.

Q30

Asking Landon

q23

Can see from table that systolic is increased over diastolic Educational objective: Arterial blood pressure is highest when the heart contracts (during systole) and lowest when the heart relaxes (diastole). When measuring blood pressure (mm Hg), the higher systolic blood pressure is recorded over the lower diastolic blood pressure.

Q6

Cilia are hair-like organelles that are found in some specialized cells. PCD in stem causes dysfunctional cilia, which would allow mucous to accumulate in the airways and result in wheezing sounds. Ciliated cells beat synchronously to sweep foreign bodies trapped in mucus up the respiratory tract to the pharynx. So, the ability to remove inhaled particulates would be impaired. Educational objective: In the upper respiratory tract, mucus-producing cells and ciliated cells work together to trap and remove potentially pathogenic bodies. Therefore, dysfunctional cilia would decrease the respiratory system's ability to remove inhaled particulates.

q25

Compared to the healthy individual's ECG, the patient's ECG shows more waves of atrial and ventricular depolarization over the same time period. This indicates an increased number of APs fired from the patient's SA and AV nodal cells, leading to more frequent atrial and ventricular contraction. Because depolarization and AP generation depend on the influx of positive ions into cells (Na+), a more rapid influx of positively charged ions into SA nodal cells may lead to increased firing of APs from the SA node and more frequent atrial contraction, as observed in the patient. concept: Heart rate is regulated by the activity of specialized clusters of self-depolarizing cells known as the SA and AV nodes. APs are initiated in the SA node and travel through the atria, stimulating atrial contraction. These APs then reach the AV node and, after a brief delay, are relayed to ventricular cells, stimulating ventricular contraction.

Question 9

Concept: The synthesis and secretion of hormones from the anterior pituitary is regulated by the release of neurohormones into the blood from neurons located in the hypothalamus. In contrast, posterior pituitary hormones are synthesized in hypothalamic neurons and undergo anterograde axonal transport to the posterior pituitary. The secretion of stored posterior pituitary hormones from the axon terminals is mediated by depolarization of the nerve terminals.

question 16

However, if WT mice fed a HFD were to show significantly improved insulin sensitivity than Alb Tg mice fed a HFD, this would fail to support the hypothesis as it would mean that miR-26a over-expression in Alb Tg mice does not improve obesity-induced insulin resistance. concept: An organism is considered insulin sensitive if only a minimal amount of insulin is needed to induce an appropriate reduction in glucose levels. In contrast, insulin-resistant organisms need substantially more insulin to take up the same amount of glucose.

q17

Hyperventilation = you are breathing in and out so much more you are gonna get a decrease in the CO2 concentration in the blood Hypoventilation = you have more CO2 in the blood Had to use CADET here too. Qstem states hyperventilation shows same O2 dissociation curves as patients with cirrhosis so if you look at fig 1 in psg you can see a left shift which means you have to look for something which would be a decrease in CADET and thats A. Another way to look at this is that hyperventilation means less CO2 in body bc more is being exhaled out so bicarb buffer system will shift to make more CO2 so H+ and carbonic acid will decrease so B is wrong cuz decreasing bicarbonate would shift to make more H+ and carbonci acid???. D is wrong bc CO2 and H+ allosterically decrease the affinity of Hb to O2 (right shift) so if there's a decrease of CO2 then you would increase affinity of Hb to O2 (left shift) Educational objective: A left shift in the oxyhemoglobin dissociation curve signifies an increase in the affinity of hemoglobin for oxygen. This affinity is inversely related to CO2 and H+ concentrations. A left shift in the ODC therefore often indicates lower CO2 and/or acid levels in the blood.

question 14

I looked at the figure in the passage and realized that the brain can synpase onto either a preganglionic neuron or an interneuron or a sensory neuron. I was bale to figure that out bc a preganglionic neuron synapses onto another neuron and that would be a postganglionic neuron. So out of the answer choices there is only preganglionic neuron and thats why i chose it. Educational objective: A reflex is an involuntary response to a stimulus that does not require input from the brain. Reflexes are mediated by reflex arcs, neuronal pathways that include a sensory neuron, an effector neuron, and possibly an interneuron.

q26

If theres less gas exchange, there will be more co2 in the blood. More co2 in the blood will produce more h2co3 and H+. If theres more H+, that means the blood would be acidic leading to a decrease in pH concept: Pulmonary gas exchange facilitates the removal of CO2 from the blood. The amount of circulating CO2 affects blood pH by shifting equilibrium of the bicarbonate buffer system. Impaired gas exchange in the lungs will decrease blood O2 levels and increase blood CO2 levels, causing respiratory acidosis as H+ concentration increases.

q21

In Experiment 2, the researchers studied how three infusions, each containing different combinations of acetylcholine and anti-nAChR antibodies, affected the tension of muscle samples when added to an electrolyte-rich bath. Infusion type was changed = independent variable Muscle tension was measured = dependent variable Things that the researchers needed to do to eliminate confounding variables: 1) muscle samples of similar weight and size bc otherwise the tension would be smaller for small muscles and larger for large muscles 2) No interaction btw ACh and anti-nAChR antibodies bc you needed ACh binding is needed for contraction aka for muscle tension Concept: An ideal experiment features an independent variable, a dependent variable, and no confounding variables.

Q5

In addition to phospholipids, animal cell plasma membranes also contain cholesterol, glycolipids (lipids with attached sugar groups), glycoproteins (proteins with attached sugar groups), and other proteins. Cholesterol, which is found in eukaryotic but not prokaryotic cell membranes, functions to regulate the fluidity of the bilayer. Due to their unique structure, cholesterol molecules decrease fluidity at higher temperatures and increase fluidity at lower temperatures. As eukaryotic cells, intestinal epithelial cell membranes contain cholesterol. Concept: The plasma membrane is a fluid lipid bilayer that contains a diverse array of proteins scattered throughout. Certain cell types have a cell wall surrounding their plasma membrane; however, animal cells do not have a cell wall. The primary component of bacterial cell walls is peptidoglycan.

q26

In the given experiment, germ-free mice were transplanted with fecal (bacteria-rich) samples from mice that had either undergone RYGB or sham surgery. The study showed that when fecal samples from postoperative RYGB mice are transplanted into germ-free mice, the recipient germ-free mice (RYGB-R) are expected to lose weight. In contrast, when fecal samples from postoperative sham mice are transplanted into germ-free mice (SHAM-R), these recipient animals are expected to show no significant percent change in body weight. The graph in Choice A is the only option that depicts these expected findings. concept: Research suggests that restructuring the composition of gut flora can influence body weight. The human gut generally contains nonpathogenic bacterial flora that symbiotically cohabit with enterocytes. These microorganisms produce vitamins (eg, vitamin K, necessary for clotting) and aid with nutrient and drug metabolism.

Question 14

In the passage they mentioned calcium in the pore so that means neg AA would interact with it which is A Concept: Ion channels allow charged atoms to pass between the extracellular and intracellular environments down their concentration gradient. The charge of the amino acids that line the inside of the channel's pore plays a key role in determining which ion can pass through.

question 6

Know the steps of spermatogenesis: Spermatogonia --> spermatocytes --> spermatids --> spermatozoa (mature sperm) spermatogenesis occurs in seminiferous tubules of the testes (male gonads). Cross-section of these tubules shows sperm cells and nurse cells which are Sertoli cells. Sertoli cells provide nourishment to sperm and regulate their development. Leydig cells are also present in seminiferous tubules but in the interstitial tissue, and they secrete testosterone in response to LH release from the anterior pituitary and stimulate sperm cell differentiation. Educational objective:Spermatogenesis is a process in which male gametes (sperm) are produced and occurs in the seminiferous tubules of the testes. Spermatogonia undergo meiotic division and become spermatocytes, which then become spermatids when meiosis is completed. Spermatids then mature into spermatozoa (mature sperm) through a series of morphological changes.

q8

Muscle fiber action potentials propagate along the sarcolemma, the plasma membrane of the muscle fiber. The sarcolemma burrows deep into the muscle fiber, forming hollow structures known as transverse (T) tubules. Action potentials propagate along T-tubules just as they propagate along the superficial sarcolemma, resulting in the rapid and complete depolarization of the muscle fiber. This promotes calcium release from the sarcoplasmic reticulum, leading to muscle fiber contraction. Educational objective: Electrical depolarization caused by action potential propagation leads to the contraction of skeletal muscle fibers. As extensions of the plasma membrane that penetrate deep into muscle fibers, transverse (T) tubules are specialized muscle fiber structures that facilitate action potential propagation throughout individual muscle fibers.

question 11

Protein channels in the cell membrane enable certain ions to move down their concentration gradient across the membrane. For example, in resting neurons, potassium leak channels help maintain the membrane potential by enabling the passive transport (without using energy) of K+ out of the cell. Because the membrane is more permeable to K+ than to Na+ (ie, selective permeability), the resting membrane potential of neurons is approximately −70 mV, which is close to the negative equilibrium potential of K+ Active transport pumps embedded in the outer membrane of neurons hydrolyze adenosine triphosphate (ATP) to provide energy to transport molecules against their concentration gradient. For example, sodium-potassium pumps (Na+K+ ATPase) transport 2 K+ into the cell for every 3 Na+ moved out of the cell. This is important for maintaining the unequal concentration of ions across the membrane; without active transport pumps, leakage of ions through the cell membrane would eventually result in equilibration and a membrane potential of 0 mV concept: The presence of protein channels in the cell membrane allows passive transport of certain ions down their electrochemical gradient. This selective membrane permeability is responsible for generating the resting membrane potential in nerve and muscle cells. Active transport pumps help maintain the concentration gradient and are critical for maintaining the resting membrane potential.

q31

Qstem says that a patient has reduced pulmonary blood flow and excess fluid accumulation in the abdomen and legs. So reduced flow from the RA to RV would mean less blood flow to the pulmonary vessels. If there's reduced blood flow through pulmonary circuit then we will have backup of fluid accumulating in extremities bc its not functioning properly and not bringing back that deoxy blood In addition, the reduced volume of blood flowing into the right ventricle and through the pulmonary circuit would cause a backup of blood in the right atrium and systemic vessels. Accordingly, blood volume and hydrostatic pressure in systemic veins would increase, causing excess fluid leakage from systemic vessels into the surrounding interstitial fluid and subsequent fluid accumulation (edema) around tissues. Educational objective: Deoxygenated blood returning to the heart fills the right atrium, which transfers blood to the right ventricle. The right ventricle pumps blood to the lungs via the pulmonary circuit.

Question 8

The question is basically asking where would you see the HU cells in the immunostaining during the treatment. Looking at the experimental protocol you have to color code to the immunostain and see that idU is done before the stress (hu treatment) so that's the only thing you will see which correlates to 5 and 6. Content -> DNA replication occurs in S phase of the eukaryotic cell cycle

question 2

The question is saying that the chromosomes distribution was not even. This is talking about nondisjunction and that occurs in anaphase I. Normally, if a diploid cell with 46 chromosomes undergoes meiosis, four haploid daughter cells should be produced with each containing 23 chromosomes. concept: During meiosis, chromosomal nondisjunction occurs when homologous chromosomes (in meiosis I) or sister chromatids (in meiosis II) fail to separate to opposite poles of the cell during anaphase, leading to extra chromosomes in some daughter cells and missing chromosomes in others.

Question 10

Tissue a is showing insulin and glucagon presence so the organ must be pancreas and B is showing GH and LH so it much be the anterior pituitary concept: Endocrine glands modulate physiological activity via the secretion of hormones. For example, the pancreas secretes insulin and glucagon to regulate blood glucose levels. In contrast, the anterior lobe of the pituitary gland secretes multiple hormones that regulate several processes, including metabolism and reproductive function.

q3

When acetylcholine (ACh) is released by the motor neuron at the neuromuscular junction, the following occur: 1) ACh binds and opens ligand-gated ion channels in the sarcolemma (the plasma membrane of the muscle cell) 2) Na+ flows down its electrochemical gradient and into the cell through the channel, resulting in depolarization of the sarcolemma and generation of an action potential that propagates along the muscle fiber in all directions. 3) At certain locations along the muscle fiber, the sarcolemma burrows deep into the cells, forming a channel known as the transverse (T) tubule, which brings depolarizing current close to the sarcoplasmic reticulum (SR). The SR is a specialized smooth endoplasmic reticulum responsible for regulating cytosolic Ca2+ levels within the muscle cell. 4) Action potential propagation through the T tubule ultimately leads to the opening of Ca2+ channels in the SR membrane. Because Ca2+ is more highly concentrated inside the SR than in the cytosol, the opening of these channels results in Ca2+ flowing down its concentration gradient and into the cytosol. 5) Cytosolic Ca2+ ions then bind to troponin, which allows the actin and myosin filaments of the sarcomere to slide across one another. The sliding of the filaments results in shortening of the sarcomere and overall muscle contraction. 6) The Ca2+ channels in the SR membrane close when the depolarizing stimulus ceases. Active transport Ca2+ pumps sequester the Ca2+ back into the SR, which allows the muscle to return to its relaxed state as cytosolic Ca2+ concentration falls.

q11

When body temperature is above normal (as would occur during HA exercise), the following processes cool the body (ie, decrease body temperature): Vasodilation (widening) of skin arterioles increases blood flow to skin capillaries and maximizes heat loss through the skin. This occurs because blood, which is warmed in the body core, transfers heat to the environment when it passes through skin capillaries. Vasodilation occurs when the smooth muscle surrounding the blood vessels relaxes Sweat, a hypotonic solution, is secreted onto the skin surface by sweat (sudoriferous) glands. Heat loss and subsequent cooling occur due to the evaporation of the water in the sweat, an endothermic process that absorbs heat from the body. When body temperature is below normal, the following processes warm the body (ie, increase body temperature): Vasoconstriction (narrowing) of skin arterioles minimizes heat loss by diverting warm blood away from skin capillaries and toward blood vessels in the interior of the body. Vasoconstriction occurs when the smooth muscle surrounding the blood vessels contracts Shivering generates heat through rhythmic involuntary contractions of skeletal muscle. concept: One major function of the skin is thermoregulation. Body temperature can be increased by vasoconstriction of skin arterioles, shivering, and (in hairier animals) piloerection. Body temperature can be decreased by vasodilation of skin arterioles and sweating.

Question 7

When homologous chromosomes fail to align properly, unequal crossing over occurs. Unequal crossing over is the exchange of unequal segments of DNA between non-sister chromatids, yielding one chromosome that contains extra copies of one or more genes (ie, gene duplication) and another chromosome that lacks those genes entirely (ie, gene deletion). According to the passage, 21-hydroxylase deficiency occurs due to insufficient activity of 21-hydroxylase, the enzyme encoded by the CYP21A2 gene on chromosome 6. If unequal crossing over involving chromosome 6 occurs, it may result in a daughter cell that contains a chromosome lacking CYP21A2, which would contribute to the development of 21-hydroxylase deficiency in offspring. concept: Crossing over is a form of genetic recombination involving the exchange of genetic material between non-sister chromatids of homologous chromosomes. Unequal crossing over can result in offspring that lack important genes.

What are transposons

are small fragments of DNA that can move between different regions of the genome. Movement of a DNA sequence, such as a transposon, from one chromosomal arm to another may alter the nucleotide sequence of each chromosome but would not reduce chromosome number in the yeast cell

Question 50

A gene's biological function can be inferred by comparing the differences in organisms with the gene knocked out (inactivated) to wild-type organisms.

Question 14

ASO increases the tail length by keeping the exon 7 in the SMA mice. So, its processing the mRNA transcript to mature mRNA Content -> Processing (which includes splicing) of the pre-mRNA transcript into mature mRNA occurs in the nucleus.

question 18

According to the passage these two procedures (peritoneal fluid sampling and endometrial tissue sampling) took place within five days of each other. Due to the substantial amount of time elapsed between the two procedures, peritoneal fluid and endometrial tissue sampling may have occurred during different phases of the menstrual cycle for some participants. This could lead researchers to incorrectly classify the menstrual cycle phase during which the peritoneal fluid was obtained. Thats why the answer is A Concept: When collecting data or samples from participants in a study, researchers should assess whether the time at which these data/samples are collected could impact the validity of the experiment, and then adjust their experimental design to compensate.

Question 6

According to the passage, 21-hydroxylase deficiency is associated with defective glucocorticoid synthesis and excessive testosterone synthesis. In female patients, excess testosterone would lead to abnormal development of the female external genitalia that can be detected on physical examination. Because testosterone is a normal contributor of male sexual development, excess testosterone would not significantly alter development of the external sex organs in males, and a physical examination would yield no abnormal results. The SRY gene on the Y chromosome region is the sex determining region Concept: Testosterone is a sex hormone that facilitates the development of male sex organs and secondary sexual characteristics.

q5

According to the passage, TNF stimulates the release of sCD18. Consequently, TNF-treated mononuclear cells would be expected to have significantly increased sCD18 concentrations compared to the untreated group. The passage describes treating cultured cells with a monoclonal antibody against TNF (anti-TNF), which will most likely prevent TNF from interacting with the cultured cells. Therefore, the cells treated with anti-TNF would be expected to have decreased levels of sCD18 because the antibody treatment inhibits the effects of endogenous TNF. The expected result of the experiment would show the highest sCD18 levels in the TNF-treated cells, the lowest levels in the anti-TNF-treated cells, and the untreated between the two because of the endogenous TNF action. concept: Monoclonal antibodies can sequester specific molecules and subsequently inhibit their molecular action. Monoclonal antibodies are derived from a single clone, and all recognize a single epitope.

q14

According to the passage, heat acclimation (HA), the process by which the body increases tolerance to environmental heat, can be induced by repeatedly exercising at elevated temperature and humidity. The researchers measured the participants' heat tolerance before and after a seven-day period of HA training by testing their ability to complete a 90-min exercise regimen. Heat tolerance testing was halted before 90 min if the internal temperature (IT) of subjects exceeded 40°C. Pre-HA, all controls were able to complete the 90-min exercise regimen, indicating that controls were able to sustain ITs below 40°C throughout HA training, as shown in Figure 1. However, in both skin graft recipient groups, some members were unable to complete the regimen, meaning that these individuals' ITs rose above 40°C. Given that all participants had the same average IT prior to beginning exercise, this signifies that graft recipients were less able to tolerate heat compared with controls and had a larger average increase in IT during exercise. Post-HA, all controls were still able to complete the 90-min regimen, but the grafted individuals showed improved performance as more were able to make it to the 90-min endpoint. Therefore, the average change in IT of the grafted groups was smaller post- than pre-HA. In addition, the passage states that HA occurs normally in healthy individuals. Therefore, control individuals would also be expected to show some increase in heat tolerance (a smaller post-HA change in IT). The table in Choice D is the only option that illustrates the expected increase in heat tolerance as HA occurred in both groups. Educational objective: The skin plays a key role in the thermoregulation of body temperature, and injury can impair its ability to tolerate significant temperature changes."

Q11

According to the passage, scientists proposed that a prolonged symbiosis period could induce gene transfer from eukaryotes to prokaryotes. The passage states that Cu/Zn SOD is typically found in the cytosol of eukaryotes; therefore, scientists concluded that SODs found in P. leiognathi were most likely acquired from ponyfish. However, the discovery of Cu/Zn SODs in other free-living bacterial species with no known eukaryotic symbiotic hosts would argue against the hypothesis of eukaryotic to prokaryotic gene transfer due to symbiosis. concept: The endosymbiotic theory explains how primitive eukaryotic anaerobes engulfed ancient aerobic prokaryotes, and consequently acquired the ability to produce energy through oxidative phosphorylation.

q19

According to the passage, sodium-dependent glucose linked transporters (SGLTs) are secondary active transport proteins. This means that SGLTs utilize the energy released by the movement of Na+ along its concentration gradient to move glucose against its concentration gradient (back into the body). For charged particles such as Na+, this movement occurs in reference to an electrochemical gradient, which incorporates both the electrical (potential) and chemical (concentration) gradients of the molecule. Accordingly, the movement of Na+ along its electrochemical gradient enables glucose reabsorption. Educational objective: Primary active transport utilizes the energy from ATP hydrolysis to move a substance against its concentration gradient. Secondary active transport utilizes the energy released by the passive transport of another molecule.

q10

According to the passage, the damage associated with a full-thickness burn extends through the epidermis and dermis and into the subcutaneous layer (hypodermis). Therefore, this type of burn involves injury to the epidermal and dermal sensory receptors, resulting in impaired touch sensation not increased touch sensation. concept: The skin functions as a physical barrier to prevent the loss of fluid from the body while simultaneously blocking the entry of pathogens or harmful chemicals. The skin also contains receptors that gather and respond to sensory information from the surrounding environment. Ultraviolet radiation that strikes the skin induces the synthesis of a vitamin D precursor.

q10

According to the psg, OA deteriorates articular (hyaline) cartilage and hyaline cartilage lines the ends of articulating bones. the passage also says OA can result in the formation of bony growths on the subchondral bone, which is the bone under the damaged articular cartilage. In other words, additional bone deposits on the epiphyses of the bones at the affected joint. Epiphyses are rounded ends that have joint surfaces covered by articular cartilage. Diaphysis is the hollow shaft (medullary cavity) filled with bone marrow. Metaphyses are the regions where the diaphysis and epiphyses meet. The epiphyseal (growth) plate, a cartilaginous structure that lies between the epiphyses and metaphyses, is present only during childhood and serves as the site of longitudinal growth. When growth ceases, the growth plate is replaced with mature bone and is referred to as the epiphyseal line. The periosteum is a thin layer of connective tissue that covers and protects the long bone. (Note: The periosteum does not cover joint surfaces.) concept: The structure of a long bone includes the epiphyses (rounded ends covered by articular cartilage), the diaphysis (shaft containing the medullary cavity filled with yellow bone marrow), and the metaphysis (where the diaphysis and epiphyses meet).

Q35

According to the question, 5 bacterial cells were incubated for 3 hours (180 minutes); these exhibited a 20-minute lag phase and a 40-minute generation time. Assuming unrestricted growth during incubation, the bacterium could only have had a log phase and a lag phase. Therefore, the number of cells that result after incubation can be calculated as follows: 1) Calculate the amount of time in which the bacterial population was in log phase: 180 min incubation time − 20 min lag phase = 160 min log phase 2) Find the number of generations in the log phase period: 160 min log phase / 40 min generation time = 4 generations 3) Multiply the original number of bacterial cells by 2n to get the final population size:5 × 2^4 = 80 cells Educational objective: Binary fission is the process by which a unicellular organism (eg, bacterium) divides into two identical daughter cells. The generation time refers to the time in which a bacterial population doubles. When a single cell divides by binary fission, the number of cells that result is calculated by multiplying the original number of bacterial cells by 2n, where n equals the number of generations

Q4

According to the western blot, there were no bands when LuxS was knocked out but in PCM all the bands were present. That means that PCM must have a way to activate the transcription of the genes involved in LuxS function (the lee genes). Thats why D is correct concept: Inducers are chemical substances (signals) that regulate gene expression. Specifically, secreted inducers can influence gene expression in other cells.

question 20

Accordingly, PTH acts to increase serum calcium and decrease serum phosphate by: Stimulating calcitriol synthesis, which then induces absorption of dietary calcium from the small intestine Promoting bone resorption (osteoclast activity), releasing calcium into the blood Increasing calcium reabsorption and phosphate excretion in the kidneys The passage explains that secondary hyperparathyroidism involves increased PTH release in response to low serum calcium. Secondary hyperparathyroidism could result from deficient synthesis of renal calcitriol (ie, reduced intestinal absorption of calcium and phosphate), which would lead to low levels of circulating calcium and require constant PTH release to correct this imbalance Concept: PTH release stimulates the synthesis of calcitriol, the active form of vitamin D, in the kidney. Consequently, calcitriol primarily functions to promote absorption of dietary calcium and phosphate from the small intestine.

Q25

Acidification of the endosomal lumen as it matures from an early endosome to a late endosome involves the transport of protons against their concentration gradient from an area of relatively low concentration (cytosol; pH ≈ 7.4) to one of relatively high concentration (endosomal lumen; pH ≈ 5.5). Therefore, protons from the cytosol can only enter the endosomal lumen if only energy is added to the system. In simple diffusion, molecules that are small or lipid-soluble travel across the cell membrane directly, without help from protein channels. In facilitated diffusion, molecules that cannot normally cross the membrane on their own due to charge or size are allowed to pass through protein channels or gates that do not require energy input to operate. concept: Molecules can cross biological membranes by active or passive transport. Active transport involves a molecule moving against its concentration gradient and requires energy input (eg, ATP). In contrast, passive transport involves molecules diffusing down their concentration gradient and requires no external energy. Passive transport can be divided into simple diffusion, in which molecules cross membranes directly, and facilitated diffusion, in which molecules must go through protein channels.

Question 11

Aldosterone, secreted from the adrenal cortex, stimulates Na+ reabsorption in the kidneys, leading to increased water retention due to osmosis. Increased water retention increases blood volume, which in turn leads to an increase in blood pressure. Norepinephrine and epinephrine, secreted from the adrenal medulla, function to mobilize the body under extreme stress and promote rapid information processing in part by maximizing blood flow to organs essential for survival. Norepinephrine and epinephrine achieve this effect by promoting the following changes in blood vessel diameter that lead to altered blood pressure: 1. Vasoconstriction (narrowing) of blood vessels supplying the intestines, kidney, and other abdominal organs decreases blood flow to these organs and conserves oxygen/nutrients for other tissues. 2. Vasodilation (widening) of blood vessels leading to the heart and skeletal muscles increases oxygen/nutrient delivery to these organs. Difference between epinephrine and norepinephrine: Epi = primarily a hormone. Also an anti-histamine. Norepi = a hormone and a neurotransmitter; inhibits insulin Concept: Each adrenal gland is composed of two anatomically distinct regions: the adrenal medulla and the adrenal cortex. Both regions secrete hormones that regulate blood pressure and allow the body to respond to stressors.

Question 5

Aldosterone, which is secreted by the adrenal cortex in response to high serum potassium (K+) and low blood pressure (via the renin-angiotensin system), acts on the kidney to promote sodium reabsorption and potassium secretion. However, more sodium is reabsorbed than potassium is secreted. Consequently, an ion gradient is generated between the filtrate and the interstitial fluid in the adjacent interstitial space, causing water to be reabsorbed into the interstitial fluid (and eventually the bloodstream) via osmosis. As a result, blood volume and therefore blood pressure increase. According to the passage, patients with congenital adrenal hyperplasia (CAH) exhibit deficient synthesis and secretion of aldosterone. After administering synthetic aldosterone, increased sodium and water reabsorption by the kidneys will result in increased blood volume and increased serum sodium content. Concept: Aldosterone is secreted by the adrenal cortex in response to elevated plasma potassium and decreased blood pressure. Aldosterone acts on the kidneys to increase the reabsorption of sodium, which in turn leads to increased reabsorption of water and, ultimately, increased blood volume and pressure. **LOOK AT THE RENIN ANGIOTENSIN PATHWAY**

question 13

All you had to know is that a negatively charged ion would hyperpolarize the cell membrane and not initiate an AP like Na+ would and cause depolarization. concept: At inhibitory synapses, pre-synaptic neurons release neurotransmitters that cause either an influx of negative ions into the post-synaptic neuron or an efflux of positive ions out of the post-synaptic neuron. This causes hyper-polarization of the membrane potential and inhibits action potential initiation in the post-synaptic neuron.

question 12

All you to know was that Ach is an NMJ NT which means that inhibition of it would not allow the contraction of skeletal muscles. Concept: At the neuromuscular junction, acetylcholine is released via exocytosis from presynaptic motor neurons. Acetylcholine binds to receptors on the motor end plate, triggering muscle contraction. Disruption of this process causes skeletal muscle paralysis.

q1

Antigens = foreign substances that trigger immune response Antibody = Y shaped proteins that bind and neutralize these antigens in the blood and lymph The variable region of the antibody binds the antigen. A single antibody can bind only one type of antigen, and this specificity is determined by the unique amino acid structure of the antibody's variable region. But antibodies bind different antigens but that means they have different variable regions. Concept: Antibodies are synthesized and secreted by effector B cells, which are part of the adaptive immune system. Each antibody consists of two identical light chains, two identical heavy chains, a variable region that interacts with a specific antigen, and a constant region that interacts with the cells (eg, phagocytes) and proteins of the body to facilitate antigen destruction.

q7

B is right because its an except question and since the q stem is asking what will not happen as the extra oxygen is used to restore muscle fibers to their pre exercise state, you will not facilitate Ca2+ from the SR to the muscle fiber bc that will cause a muscle contraction to take place. Instead the Ca2+ would be sequestered into the SR. concept: Oxygen intake remains elevated after exercise due to the increased demand in muscle cells for oxygen to replenish ATP, creatine phosphate, and glycogen stores and restock myoglobin with oxygen. The amount of additional oxygen consumed is called the excess post-exercise oxygen consumption, or oxygen debt.

q12

B is the answer because chondrocytes make up the cellular component of cartilage not bone A shell of cortical (compact) bone generally surrounds cancellous (spongy) bone. Compact (hard) bone is organized into structural units called osteons, or haversian systems, which are made up of lamellae (concentric rings of bone matrix) that surround a central haversian canal, a cylindrical channel that runs parallel to the long axis of bone and through which blood vessels and nerves traverse. Volkmann canals, which run perpendicular to the long axis of bone, allow the passage of blood vessels and nerves between different haversian canals Osteoblasts continue to build successive concentric layers of bone, but as the osteoid mineralizes, some osteoblasts become trapped within lacunae (spaces) in the lamellar matrix and are known as (mitotically inactive) osteocytes at this stage. Within each osteon of compact bone, lacunae connect to one another via microscopic channels called canaliculi, which allow osteocyte waste exchange and nutrient delivery concept: Compact bone is organized into concentric rings of bone matrix called lamellae. The entire unit of concentrically arranged lamellae surrounding a central haversian canal is known as an osteon, or a haversian system. Within each osteon, lacunae (spaces containing osteocytes) connect to one another via microscopic channels called canaliculi, which allow osteocyte waste exchange and nutrient delivery.

q11

Based on the passage, Caco-2 cells are derived from the colonic segment of the human gut and, accordingly, this segment absorbs water. Educational objective: The colonic segment of the large intestine functions to absorb water and electrolytes from undigested waste material that is left over from digestion and absorption in the small intestine.

question 8

Based on the passage, G6PD converts BCB to a colorless compound and is downregulated as the oocyte matures. Mature (competent) oocytes would have a blue cytoplasm because G6PD activity would be low and the enzyme would not be sufficiently active to convert BCB into a clear compound. In contrast, an immature (noncompetent) oocyte would show high levels of G6PD activity, and when stained with BCB, would show a colorless cytoplasm because G6PD would convert BCB into a clear compound. Concept: Oocytes can be competent or noncompetent based on their maturation status. Oocytes that are competent are mature and most likely result in viable progeny after being fertilized by sperm.

q12

Based on the passage, HIF-1 is a transcription factor that controls the expression of genes that establish and maintain epithelial barrier integrity under normal oxygen conditions. Psg states that HIF1 will be more activated in hypoxic conditions and since HIF1 is a transcription factor it will be found in the nucleus, initatiating transcription. Transcription factors are proteins that are made in the cytoplasm but act in the nucleus to control gene expression. Educational objective: Transcription factors are translated in the cytoplasm but act in the nucleus to control gene expression. They contain a nuclear localization sequence that allows nuclear import proteins to direct them back to the nucleus to alter gene transcription.

q20

Based on the passage, anti-nAChR antibodies bind nAChR reversibly to block the interaction between ACh and nAChR, resulting in muscle weakness. Accordingly, administration of these antibodies in a wild-type mouse would temporarily prevent binding of ACh to its receptor. However, if a mutant mouse were to exhibit normal muscle contraction despite being treated with anti-nAChR antibodies, the most likely reason in this scenario is that ACh degradation within the mutant's synapse would be decreased compared to the wild-type. If ACh degradation is decreased, more ACh molecules will remain in the synapse for a longer time period and be able to bind nAChRs after these are no longer bound to anti-nAChR antibodies. This would result in muscle contraction that is delayed but still normal. In contrast, the following changes would magnify the inhibitory effect of anti-AChR antibodies on muscle contraction by decreasing the likelihood of ACh binding to nAChR: Fewer nAChR being present within the postsynaptic membrane decreases the binding sites available to synaptic ACh, resulting in muscle weakness (Choice B). Following motor neuron action potentials, reduced release of ACh from the presynaptic vesicles and decreased recycling of ACh breakdown products would decrease the quantity of ACh released into the synapse, also leading to muscle weakness (Choices C and D). Educational objective: Acetylcholine (ACh) degradation within the neuromuscular junction is one process by which muscle fiber contraction is terminated. However, decreased ACh degradation would allow more ACh molecules to remain in the synapse and prolong stimulation of nAChR and subsequent muscle contraction.

question 18

Based on the passage, secondary hyperparathyroidism is characterized by increased PTH secretion. PTH release increases bone resorption (breakdown) by osteoclasts, bone-resorbing cells, and causes decreased bone mineralization. The majority of calcium in the body is stored as hydroxyapatite, a mineral found in the bone matrix that primarily contributes to bone strength and hardness. Osteoclast-mediated bone destruction causes the calcium stored in the matrix to be released into the blood, dissolving mineralized bone and increasing plasma calcium levels. Concept: The net effect of parathyroid hormone (PTH) activity is to increase serum calcium and decrease serum phosphate concentration. PTH secretion from the parathyroid gland is triggered by decreased plasma calcium. PTH indirectly stimulates osteoclast activation, promoting bone breakdown and the release of ionic calcium into the blood.

q8

Based on the passage, some fish species have glomerular kidneys (eg, sculpin) and some have aglomerular kidneys (eg, goosefish). The researchers' claim is that the glomerular kidney is 500 times as efficient as the aglomerular kidney in eliminating nontoxic substances. So you had to multiply the goosefish value by 500 to get the sculpin's value. Educational objective: As the functional unit of the kidney, the nephron consists of a ball-like network of capillaries called the glomerulus. The glomerulus filters blood plasma into the tubular segments of the nephron.

q22

Because the left ventricle must pump blood to a larger structure (all body tissues except the lungs) across longer distances, it must pump blood at a higher pressure. To generate the high pressure necessary to propel blood through the body's entire circulatory system, the walls of the left ventricle must be extremely muscular and powerful (ie, thicker). Thats why the right ventricle has thinner walls compared to the left ventricle because it needs to pump blood at a lower pressure to propel blood to the lungs only. concept: The right ventricle has thinner walls than the left ventricle because it needs to pump blood at a lower pressure to reach only the lungs. In contrast, the thicker walls of the left ventricle allow it to pump blood at a higher pressure to reach all other tissues in the body.

question 2

Biologist 1 axons had more myelin than biologist 2 axons so looking for answer that would indicate that. Voltage gated ion channels would only be clustered at specific locations (nodes of ranvier) on a myelinated axon Educational objective: The myelin sheath increases the speed of action potential (AP) propagation by acting as an electrical insulator that prevents dissipation of charge across the membrane. APs in myelinated axons travel via saltatory conduction.

CADET face RIGHT mnemonic

C - increase in Co2 A - Acidic (higher H+) D - increase in 2,3 BPG E - increase in exercise T - increase in temperature All this causes a right shift on the oxygen dissociation curve which means lower affinity for oxygen = more oxygen being unloaded

q13

C is correct bc ligaments and tendons are important connective tissue structures at joints. Ligaments connect bone to bone (ligabone) and tendons connect bone to muscle. Know the 3 types of cartilage Also, hyaline cartilage is predominantly made up of collagen concept: Cartilage is a firm but flexible connective tissue that lacks blood vessels and nerves. Chondrocytes (cartilage cells) secrete chondrin, which is the specialized extracellular matrix that makes up cartilage. The most common type of cartilage is hyaline cartilage, which plays a role in bone development and lines the ends of articulating bones.

q2

Cells originating in the bone marrow begin as multipotential hematopoietic stem cells and can differentiate into lymphoid or myeloid progenitor cells. Lymphoid progenitor cells go on to become B cells, T cells, and natural-killer (NK) cells. Myeloid progenitor cells in the bone marrow further differentiate into erythrocytes, platelet-producing megakaryocytes, neutrophils, basophils, eosinophils, monocytes, or mast cells. Psg states that cells with LFA-1 are Leukocytes (WBCs) and all blood cells originate from the bone marrow Educational objective: Hematopoietic stem cells originate in the bone marrow and differentiate into myeloid or lymphoid progenitor cells. Lymphoid progenitor cells go on to become T cells, B cells, or NK cells whereas myeloid progenitor cells differentiate into erythrocytes, megakaryocytes, neutrophils, basophils, eosinophils, monocytes, or mast cells.

Q3

Chemoreceptors which are sensitive to the pH of the blood measure the regulation of respiratory rate. Blood pH is determined by H+ which is dependent on the bicarb buffer system. The equilibrium of the bicarb system depends on the concentration of CO2 and since these chemoreceptors directly measure the H+ concentration in the blood, they therefore measure the partial pressure of co2 in the blood concept: The regulation of respiratory rate primarily depends on the pH of the blood as measured by central and peripheral chemoreceptors. These receptors directly detect the [H+] in the blood, which is dependent on the partial pressure of CO2 in the blood through the bicarbonate buffer system.

q1

Circulating blood enters the glomerulus through the afferent arteriole and exits through the efferent arteriole. Blood is filtered in the glomerulus by high hydrostatic (blood) pressure, which forces excess fluid and waste products (filtrate) across the porous endothelium and into Bowman's space. Proteins and cells are too large to pass through the pores. The volume of fluid filtered through the kidney per unit time is known as the glomerular filtration rate (GFR). Narrowing of the afferent arteriole would decrease the hydrostatic pressure and so the blood pressure reaching glomerulus would decrease. concept: The volume of fluid filtered through the kidney per unit time is known as the glomerular filtration rate (GFR), which can be modulated by controlling blood flow through the glomerulus. Higher hydrostatic (blood) pressure in the glomerulus increases GFR, and lower hydrostatic pressure decreases GFR.

q13

Control cells = under normal oxygen so they would have high TEER (measures barrier function) Hypoxia = low oxygen levels would have low TEER bc the epithelial barrier is destroyed hypoxia + emodin = emodin prevents barrier damage so it would inhibit hypoxia function on the barrier and so the TEER level would increase hypoxia + emodin + inhibitor = the inhibitor inhibits emodin function so the barrier would be damaged and the you would have low TEER value concept: Inhibitors prevent chemical reactions, block gene/protein expression, or reduce the activity of a particular substance (eg, enzyme, reactant). Transcription factors can be inhibited by certain drugs or substances.

q7

Desmosomes provide tensile strength to epithelial cell sheets by anchoring the cytoskeletons, specifically the intermediate filaments, of two cells together. Creates a continuous cytoskeletal network that spans the entire epithelial sheet, one through which mechanical stress (eg, pulling, stretching, tension) can be distributed. Found in tissues that have a lot of mechanical stress like muscle tissue and it prevents tissue tearing Gap junctions are cell-cell junctions that mediate communication between cells. Protein channels (connexons) in one cell align with complementary channels in another cell to form pores that facilitate the passive and bidirectional exchange of ions and small solutes. Gap junctions are found in cell populations that depend on coordinated activity, such as smooth muscle, cardiac muscle, or neural tissue. Tight junctions are cell-cell junctions that prevent water and solutes from diffusing between cells and across the epithelial cell layer. These junctions form a watertight seal that fully encircles the apical end of every cell in the basal epithelial sheet. Tight junctions serve as a barrier and separate tissue space; they are found in a number of tissues, including skin, gastrointestinal tract, and testis. concept: Desmosomes, gap junctions, and tight junctions are cell-cell junctions that provide tissue with mechanical strength, cytoplasmic continuity, and watertight seals, respectively.

q17

During urination, activity within stretch receptors in the bladder leads to: Contraction of the detrusor muscle, which pushes urine out of the bladder and into the urethra. Relaxation of the IUS, which opens the urethra and allows urine to pass. Nerve damage that impairs the contraction of the detrusor muscle of the bladder would impair emptying of the bladder and lead to urinary retention. concept: Urine flows from the kidney to the bladder through the ureter. Once the bladder is full, urine exits the body via the urethra through the process of urination. Urination is controlled by the detrusor muscle (smooth muscle lining of the bladder), the internal urethral sphincter, and the external urethral sphincter.

Q29

Enveloped viruses - sensitive to heat, detergents, and changes in moisture. Non-enveloped/naked viruses - have only a capsid as a protective outer layer and are more resistant to heat, detergents, and changes in moisture. In the last paragraph, the researchers determined whether hepatitis viruses were enveloped or non-enveloped by exposing each virus to fluorescent antibodies designed to bind viral capsid proteins directly. Therefore, depending on the presence or absence of fluorescence, researchers would be able to determine whether the virus is non-enveloped or enveloped, respectively. The fluorescence assay revealed that capsid proteins were detected only in media containing HAV or HEV. Because the antibodies were able to bind directly to the capsid proteins of both viruses, it can be determined that HAV and HEV are non-enveloped viruses (ie, they do not have a phospholipid bilayer enveloping the capsid). Educational objective:All viruses contain a protective protein coat known as the capsid. Viruses that contain only a capsid as an outer layer are known as non-enveloped or naked viruses, and are able to survive in harsh conditions. Viruses with a phospholipid bilayer surrounding the viral capsid are referred to as enveloped viruses; these are more susceptible to changes in environmental conditions.

Question 16

FL-SMN and snRNA have the same level of expression bc one enhances the expression of the other. The question asks what will happen in a SMA patient so the expression of both FL-SMN and snRNA would be low. Content -> Northern blots detect target RNA in a sample. The labeled RNA can be visualized as bands, where band intensity denotes the quantity of RNA expression and band position denotes the size (smaller molecules appear lower than larger ones).

question 10

Fig 1 graph shows that CSF coverage is greater during sleep than when you're awake within the first 60mins. Need to know stages of sleep: BATS Drink Blood (higher to lower freq brainwaves) so had to look for graph where CSF coverage increases as brain wave frequency decreases within the first hour of sleep. Educational objective: Brain frequencies observed during most sleep stages are lower than brain frequencies observed during wakefulness.

question 38

Figure 1 shows that the heterozygosity of wild-born male cheetahs has decreased over time. Decreased heterozygosity is associated with inbreeding; therefore, the conclusion can be drawn that free-ranging African cheetahs are inbred. In contrast, the heterozygosity of captive-born male cheetahs has increased. These results indicate that conservation groups are selectively outbreeding captive cheetahs with individuals from distinct populations. Concept: Inbreeding results in decreased heterozygosity (genetic diversity), reduced fecundity, and reduced fitness. Species that mate with nonrelatives (outbreed) increase their fitness because the introduction of new genetic material results in increased heterozygosity.

question 14

Figure 1A shows that blood glucose levels decrease between 30 and 60 minutes after the glucose injection in WT (normal) mice. This occurs because the glucose injection raises blood glucose levels, leading to increased release of insulin to lower blood glucose. Insulin decreases blood glucose in the 30-60 minute interval by promoting glucose uptake by the liver for glycogen synthesis, which increases hepatic glycogen stores. Insulin release also decreases glucagon release (Choice A), decreases plasma amino acids by stimulating protein synthesis (Choice B), and decreases hepatic gluconeogenesis (Choice D). concept: In the setting of high blood glucose, insulin is released in response and functions to decrease blood glucose levels. In contrast, glucagon is released from pancreatic alpha cells in response to low blood glucose and functions to increase glucose levels.

question 27

Figure 2 shows the negative feedback pathway that influences the synthesis and secretion of cortisol. The hypothalamus secretes CRH, which then induces the anterior pituitary gland to release ACTH. In turn, ACTH stimulates the adrenal glands to synthesize and secrete the glucocorticoid hormone, cortisol. In addition to mediating the stress response, cortisol also inhibits CRH release from the hypothalamus and ACTH release from the pituitary gland. By inhibiting the secretion of CRH and ACTH, secreted cortisol increases negative feedback such that the adrenal cortex receives less stimulatory input and the synthesis and secretion of additional cortisol decrease.

question 19

Figure 3 shows that as AMG 416 dosage increases, PTH levels decrease. This graph depicts PTH concentration immediately and at four time points following administration of three different AMG 416 dosages to 5/6 Nx mice. PTH levels return to the 100% baseline eight hours after 0.3 mg/kg AMG 416 administration. However, treatment with higher AMG 416 dosages of 1 mg/kg and 3 mg/kg result in suppressed PTH levels at 50% and 25% of the baseline, respectively. PTH is most suppressed at the highest AMG 416 dosage of 3 mg/kg after eight hours. The net result of PTH suppression will be decreased circulating calcium and increased phosphate (ie, the opposite of PTH release). Less PTH due to higher AMG dosages causes greater retention of serum phosphate. Concept: PTH release causes increased plasma calcium but decreases plasma phosphate. Decreased PTH availability or diminished physiological response to PTH will increase phosphate levels in the blood.

q20

Flow through the nephron concept: Fluid that is filtered by the kidney first enters the Bowman's capsule before passing into a long tubule, where water and solutes are selectively reabsorbed. Waste and metabolites are excreted as urine.

question 3

For this q you had to look at the x axis and see the time it took for the synapse to go to the postsynaptic cell. Chemical synapses are slower than electrical synapses because the signal has to be modified and must cross a larger intercellular distance. Looking at the graph, set A had a delay and set B did not. thats why Set A is chemical Chemical synapses - info transfer = unidirectional, cytoplasms of the 2 neurons are not continuous, neurotransmitters diffuse across the synaptic cleft Electrical synapses - info transfer is like an ionic current that passes directly from 1 neuron to another via gap junctions, pre/post-synaptic neurons are continuous so info flow bidirectional concept: Electrical synapses transfer information from one cell to another via passive ionic current flow through gap junctions. In contrast, chemical synapses use neurotransmitters to transfer information, which is a slower process.

question 12

Given that T1DM is caused by a failure in insulin production, affected patients have impaired endocrine function of pancreatic beta cells. Paracrine function: secrete substances that exert effects on neighboring cells Exocrine: secretes substances thru a duct and onto an epithelial surface (saliva, sweat, enzymes) Endocrine: secrete hormones into the bloodstream to cause an effect in a different part of the body Beta cells produce insulin, a hormone that promotes glucose uptake from the blood when blood glucose is high. In this setting, beta cell secretions inhibit neighboring alpha cell function and therefore also exhibit paracrine function. Alpha cells produce glucagon, a hormone that promotes glucose release into the blood when blood glucose is low. Alpha cells also exhibit paracrine function, as they are able to inhibit beta cell function in the setting of low blood glucose. Delta cells produce somatostatin, a hormone that has a generalized inhibitory effect on digestive function and has been shown to suppress insulin and glucagon release. concept: In the setting of high blood glucose, beta cells release insulin to promote glucose uptake from the blood and inhibit alpha cell function. In contrast, when blood sugar levels are low, alpha cells produce glucagon to promote glucose release into the bloodstream and inhibit beta cell function.

Question 42

Had to interpret the graph correctly. The question is asking which one will contribute to cancer the most. So, when a gene is knocked out from the cluster and the cluster's % apoptotic count is high without that gene but low with that gene that means that gene knocked out gene is contributing to cancer bc when cancer present = less apoptotic cells Content: Apoptosis is the programmed and controlled death of aged, unnecessary, or damaged cells. Cancerous cells exhibit rapid and uncontrolled proliferation, which is caused by cell growth that outpaces normal apoptosis or by dysfunction of the apoptotic process itself.

q20

Had to know that endothelial cells are a type of epithelial cells that line teh interior of the cardiovascular system. It is in direct contact with the blood in the lumen. The endothelium also acts as a barrier that regulates the exchange of substances between the blood and nearby tissues. For example, in response to inflammatory chemicals released during an infection, endothelial cells promote the movement of white blood cells out of the bloodstream to the affected tissue. Educational objective: The endothelium is a single-cell layer that makes up the innermost surface of the cardiovascular system (heart and blood vessels). The endothelium promotes blood fluidity and functions as a selective barrier between the blood and surrounding tissues.

q8

Had to know that lymph nodes filter lymph and spleen filters blood, bacteria circulating through lymph are removed in the lymph nodes and bacteria circulating through blood are removed in the spleen. Educational objective: In the lymph nodes, white blood cells remove and mount immune responses against pathogens from the lymph, whereas in the spleen, white blood cells remove path

Q34

Had to know the difference between lytic and lysogenic cycles Steps for lytic cycle: 1) Attachment: The bacteriophage comes in contact with the bacterial cell wall and attaches to the host bacterium using its tail fibers. 2) Viral genome entry: The phage uses its tail sheath to inject its genome into the cytoplasm of the bacterial host 3) Host genome degradation: Viral enzymes degrade the host genome into its nucleotide components to provide the building blocks for replication of the viral genome 4) Synthesis: Loss of the bacterial chromosome ends synthesis of host molecules. As a result, the host machinery (eg, ribosomes), now under the control of the viral genome, begins to synthesize the components needed for new viral progeny, which then assemble inside the host cell. 5) Release: Many newly assembled viral progeny (virions) are released as the bacterium disintegrates (lysis) due to the action of lysozymes on the host cell wall. Concept: Bacteriophages use the host cell's machinery and resources to replicate their genome and synthesize viral proteins to form new virions. Phages with a lytic life cycle replicate rapidly and release progeny via lysis of the host cell. In contrast, phages with a lysogenic life cycle integrate their genome with the host genome and replicate as the cell divides.

Q36

Had to know the domains of Archaea, bacteria and eukarya Educational objective: Prokaryotes are unicellular organisms that can be further classified into the domains of Archaea or Bacteria. Organisms in these domains lack a nucleus and membrane-bound organelles but have circular chromosomes and are able to reproduce asexually through binary fission. However, bacteria (not archaea) have peptidoglycan in their cell wall as a distinguishing feature.

q15

Had to know the function of an endoskeleton and an exoskeleton Educational objective: The vertebrate endoskeleton is composed of both bone and cartilage and provides an internal scaffold that facilitates mobility while protecting and supporting internal organs. In contrast to vertebrates, many invertebrate animals possess an external skeleton (exoskeleton), which is a rigid outer covering that serves to protect the soft tissues underneath.

Q11

Had to know the functions of a spleen: - filters aged/damaged RBCs (liver also involved in the removal and reuse of aged RBCs) - Reservoir for Blood - Immune response (B cell activation site, housing macrophages) **ALL blood cells come from the bone marrow** Spleen - largest lymphoid organ Educational objective: The spleen is involved in immune function, blood storage, and the filtration and removal of senescent red blood cells from the blood.

question 7

Had to know the functions of the glial cells in the CNS and PNS The nervous system is composed of neurons and glial cells. While neurons conduct electric impulses, glial cells serve a variety of vital support functions. Glial cells in CNS: - microglia are the primary immune cells of the central nervous system (CNS) and act as macrophages by phagocytizing pathogens, damaged cells, and other waste materials. - Oligodendrocytes form myelin sheaths around axons to reduce ion leakage, decrease capacitance, and increase the speed of action potential propagation along the axon. Each oligodendrocyte myelinates segments of multiple adjacent axons in the CNS. - Astrocytes make extensive contact with blood vessels and regulate blood flow in coordination with synaptic activity and chemical changes. Astrocytes are important for maintaining the chemical homeostasis of the interstitial space, including regulation of fluid and ion balance, pH, and neurotransmitter concentrations. They are also thought to play important roles in neuron development and structural maintenance, as well as coordination between neurons and other glial cells. - Ependymal cells are epithelial cells that line the compartments of the CNS and secrete CSF Functions of glial cells in the PNS include: - Schwann cells wrap the axons of some neurons with myelin to increase conduction speed. Unlike oligodendrocytes in the CNS, each Schwann cell forms a myelin sheath for a single neuron in the PNS - Satellite cells provide structural support and supply nutrients to neuron cell bodies in sensory, sympathetic, and parasympathetic ganglia (groups of cell bodies). They are thought to play roles similar to those of astrocytes in the CNS. Educational objective: Glial cells are the support cells of the nervous system. The PNS contains satellite cells (support) and Schwann cells (myelination). The CNS contains oligodendrocytes (myelination), astrocytes (support, blood-brain barrier, interstitial space), microglia (macrophages), and ependymal cells (CSF, compartments).

question 25

Had to know the germ cell derivatives (in picture). The passage states that the notochord in a mesodermal structure. You had to know that out of the ans choices only osteocytes (bone cells) are also derived from the mesoderm. - Endoderm (innermost layer): gives rise to accessory digestive organs (eg, liver, pancreas) as well as to the lining (epithelium) of the digestive and respiratory tracts. - Mesoderm (middle layer): gives rise to the circulatory system, the musculoskeletal system, and parts of the urinary and reproductive systems. - Ectoderm (outermost layer): gives rise to the nervous system (neurulation) and develops into the integumentary system, which includes hair, skin, nails, and the lining of the mouth, nostrils, and anus. Educational objective: The three primary germ layers (endoderm, mesoderm, ectoderm) form during gastrulation. Each gives rise to particular cell types in specific organ systems.

q9

Had to know what constitutes connective tissue: bone, blood, fat (adipose), tendons/ligaments, cartilage the only answer is muscle tissue also bc its another main type of tissue (CMEN) Educational objective: The body consists of four types of tissue: epithelial, muscle, connective, and nervous tissue. Connective tissue provides protection and support for organs. Types of connective tissue include bone, blood, and adipose tissue.

q2

Had to look at fig 1 for this and i used POE cuz through the RANK/RANKL signaling pathway you can see that A,B, and C are all going to happen if there is elevated RANKL signalling so the only one that wouldn't be true is D. D bc osteoclast activation by the RANK/RANKL cascade depends on the interaction between RANKL and its receptor RANK. Accordingly, the observation of increased RANK expression in the absence of information on its interaction with RANKL would not support the scientists' hypothesis that increased RANKL signaling leads to decreased bone density in osteoporosis patients. Educational objective: Decreased activity of osteoblasts (bone-depositing cells) or increased activity of osteoclasts (bone-resorbing cells) can lead to reduced bone mass and density

Q15

Had to use CADET shift right. Q-stem asking abt healthy patients after exercise (HC after exercise on graph) and its asking for what reason will cause a right shift from HC at rest. A and B can be eliminated cuz an increase in CO2 and exercise would cause a right shift. C incorrect bc Hb binds Oxygen less tightly during exercise due to a decrease in blood pH bc muscles need O2 so Hb is more willing to give it up. D is correct bc Lactic acid increase causes Acid in cadet to increase which causes right shift Educational objective: Acidic byproducts of glycolysis produced during exercise cause a right shift in the hemoglobin oxygen dissociation curve. This phenomenon, known as the Bohr effect, provides muscles with a means to receive more oxygen from the blood during activity.

question 26

Have to know definitions of totipotent, pluripotent, multipotent Totipotent stem cells, which are found only in the zygote and up to the eight-cell stage of its mitotic division, have the greatest potency as they alone can autonomously give rise to an entire organism. Consequently, totipotent cells can differentiate into any cell type from either an embryonic or extraembryonic (placental) lineage pluripotent cells are able to differentiate into any cell found in the body. Pluripotent cells can give rise to any of the three primary germ layers found in the embryo but not the cells found in placental structures. Therefore, pluripotent cells are more specialized (differentiated) than totipotent cells. multipotent cells can differentiate into cells with many specialized functions but are limited in that they are "committed" to a specific lineage. For example, multipotent neural stem cells can give rise to the various cells of nervous tissue but cannot give rise to cells found in other tissue types. Educational objective: Totipotent stem cells are the least specialized cells and can give rise to both placental and fetal cells. Pluripotent stem cells can give rise to only fetal cells (ie, all cell lineages from the three germ layers). Multipotent cells are able to differentiate only into the specialized cells of certain tissues and are also found in adults.

Question 4

Have to know the similarities and differences of spermatogenesis and oogenesis. Primary spermatocytes become mature sperm via meiosis which involves two rounds of cell division (meiosis I and meiosis II). In contrast to spermatogenesis, oogenesis (oocyte production) begins before birth, not at puberty, in the female gonads (ie, ovaries). 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. 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.

q9

Helper T cells are T cells that recognize foreign antigens displayed by MHC proteins of other immune cells, such as B lymphocytes and macrophages and dendritic cells. In this scenario, a patient with an abnormally low number of helper T cells contracts an infection. Consequently, a low count of helper T cells would affect the activation of cytotoxic T cells and B lymphocytes (which produce antibodies against the bacterium) (Choices B, C, and D). However, because expression of MHC proteins is solely dependent on a cell's transcriptional and translational machinery, cells would still be able to display bacterial antigens on their MHC proteins regardless of helper T cell count. Educational objective: Helper T cells bind foreign antigens presented by other immune cells and release signaling molecules that enhance immune responses, such as cytotoxic T cell activation and antibody production by B lymphocytes.

What is lymph?

Hydrostatic pressure (the force exerted by blood against vessel walls) forces some liquid and smaller molecules to leak out of the capillaries. This liquid, at this point known as interstitial fluid, is present in the space between the blood vessels and surrounding cells. The interstitial fluid is then collected by the lymphatic vessels. The fluid, now called lymph, is filtered through lymph nodes (pockets on the lymphatic vessels containing white blood cells) and returned to the circulatory system.

q27

Hydrostatic pressure is the pressure that forces fluid out into the interstitial space from the vessel. And osmotic pressure draws the fluid back into the vessel (this is exerted by the proteins). As blood flow through a capillary increases, hydrostatic pressure exerted on the vessel walls also increases, leading to a higher rate of fluid leakage from the capillary into the interstitial space. If the rate of fluid leakage from capillaries exceeds the rate at which fluid is taken up by surrounding blood and lymph vessels, a higher volume of leaked fluid can build up in associated tissues (ie, edema). The q stem says that theres extra lungs in the fluid which means that fluid is outside of the vessels and thats caused by an increase in the hydrostatic pressure exerted on the capillary walls. concept: Blood flowing through capillaries exerts hydrostatic pressure on the vessel walls that forces some fluid from the vessel into the interstitial space. In contrast, circulating proteins in the bloodstream cause osmotic pressure in the blood to be higher than that of the interstitial fluid, drawing fluid from the interstitial space into the capillary.

question 17

I got where leptin and ghrelin is secreted from. Leptin is secreted by white adipocytes to suppress appetite and ghrelin is secreted by the gastric cells to trigger feelings of hunger Concept: In an energy-rich state (eg, after a meal), leptin is released by white adipocytes to trigger appetite suppression via the hypothalamus. In contrast, in an energy-poor state, ghrelin is released by stomach gastric cells to trigger hunger and food-seeking behavior via the hypothalamus.

Question 2

In glucocorticoid deficiency, an insufficient amount of glucocorticoid hormones in the blood leads to diminished binding of these hormones to their receptors. This condition would compromise the normal function of the various cells that express glucocorticoid receptors. Concept: Hormones are a diverse group of molecules that participate in endocrine signaling, a form of long-distance cell-to-cell communication. Any cell type that expresses a hormone receptor will respond to serum levels of the hormone.

Question 1

In response to a stressor, the adrenal medulla releases catecholamines (norepinephrine and epinephrine) that induce the following physiological changes: Redirection of blood flow throughout the body via vasoconstriction and vasodilation Vasoconstriction (narrowing) of blood vessels and reduced blood supply to organs that carry out nonessential functions (eg, stomach, intestines, kidneys) conserve oxygen and nutrients for organs that are necessary for immediate survival. Vasodilation (widening) of blood vessels increases blood flow to organs essential for immediate survival (eg, heart, skeletal muscle). Increased heart rate and cardiac muscle contractility promote increased blood flow to the brain, lungs, and skeletal muscles, allowing an organism to process and execute a response to the stressful stimulus. Dilation of airways (bronchioles) enables increased respiratory function and oxygen delivery to tissues. Concept: During the stress response, catecholamines secreted from the adrenal medulla act to promote the "fight-or-flight" response. Some effects of catecholamines include redirected blood flow to maximize the delivery of oxygen and other nutrients to organs essential for immediate survival (eg, brain, lungs, skeletal muscles), increased heart rate, and dilated airways.

question 27

In the experiment, the researchers removed undifferentiated gastrula cells that were presumed to give rise to neural tissue from one embryo (donor) and transplanted them onto a different gastrula (host) at a non-neural location. The researchers wanted to determine whether the prospective donor neural cells would continue to develop independently into nervous tissue or if they would assume a different tissue type or cell fate due to communication with surrounding cells. The transplanted donor cells developed into normal epidermal tissue in the host, suggesting that signals from the surrounding skin-destined cells influenced the fate of the transplanted cells. Cell determination - specification of cell fate Cell differentiation - acquisition of unique/specialized biochemical and structural cellular characteristics concept: Cell fate is influenced by inductive signaling between cells in the early embryo. In inductive signaling, an inducer (the signaling cell) releases chemical signals that act on competent neighboring cells by regulating the expression of specific genes required for cell determination.

q4

In the given scenario, patients with osteoporosis due to increased RANK-RANKL signaling were split into the following two groups during a clinical trial: an untreated group and a treated group, which received a drug that inhibits cathepsin K, a proteolytic enzyme secreted by osteoclasts to break down the bone matrix When untreated patients were given the saline (placebo), their heightened RANK/RANKL signaling most likely remained unchanged. As a result, RANK/RANKL-induced OPC differentiation into osteoclasts continued in untreated patients. This allowed osteoclasts to increase in number and secrete cathepsin K to break down bone, leading to the characteristic of decreased bone density seen in untreated osteoporosis. In contrast, when the treated group was given the cathepsin K inhibitor, this inhibitor prevented cathepsin K from breaking down bone in these patients, increasing their bone density over time. Accordingly, cathepsin K activity would be decreased in patients treated with the inhibitor but remain overactive due to increased RANK/RANKL signaling in untreated patients with osteoporosis. As a result, after the trial, bone density would be decreased in the untreated group compared to the treated group. concept: During bone resorption, osteoclasts release proteolytic enzymes that break down the organic bone matrix. In osteoporosis patients, overactivity of osteoclasts and increased release of these enzymes can lead to decreased bone density across time.

Q37

In the given scenario, scientists must identify a characteristic unique to either prokaryotes or eukaryotes that will allow them to classify the pathogen into one of the two groups. However, glycolysis is a process that occurs in the cytoplasm of both prokaryotic and eukaryotic cells, and results in the synthesis of pyruvate. Therefore, the presence of pyruvate in the pathogen's cytoplasm would not be a defining characteristic that would help determine whether it is prokaryotic or eukaryotic Concept: Eukaryotic cells contain membrane-bound organelles and linear genomes with multiple origins of replication. In contrast, prokaryotic cells lack membrane-bound organelles and have circular genomes with a single origin of replication. However, both cells have ribosomes (80S in eukaryotes and 70S in prokaryotes) and can perform glycolysis

question 5

In the passage (IVF) was used to introduce sperm cells into BCB− and BCB+ oocytes. IVF involves introducing numerous sperm into a dish containing an oocyte. Fully developed sperm generally have little to nonexistent cytoplasm; so saying that the cytoplasm is reduced in sperm used for IVF would be a normal finding that is unlikely to impact fertilization rate. concept: A human sperm cell is composed of a head (contains the acrosome and nucleus), a midpiece (contains mitochondria that generate the ATP required for flagellum-driven sperm motility), and a tail (ie, the flagellum necessary for motility through a fluid environment).

Q12

In the passage they say that P. leiognathi is a bacterium which is a prokaryote. So, the only place the secretion could go to is the plasma membrane. All the other structures are eukaryotic concept: Prokaryotic cells lack membrane-bound organelles and utilize specialized channels in the plasma membrane to secrete proteins. In contrast, eukaryotic cells contain membrane-bound organelles (eg, nucleus, Golgi body, mitochondria); the rough endoplasmic reticulum and Golgi body are involved in eukaryotic protein secretion.

Q10

In the passage they say that iron-bound transferrin (Tf) is referred to as holotransferrin (holoTf), and non−iron-bound Tf is known as apotransferrin (apoTf). holoTf first binds the cell surface transferrin receptor (TfR), forming the holoTf-TfR complex and triggering vesicle formation. holoTf first binds the cell surface transferrin receptor (TfR), forming the holoTf-TfR complex and triggering vesicle formation. Psg says that acidification of the endosome vesicle makes this complex release iron which means that holoTf becomes apoTf. Afterwards the apoTf is going to leave the cell to bind more iron while TfR will stay in the cell membrane to bind holotf. Concept: Endocytosis is a mechanism by which extracellular contents are transported into the cell within a vesicle formed from the plasma membrane. During receptor-mediated endocytosis, specific extracellular targets bind their corresponding receptors and are incorporated into intracellular vesicles that mature into endosomes.

question 15

In the passage they say that the gustatory-salivary reflex pathway is part of the parasympathetic division of the NS. So it must be a part of the peripheral NS and the autonomic NS CNS - responsible for the integration of information and includes nuclei (bundles of cell bodies) and tracts (bundles of axons) within the brain and spinal cord. PNS - composed of a sensory division and a motor division. concept: The nervous system can be divided into two major branches, the central and peripheral nervous systems. The peripheral nervous system can be further divided into the sensory (afferent) and motor (efferent) divisions, with the motor division being composed of the somatic and autonomic nervous systems. Furthermore, the autonomic nervous system is divided into the parasympathetic and sympathetic branches.

q18

In the psg, they say that vitamins help with RBC production but they also cause oxidative stress that results in hemolysis. Since that extract was administered to counteract the side effects of the treatment, its safe to assume, it was done so to neutralize the ROS. concept: Oxidative stress, caused by reactive oxygen species, can cause hemolysis. Compounds that remove reactive oxygen species can relieve the adverse effects of oxidative stress.

q29

In this scenario, a patient with a cut experiences excessive bleeding, indicating an impaired ability to form clots and prevent blood loss from damaged vessels. This symptom could have several causes, including reduced liver function, which would likely decrease clotting factor production, and reduced platelet proliferation. In addition, because antibodies mark cellular materials for destruction by immune cells, heightened levels of serum antibodies against clotting factors would promote destruction of these factors, impairing clot formation. (Choice B) Macrophages, phagocytic immune cells, engulf and destroy foreign pathogens and damaged cells. Although these cells may destroy pathogens entering the body through the cut, they would not inhibit clot formation and cause the excessive bleeding observed in the patient. Educational objective: Platelets and specialized proteins (ie, clotting factors) work synergistically in response to blood vessel damage, binding to the damaged portion of the vessel and forming a blockage that prevents blood loss.

Question 19

Incomplete dominance gives the intermediate phenotype between the parents Concept: Incomplete dominance between alleles results in the expression of phenotypes in heterozygous offspring that are intermediate to the phenotypes of homozygous parents.

Q5

Inhalation: Negative Pressure breathing Active process Diaphragm and external intercostal muscles contract High intra-pleural space, high thoracic cavity, low pressure High lung volume, low lung pressure Exhalation: Passive process Muscle relax low lung volume, high lung pressure Active Exhalation: When your internal intercostal muscles and abdominal muscles help force air out concept: The intrapleural space is the thin space between the lungs and the thoracic wall. The contraction of the diaphragm and the elevation of the rib cage decrease the pressure in the intrapleural space, causing the lungs to expand. Negative pressure breathing refers to the reduction of intrapleural pressure during inspiration.

q15

Injection of inactivated antigen A into a patient will allow B lymphocytes to bind the antigen, leading to the proliferation of both plasma cells and memory B cells. If future exposure occurs, these memory B cells will likely respond more rapidly to the antigen, increasing antibody production. The saline-treated patient is not exposed to inactive antigen A, and therefore never produces memory B cells following the saline injection. However, upon the first post-injection exposure to antigen A, this patient will also produce antibodies against antigen A, but this response will be slower than in the patient initially injected with inactivated antigen A. This first-time exposure to antigen A will lead to the proliferation of memory B cells that can more rapidly produce antibodies following a second exposure concept: Upon binding foreign antigens, activated B lymphocytes divide and differentiate into antibody-secreting plasma cells and long-lived memory B cells that can more rapidly respond to foreign antigens in the event of future infections.

question 24

Innervation of the adrenal medulla by the autonomic nervous system is unique in that the adrenal medulla is only innervated by the sympathetic nervous system. The passage states that norepinephrine and epinephrine release is stimulated by ACh-mediated signaling in the sympathetic component of the autonomic nervous system. Accordingly, for a drug to trigger an increased secretion of norepinephrine and epinephrine by the adrenal medulla, the drug must generate effects that mimic the activity of the sympathetic nervous system. concept: The autonomic nervous system is divided into the parasympathetic and sympathetic divisions, which are generally antagonistic in function. Although most tissues are innervated by both sympathetic and parasympathetic motor fibers, the adrenal medulla is unique in that it is only innervated by the sympathetic nervous system.

Q9

Interpretation of the results in the last paragraph They say that NTBI (non-transferrin bound iron) decreased over the 15 days period so on the table you would see a lower number for NTBI at day 15 compared to 0 days. They also say that NTBI is decreased bc of the increase in urinary holoTf excretion. So, on the table you need to look for an increase in the urinary holoTf number at day 15 compared to at day 0. concept: In the case of experiments that involve assessment of a dependent variable before and after administration of a particular treatment, a proposed hypothesis can be supported or not supported by assessing how the dependent variable changes before and after treatment administration.

Question 49

It said it in the passage Content: During cDNA synthesis, reverse transcriptase uses mRNA as a template to synthesize a new single-stranded DNA molecule. Because mRNA has the same sequence as the sense strand of genomic DNA (using uracil instead of thymine), cDNA will have the same sequence as the antisense strand. Accordingly, cDNA hybridizes with single-stranded DNA that corresponds to the sense strand of genomic DNA.

q4

It's a least likely Q and says that patinet consumes 5x the amt of food but can't absorb the nutrients so he would not exhibit and increased activity of nutrient transporter proteins in the SI Small intestine: (Don't Judge Idiots: duodenum, jejunum, ileum) digestion of macromolecules primary site of water absorption Large intestine: (cecum, colon, rectum) Absorbs remaining water and electrolytes concentrates and stores feces bacterial flora (metabolize carbs into fatty acids and synthesize vitamins) Educational objective: Nutrient absorption in the gastrointestinal tract is affected by the diversity of intestinal bacteria (gut flora), the surface area of the small intestine, and the functions of intestinal proteins (digestive enzymes, nutrient transporters, structural proteins).

q14

Know that miRNA regulate gene expression by silencing gene expression so and in this case they regulate Sox9 expression. Binding of miRNA to the 3'-UTR region of an mRNA transcript will generally prevent translation of the entire transcript. So if Group A and B has the transfected vector but Group B has the miR-145 mimic then you would not observe bioluminscence in group B but you would observe it in Group A cuz there is nothing blocking translation of it. Educational objective: A reporter gene allows researchers to study the regulation or expression of other genes. The luciferase gene is a commonly used reporter gene whose protein product catalyzes a reaction that results in bioluminescence, which allows easy expression quantification of the target gene.

q28

Leaked excess fluid from the capillaries, now also known as interstitial fluid, must be returned to the bloodstream to maintain the proper blood volume and protein concentration. To reenter the bloodstream, the fluid must first pass from the interstitial space and enter a network of vessels and nodes known as the lymphatic system. Once in the lymphatic system, this fluid is now known as lymph and is ultimately drained into two large veins near the heart. The lymphatic system also collects large lipid droplets absorbed by the small intestine. These lipid droplets are transported through the lymphatic system and into the bloodstream as follows: 1) Lipid droplets are released from the epithelial cells into the interstitial fluid. 2) Lymph capillaries collect lipid droplets from the interstitial fluid. 3) Lymph containing the lipid droplets then flows from the capillaries into increasingly larger lymph vessels. 4) Lymph is then transported into a large tubule structure called a lymph duct, which drains into a large vein near the heart. As a result, the lipid droplets within the lymph enter the bloodstream and circulate throughout the body concept: The lymphatic system collects protein-containing fluid leaked from blood capillaries and transports it back to the bloodstream. In addition, large lipid droplets absorbed by the small intestine are also transported to the bloodstream via lymph vessels. Fluid is collected by lymph capillaries, flows into larger lymph vessels, and is transported into lymph ducts that drain into veins near the heart.

q24

Manipulation of this equation gives the relationship of VR to ΔP and CO: ΔP/CO = VR So, an increase in ΔPor a decrease in CO would increase VR. CO is dependent on both heart rate (number of times a person's heart beats per minute) and stroke volume (blood volume expelled per left ventricular contraction). Stroke volume depends on the amount of blood loaded into the left ventricle prior to contraction. Reduced ventricular filling would lead to less blood being expelled from the ventricle with each contraction and decreased CO. Given the relationship of VR to CO, decreased CO (in the absence of other physiological changes) would ultimately increase VR. Concept: Blood flow through the circulatory system is influenced by blood pressure, vascular resistance (VR; force opposing blood flow through a vessel), and cardiac output (CO; blood volume expelled from the ventricles per unit time). The relationship of these factors in the systemic circulatory system is given by: ΔP = CO × VR.

q5

Marine teleosts lives in a higher concentration of salt envi and a lower concentration of water molecules than their internal environment. It is thermodynamically favorable for salt ions at high external concentration to diffuse through the skin into the low-salt internal environment of the fish. Water will traverse the opposite path via osmosis, moving from the high-concentration environment inside the fish to the lower concentration environment of the surrounding salt water. Osmoregulation uses external energy to reverse these processes and maintain constant salt and water concentrations within the fish. Freshwater teleosts live in an environment where there is more water and less salt outside their bodies than inside. As a result, salt ions are at relatively high concentration within the teleost and will tend to move to the lower concentration environment of the freshwater. Water molecules will move from the higher concentration of the external environment to the lower concentration inside the fish. Thats why the freshwater teleosts must use osmoregulation to combat their natural tendency to lose salt and absorb water. concept: In the absence of external energy input, molecules in solution diffuse from areas of high concentration to low concentration. Osmoregulation provides the energy needed to reverse this process and maintain constant salt and water concentrations within an organism.

Q26

Microtubules involved in intracellular transport originate near the nucleus on small organelles called centrioles and radiate out toward the plasma membrane The passage indicates that endosomes containing AAV2 move along microtubule tracks toward the nucleus. Because AAV2 enters the cell at the plasma membrane and moves along microtubules toward the nucleus, the Endosomes that carry it must be moving toward the centrioles near the nucleus. concept: the cytoskeleton consists of actin filaments (microfilaments), intermediate filaments, and microtubules. Microtubules are involved in intracellular transport. The microtubules involved in this process originate near the nucleus on organelles called centrioles and radiate out toward the plasma membrane.

q6

Needed to know the distinction btw the different types of muscle fibers. To perform aerobic respiration more efficiently, endurance-trained type 1 fibers would likely exhibit: 1) increased mitochondrial density to improve metabolic capacity 2) increased capillary supply to enhance oxygen delivery to the fiber 3) increased myoglobin content to ferry oxygen within the fiber more efficiently concept: Skeletal muscle fibers can be classified as type 1 (slow oxidative), 2A (fast oxidative-glycolytic), or 2X (fast glycolytic). The speed of contraction (slow or fast) is determined by the catalytic speed of the myosin ATPase and the activity of the fiber's associated motor neuron. Oxidative fibers are fatigue resistant and synthesize ATP through aerobic respiration (oxygen-requiring pathways). Glycolytic fibers are easily fatigable and generate ATP through anaerobic glycolysis (in the absence of oxygen).

q2

Needed to know the functions of the liver cells: A. Ketogenesis: Occurs mainly in the liver and plays a critical role in energy metabolism when glycogen stores are depleted and blood glucose levels become low. During times of fasting, mitochondria within liver cells convert acetyl coenzyme A into ketone bodies, which are then used by the brain to make ATP B: Detoxification of drugs: Drugs are often exogenous compounds that may have toxic effects if allowed to accumulate in the body. The liver is a critical organ for the breakdown and detoxification of many drugs C: Synthesis of plasma proteins: Oncotic pressure causes a "pulling force" inside capillaries, which balances the pushing force of hydrostatic pressure, helping keep fluid in the vasculature. The majority of plasma proteins such as albumin are synthesized in the liver. Therefore, cirrhosis is expected to impact the production of plasma proteins, which in turn would affect oncotic pressure in the capillaries D is correct bc digestive proteolytic enzymes, which hydrolyze polypeptides, are synthesized and secreted from the pancreas, stomach, and small intestine. Cirrhosis would not affect that concept: The liver has various functions, including storage of glycogen, regulation of blood glucose levels, detoxification of drugs, synthesis and secretion of bile, production of serum proteins (eg, albumin, clotting factors), and synthesis of ketone bodies.

q4

Normally the immune system is said to be self-tolerant bc it does not attack itself but protects against foreign substances. 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. 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. Educational objective: Autoimmunity disorders occur due to the immune system's failure to identify and destroy immune cells that recognize self-antigens. Self-reactive B cells and T cells are normally eliminated in the bone marrow and thymus, respectively.

q13

Number I: The outermost layer of the epidermis, the stratum corneum, is composed of 20-30 layers of these dead keratin-filled cells and functions as a physical barrier to protect the organism against pathogens, ultraviolet light, water loss, and injury due to abrasion or puncture. As stated in the passage, a superficial burn damages only the epidermis; as a result, stem cell production would be affected Number II: Epidermal Langerhans cells are immune (dendritic) cells that recognize and ingest antigens before migrating to nearby lymph nodes to present these antigens to T cells, activating the adaptive immune response. The adaptive immune response is highly targeted to eliminate a specific pathogen, whereas the innate immune response is generalized to attack any foreign substance. Number III: The dermis (middle layer) is composed of connective tissue and contains blood vessels, sensory receptors, hair follicles, and sweat (sudoriferous) and oil (sebaceous) glands. The hypodermis, or the subcutaneous layer beneath the dermis, is composed primarily of adipose tissue that protects the internal organs by acting as a shock absorber and insulator concept: The skin is divided into the epidermis (outermost layer), dermis (middle layer), and hypodermis (innermost layer). The epidermis acts as a physical barrier that separates the organism from the external environment. The dermis contains blood vessels, immune cells, sensory receptors, sweat and oil glands, and hair follicles. The hypodermis is composed of insulating and shock-absorbing adipose tissue.

Question 20

ONLY mom passes the mitochondrial DNA so the only offsprings of affected mothers are affected. Look at the image Educational objective: Mitochondria have their own genome, known as mitochondrial DNA, which is inherited in a maternal fashion (ie, no paternal contribution). Mitochondria within sperm are not transferred into the ovum during fertilization; therefore, males never pass on their mitochondria.

question 9

Oogenesis begins in utero at approximately 4 weeks gestation. In the female embryo, oogonia (germ cells) are diploid stem cells that first multiply quickly via mitosis and become primary oocytes. The primary oocytes then go through the first meiotic division but become arrested at prophase I. From infancy to puberty, the ovaries are functionally inactive and the primary oocytes remain stalled in prophase I of meiosis. The chromosome pairs are arranged in a tetrad during this phase, and their proximity allows for genetic recombination. At puberty, ovulatory cycles begin, and the female is capable of reproduction. During each menstrual cycle, stimulation by follicle-stimulating hormone (FSH) followed by a surge in luteinizing hormone (LH) causes some primary oocytes to resume meiosis I. It eventually becomes a secondary oocyte that begins meiosis II but halts in metaphase II. The secondary oocyte remains frozen in metaphase II until fertilization occurs, at which point it completes its second meiotic division (telophase II) into an ootid and second polar body. In the Q, a BCB+ oocyte is similar and if its arrested in metaphase II then it needs to be fertilized by a spermatozoon to complete meiosis II. Educational objective: Primary oocytes are present in female embryos and remain arrested in prophase I of meiosis from birth to puberty. During puberty, normal menstrual cycle hormones stimulate some primary oocytes to resume differentiating into a secondary oocyte. However, prior to fertilization, secondary oocytes are arrested in metaphase II of meiosis.

Question 15

Process of GPCR: The G protein "attached" to the transmembrane cell surface receptor is composed of three subunits: alpha, beta, and gamma. When inactive, the alpha subunit (Ga) is bound to GDP, which is then replaced with GTP on ligand-receptor binding. The GTP-bound Ga dissociates from the beta and gamma subunits and proceeds to activate adenylate cyclase, an enzyme that catalyzes the conversion of ATP to cAMP, a second messenger. cAMP then activates protein kinase A, which subsequently phosphorylates the proteins necessary to produce ligand-specific physiological effects (eg, glucagon: induction of glycogenolysis and gluconeogenesis). here bc GPCR is not working the only ans that made sense was D Concept: In the G protein-regulated cAMP signaling pathway, a ligand binds the transmembrane G protein-coupled receptor and activates the GDP-bound alpha subunit of the G protein by replacing GDP with GTP. The activated G alpha subunit activates the enzyme adenylyl cyclase, which catalyzes the conversion of ATP into cAMP. Elevated cAMP leads to the activation of protein kinase A and subsequent signaling effects.

Q14

Prokaryotic cells have no nucleus or nuclear envelope, and due to the lack of this physical barrier between DNA and the ribosomal machinery, transcription and translation occur simultaneously in the cytoplasm. Qstem states that P. leiognathi (prokaryote) cells have a greater number of ribosome/SOD mRNA complexes compared to ponyfish (eukaryotic) cells. This rapid protein synthesis is likely due to the fact that all SOD mRNA transcripts in P. leiognathi quickly interact with ribosomes before being fully transcribed (ie, transcription and translation are coupled). (Choice A) Both eukaryotic and prokaryotic mRNAs can be bound by multiple ribosomes in the cytoplasm. (Choice B) Ponyfish mRNA is translated by the 80S ribosome whereas P. leiognathi mRNA is translated by the 70S ribosome. In this scenario, no information is given to suggest that ribosome affinity is a factor for SOD synthesis. (Choice C) In eukaryotic cells (eg, ponyfish), transcription of mRNA and its modification (eg, addition of the 5′ cap, 3′ poly-A tail, splicing) typically occur in the nucleus. The modifications, which result in a mature mRNA molecule, increase the stability of the transcript and prevent its degradation in the cytoplasm. Because the nucleus is separated from the cytoplasm by a nuclear envelope, mRNA must first be transported through nuclear pores to the cytoplasm, where it is translated by ribosomes. Educational objective: Prokaryotic cells have no nucleus; therefore, transcription and translation occur simultaneously in the cytoplasm (ie, translation begins before the mRNA is fully transcribed). By contrast, in eukaryotic cells transcription and post-transcriptional modifications occur in the nucleus, but translation occurs in the cytoplasm.

Q8

Protozoa and fungal are eukaryotic Bacterial is prokaryotic so it also has a phospholipid bilayer. Only viruses can be either enveloped or non-enveloped. Enveloped viruses have a membrane, or phospholipid bilayer, generally derived from the cell membrane of the host. The phospholipid bilayer of enveloped viruses surrounds the capsid and often contains host-derived proteins, allowing the virus to better evade the immune system and gain entry into the host cell. In contrast, non-enveloped (naked) viruses lack a phospholipid bilayer. Concept: Viruses possess DNA or RNA genomes surrounded by a protein capsid and are unable to reproduce outside of a host. Viruses can also be classified as either enveloped (have a phospholipid bilayer as the cell membrane) or nonenveloped (no phospholipid bilayer). In contrast, all prokaryotic and eukaryotic cells are enclosed by a phospholipid bilayer.

Q2

Psg stated that asthma causes bronchioles to narrow and says it has a decreased airway diameter which increases resistance to flow. This means that the volumetric rate of forcibly exhaled air would decrease. FEV1 in psg is the volume of air forcibly exhaled in the first second. So, FEV1 would decrease and the volume exhaled at 1 second on the patient's spirometer should be lower. Forced vital capacity (FVC) is defined in the passage as the total volume of air exhaled in a single breath. The passage states that less air can be exhaled in asthma due to air trapping in the lungs. Therefore, FVC would decrease, and the volume exhaled at the end of the patient's spirometry tracing should also be lower Educational objective: The narrowing of respiratory airways in asthma increases resistance to airflow, which would decrease the volumetric rate of forcibly exhaled air (FEV1 decreases). The narrow airways also trap air in the lungs, decreasing the total volume of air that can be exhaled (FVC decreases).

q7

Psg states that eels have glomerular kidneys and goosefish have aglomerular or tubular kidneys. Qstem says that if marine teleost urine is independent of glomerular function what would you likely have to see. To confirm that the inorganic composition of urine is independent of glomerular function, scientists would have to find similar concentrations of inorganic substances in the urine of both glomerular and aglomerular fish. This finding would signify that inorganic urine composition is the same regardless of whether the fish possess a glomerular kidney. Because divalent ions (eg, Mg2+ and SO42−) are primarily excreted in the urine and monovalent ions (Na+ and Cl−) are primarily excreted through the gills of marine teleosts, researchers would have to find equally high urinary concentrations of Mg2+ and SO42− ions in the glomerular eel and the aglomerular goosefish. Educational objective: To claim that a particular function is independent of an anatomical structure, researchers must show similar function-related findings in organisms with and without the anatomical structure of interest.

Q21

Q gives 3 antibiotic disks, one with G drug, one with chloramphenicol C, and one A drug. The only one imp is C. Q asks which of the following represents what a strain of salmonella would look like if it was transformed with the plasmid in passage. The passage indicates that plasmids containing lac genes also have chloramphenicol resistance genes. Therefore, transformed bacteria should be able to grow near the chloramphenicol disk but not the other antibiotics. Antibiotic resistance is often investigated using the agar diffusion test in which disks impregnated with different antibiotics are placed on a bacterial culture. Susceptible bacteria normally form a growth-free halo called a zone of inhibition around the disks due to the high antibiotic concentrations. However, resistant strains can grow near that antibiotic and do not form a zone of inhibition. Educational objective:Antibiotic resistance genes in plasmids are used to select bacteria with a gene of interest. In an agar diffusion test, bacteria that are susceptible to an antibiotic will form a zone of inhibition around the antibiotic disk where they cannot grow. Bacteria with resistance genes can grow close to the disk.

Q24

Q is asking what cell compartments do the AAV2 binding proteins pass through before reaching the plasma membrane. Proteins that are embedded in the plasma membrane traveled through the secretory pathway to arrive there. Translation of mRNA to proteins always begins in the cytosol, but proteins intended for the secretory pathway have an N-terminal sequence called a signal sequence. Once the signal sequence is recognized, the ribosome is transported to the rough endoplasmic reticulum (RER) where the protein is either translocated across the RER membrane or embedded into the membrane. Then it goes through some post-translational modifications may occur, including glycosylation, disulfide bond formation, phosphorylation, and protein cleavage. The RER then packages proteins into vesicles and sends them to the Golgi apparatus, where they are further processed. Finally, the Golgi packages proteins destined for the plasma membrane into secretory vesicles, which fuse with the plasma membrane. Proteins that were initially embedded in the RER membrane frequently end up embedded in the plasma membrane as receptor proteins. The peroxisome is not part of the secretory pathway and is not involved in protein transport to the cell surface. Instead, peroxisomes contain various oxidoreductase enzymes and function to help maintain the proper oxidation state within cells. Proteins in the peroxisomal lumen were translated entirely in the cytosol and imported. Educational objective: The secretory pathway involves the processing of proteins as they go through the endoplasmic reticulum and the Golgi apparatus. After processing in the Golgi, proteins destined for the cell membrane are packed into secretory vesicles, which fuse with the plasma membrane. Other organelles, such as the nucleus, mitochondria, and peroxisomes, are not involved in the secretory pathway, and most of their proteins are translated in the cytosol.

Q27

Q is asking what happens to the virus particle that does not escape the acidified endosome in fig1, step 6? The passage states that AAV2 inhibits pathways other than the endosomal pathway. Therefore, AAV2 particles that fail to escape acidified endosomes in Step 6 cannot enter the Golgi or other secretory pathway organelles to undergo exocytosis. The only other available fate is the lysosome (the last step of the endosomal pathway), where AAV2 particles will be degraded. The endosomal pathway begins at endocytosis where, following internalization of extracellular materials into vesicles, the vesicles containing these materials mature to early endosomes, late endosomes, and finally, lysosomes. The lysosome is a membrane-enclosed organelle that serves as the "digestive system" of the cell; it contains an acidic environment (pH ≈ 4.5) and various hydrolytic enzymes that facilitate the degradation of various biomolecules (eg, proteins, carbohydrates, nucleic acids, lipids). Some endosomes participate in alternative pathways such as trafficking through the Golgi apparatus. This pathway allows the cargo within the vesicles to enter the secretory pathway and be expelled from the cell by exocytosis (ie, bypassing degradation by the lysosome). Educational objective: After internalization of extracellular materials into vesicles, the vesicles containing these materials become endosomes, which then mature from an early stage to a late stage, ultimately becoming lysosomes. Endocytic cargo that does not enter alternative pathways (eg, entry into the secretory pathway) and fails to escape from endosomes will end up in the lysosome to be degraded.

question 23

Q is asking what is the function of vasodilation. Vasodilation increases blood vessel volume and decreases BP bc more area for blood to travel. It also increases blood flow so it can transport oxygen and nutrients to tissues supplied by that blood vessel. Ex. during exercise, vessels dilate so more O2 and nutrients can head to muscle. Digestion is promoted as blood vessels supplying the abdominal organs dilate following a meal. Vasoconstriction reduces blood vessel volume, resulting in increased blood pressure (BP; the pressure of blood against vessel walls). Consequently, vasoconstriction functions to maintain BP following sudden blood pressure drops (eg, following fluid loss). Vasoconstriction also reduces blood flow (the volume of blood travelling through a vessel per unit time), and can therefore redirect blood flow to other tissues. Educational objective: Vasodilation (blood vessel widening) decreases blood pressure and increases blood flow to the tissues supplied by that blood vessel. Vasoconstriction (blood vessel narrowing) increases blood pressure and decreases blood flow to the tissues supplied by that blood vessel.

Q33

Q is asking what vector can introduce its genomic content into bacterial cells without fully gaining entry into the bacterial cytoplasm? The answer is a bacteriophage bc they are viruses that exclusively infect bacteria but do not enter host cells to replicate their genetic material. Instead, they use their tail sheath, a structure that injects the phage genome into a bacterium. The remaining viral structures of the phage, such as the tail fibers, the capsid, and the tail sheath, are left outside the bacterium. Prion is a misfolded protein that acts as an infectious agent by inducing other normal proteins to change their secondary structure and become misfolded. These less soluble misfolded proteins aggregate and can cause disease. Prions do not contain genetic material and cannot transform bacteria. Viroids are not viruses; instead, they are pathogenic, circular, single-stranded RNA molecules lacking protein coats and primarily affect plants. They typically silence the expression of specific genes and inhibit protein synthesis by binding RNA sequences. Viroids enter cells by hiding inside viruses or through damaged tissue; they do not use the mechanisms described in the question. Educational objective: Bacteriophages are viruses that exclusively infect bacterial cells. They contain tail fibers that allow them to recognize and attach to the cell membrane, and a tail sheath that injects the viral genome into the bacterium. Viroids and prions are considered subviral particles because they are smaller in size and complexity compared to viruses.

Q18

Q is asking which conclusion is the most accurate regarding ST virus? All the info you needed was in table 1, you had to look at the p-value, the only section that there is significance is within the flagellar genes meaning that the answer has to have something to do with inhibition of movement. The chemosensation genes are not significantly affected by the presence of the lac genes, so the bacteria are likely still able to detect the chemical signals produced by epithelial cells. However, impaired flagella would inhibit motion toward the signal's source. Chemotaxis is the process by which mobile cells both sense and move toward or away from an increasing concentration of a specific chemical. Many pathogens find host cells to invade by sensing the chemicals that they release and moving toward them. Impaired ability to either sense or respond to chemical signals can reduce virulence. Educational objective: Bacteria respond to chemical stimuli in the extracellular environment by moving away from or toward increasing concentrations of signal molecules. This movement, known as chemotaxis, allows bacteria to adjust the direction of their movement toward target cells and is an important part of infection. Impairing the ability to detect or respond to chemical signals can reduce virulence.

Q16

Q is asking which of the following will bring back the virus effect of ST after being treated with lac+. You know from passage that lac+ inhibits the viral parts of ST. So look at fig 2 and x-axis states transcript level relative to WT and the colors represent the flagellar genes that are getting affected (meaning that it is probs what is giving the virus its viral properties). You can see that LacY and LacZ seem to have minimal effects on the flagella so they can't be the one's inhibiting virus. If you look at lac+ operon, it decreases flagellar transcript levels by the same amount as LacA alone meaning if you excise lacA then you will get viral properties back. Educational objective: Expression of one gene can affect the expression level of other genes and can alter activities such as virulence. Some genes do not significantly alter levels of other genes and do not affect their activities. The effect of one gene on others can be tested empirically.

question 14

Q is asking which stage of the cell cycle would a drug that inhibits fast-growing tumors NOT act on? Checkpoints (restriction points) in a cell control cell division. Tumors are masses of abnormal cells. Because the cells within these tumors are actively dividing, it can be assumed that the majority of the cells are in interphase and not arrested in G0. Therefore, to treat these aggressive cancers, clinicians may use anticancer drugs that try to destroy actively dividing cells (phases G1 to M) and not those in G0. Educational objective: Most cells in the human body are arrested in G0. However, cellular transition into G1 prepares a cell for division and DNA synthesis (S phase). In the G2 phase, DNA is checked for errors and the cell ensures that sufficient organelles and cytoplasm are available for cell division. Subsequently, the cell divides in the M phase via mitosis and cytokinesis. Compounds that inhibit cell division typically target the cell cycle in phases G1 to M.

Question 20

Q is basically asking you to identify the part of the flagellum that generates torque or rotation. You have to know the structure of a flagella for this Flagella in prokaryotes consist of three basic parts: a basal body, a hook, and a long helical tube called a filament. The component that generates torque is the basal body, a transmembrane unit that works as a molecular motor. Cells that are unable to generate torque likely have a defect in flagellar basal bodies. Educational objective: Bacterial flagella are composed of a basal body, a hook, and a hollow filament filled with flagellin. The hook connects the filament to the basal body, and the basal body serves as the motor that generates motion.

question 4

Q is just asking what will cause an AP in a postsynaptic neuron. In most cases when a specific ligand (neurotransmitter) binds to its corresponding ligand-gated ion channel on the postsynaptic neuron, the channel changes its conformation to the open state. Depending on the channel's specificity, ionic movement across the membrane results in either postsynaptic depolarization (excitation) or hyperpolarization (inhibition). Because the concentrations of Ca2+ and Na+ are higher outside the neuron than inside, if a ligand-gated ion channel permeable to both these ions were to open in response to neurotransmitter release, both Ca2+ and Na+ would rush into the cell and the membrane would depolarize. This depolarization, if summed with other excitatory signals, would contribute to the generation of a postsynaptic AP. (Choice C) Improper clearance of the neurotransmitter from the synapse (due to reuptake inhibition or enzymatic degradation) would prolong the neurotransmitter's effect on the postsynaptic cell. However, this does not specify whether the neurotransmitter's effect is excitatory or inhibitory. Educational objective: Although trends in neurotransmitter function exist, the postsynaptic response is ultimately determined by the type of receptor-activated. Receptors may be either ligand-gated ion channels or G-protein-coupled receptors.

Q31

Q is stating that individual with hepatitis is either exposed to henselae (0.5µm) or HCV (0.055µm). The double membrane used in diagram was used to separate henselae and HCV based on size. you just had to notice that HCV and henselae could pass through from chamber 1 to 2 to infect healthy hepatocytes and only HCV could pass through from chamber 2 to 3 to infect healthy hepatocytes. To answer if you can observe the agent that caused the patient's hepatitis in chamber 3, the answer is no because viruses cannot be seen under a light microscope, they are too small but bacteria can be. Eukaryotes - 10-100 µm diameter Prokaryotes - 0.5-2 µm diameter Viruses - 0.02-0.3 µm diameter Educational objective: Prokaryotic cells such as bacteria are approximately 10 times smaller than eukaryotic cells. Viruses are about 100 times smaller than bacteria and 1,000 times smaller than eukaryotic cells. Unlike bacteria, viruses cannot be observed with a light microscope because viral particles are smaller than the microscope's maximum resolution.

question 1

Q states that nuclear envelope is stained green so when would the green fluorescence be the most prominent? You had to know that the nuclear envelope breaks down during prophase, which allows both metaphase and anaphase to occur in the middle of the cytoplasm. The nuclear envelope reforms again during telophase which is right before cytokinesis (the division into 2 identical daughter cells). Educational objective:Mitosis in eukaryotic cells includes four distinct phases: prophase, metaphase, anaphase, and telophase. The nuclear envelope breaks down during prophase, allowing both metaphase and anaphase to occur in the cytoplasm. During telophase, the nuclear envelope reforms around the chromosomes prior to cytokinesis, the cytoplasmic division of the parental cell into two identical daughter cells.

question 20

Q states that progesterone is low in proliferative phase and highest in secretory phase and that VEGF may be inversely correlated with progesterone levels. It's asking do you see this represented in the data? If you look at fig 1, you can see that in normal ppl (w/o endometriosis) the proliferative phase has a higher VEGF conc and the secretory phase has a lower VEGF conc which is what we expect if its inversely correlated to progesterone. We also see the same trend in the endometriosis results meaning that it is represented in the data. Educational objective: Changes in the serum levels of estrogen and progesterone cause the physiological changes associated with the menstrual cycle, the female reproductive cycle that repeats every 24-36 days.

q3

Qstem says that camels live in water deficient areas and have adaptation where they can absorb more water than their ancestors could so would they have more cortical nephrons or juxtamedullary based off of diagram given. Ans is juxtamedullary bc they have a longer loop of henle which would allow for more water reabsorption back into the blood in the descending part of the loop of henle. Educational objective: In the loop of Henle, the active transport of NaCl from the ascending limb maintains the high salt concentration in the kidney's medulla, facilitating the reabsorption of water from the descending limb.

q21

Qstem states mice have a sodium transporter mutation that causes impaired active transport of NaCl from filtrate w/in Loop of Henle to medulla. It asks the mutatnt mice would exhibit all of the following EXCEPT ____. So you had to knwo that the thick tubule of the ascending LOH is permeable to NaCl and its actively transported from the filtrate to medulla. You also had to knwo that water follows solute. So now, if NaCl stays in filtrate then more water stays in filtrate so there would be an increase in urine output Educational objective: In the loop of Henle, active transport of NaCl from filtrate in the ascending limb maintains the saltiness of the renal medulla, facilitating passive water reabsorption from filtrate in the descending limb and collecting duct.

q5

RANK in the passage has osteoclast activity so it would have the same function as parathyroid hormone concept: To regulate calcium homeostasis, parathyroid hormone (PTH) and calcitonin act antagonistically to each other. PTH is secreted in response to low blood calcium levels and stimulates bone resorption by osteoclasts, causing an increase in blood calcium. In contrast, calcitonin is secreted in response to high blood calcium levels and decreases osteoclast activity, ultimately decreasing blood calcium.

question 21

Removal of the ovaries mimics what happens after a women enters menopause. During menopause, production of estrogen and progesterone within the ovaries decreases, leading to numerous physiological changes. For example, the vaginal epithelium dehydrates (causing discomfort and impairing sexual activity), changes in mood (eg, irritability, depression) may occur, and the reproductive organs and breasts shrink (atrophy). Concept: The ovaries are female reproductive organs that contain oocytes and secrete female sex hormones (ie, estrogen, progesterone). These hormones influence female reproductive function, and are directly responsible for the development and maintenance of female sex characteristics.

question 15

The Q is saying that mitochondria are transported from the cell body toward the presynaptic terminal which shows that the direction is away from the nucleus which means kinesin is doing anterograde transport. Kinesin and dynein both move on microtubules for intracellular transport. Educational objective: The intracellular scaffolding of a eukaryotic cell is composed of three families of protein filaments: Microfilaments, intermediate filaments, and microtubules. Intracellular transport of cargo (eg, organelles, vesicles) is mediated primarily by two microtubular motor proteins (kinesin and dynein). Kinesin mediates anterograde transport (ie, away from the nucleus) whereas dynein mediates retrograde transport (ie, toward the nucleus).

question 9

The drug is similar in structure to a steroid hormone (ie, relatively small and lipophilic). Therefore, this drug will cross the blood-brain barrier easily and cause the concentration in the CSF to rise during the hours following ingestion. However, increasing interstitial volume during the night would decrease the rate of change in drug concentration within the CSF. As the passage states, increased interstitial volume leads to increased clearance, which would counter the rise in concentration. concept: The blood-brain barrier is composed of endothelial cells held together by tight junctions, which limit paracellular transport. Carrier-mediated transport allows the passage of glucose, amino acids, and nucleosides into the CSF. Small, lipophilic molecules pass easily into the CSF through transcellular diffusion.

question 26

The endoplasmic reticulum, a cytosolic organelle composed of a network of membranous tubules and located near the nucleus, is present in all human cells except red blood cells and spermatozoa. The endoplasmic reticulum can be divided into two components, rough endoplasmic reticulum (RER) and smooth endoplasmic reticulum (SER). The RER has ribosomes located along its surface that translate the mRNA of proteins destined for the secretory pathway. The SER contains no ribosomes along its surface. Instead, the interior compartment (lumen) of the SER contains enzymes that synthesize lipids, including cholesterol and cholesterol-derived molecules (eg, steroid hormones). ACTH is released by the anterior pituitary and acts on the adrenal cortex to promote the synthesis and secretion of cortisol, a steroid hormone produced from cholesterol. However, the actual synthesis of this steroid hormone depends on the production of cholesterol by SER proteins, not ACTH. Consequently, defective SER proteins may compromise cortisol synthesis within the adrenal cortex regardless of ACTH stimulation. Accordingly, patients with adrenal insufficiency caused by defective SER proteins would not respond to ACTH administration as a treatment concept: The endoplasmic reticulum is a cellular organelle with rough and smooth components. Synthesis of lipids and steroid hormones takes place within the smooth endoplasmic reticulum.

Q16

The enzyme bisphosphoglycerate mutase (BPG mutase) converts 1,3-BPG to 2,3-BPG by transferring a phosphate from carbon 1 to carbon 2 within a bisphosphoglycerate molecule. Up-regulation of this enzyme would therefore increase the concentration of 2,3-BPG. Hb's affinity for oxygen is decreased in the presence of 2,3 BPG (right shift on the curve) meaning tissues are getting the oxygen. Thats why here up regulation of BPG mutase will increase the 2,3 BPG meaning oxygen release into the tissues of cirrhotic patients would take place. concept: Bisphosphoglycerate mutase converts 1,3-bisphosphoglycerate into 2,3-bisphosphoglycerate, which allosterically regulates hemoglobin. It decreases oxygen affinity by stabilizing the deoxyhemoglobin conformation. Increasing concentrations of 2,3-bisphosphoglycerate produce a right shift in the oxyhemoglobin dissociation curve and favor oxygen delivery to tissues.

question 28

The figure shows that the concentrations of sonic hedgehog (Shh) and bone morphogenic protein (BMP) vary along the embryonic dorsal-ventral axis, suggesting that both molecules are opposing morphogens that determine cell fate based on their relative concentration gradients. In other words, certain combinations of morphogen levels signal for the expression of different genes. For example, high BMP/low Shh concentrations result in Pax7 expression to define the dorsal end of the neural tube, and low BMP/high Shh concentrations result in Pax6 expression to define the ventral end. At intermediary (mixed) BMP/Shh concentrations, other genes (Dbx1, Dbx2, Irx3) are activated to induce proper differentiation along the body axis. Choice B is just saying that Shh is increasing along a concentration gradient to differentiate from dorsal to ventral concept: Morphogens are signaling molecules that influence cell differentiation in the embryo. They are released by signaling cells and diffuse outward to alter gene expression in competent cells in a concentration-dependent manner.

structure of a sperm

The head contains an acrosome and the nucleus. The acrosome is a flattened structure that encapsulates the tip of the nucleus and is rich in lysosome-like enzymes specialized for piercing the outer shell of an oocyte during fertilization. The mid-piece section is packed with mitochondria, essential organelles that produce the ATP required for flagellum-driven sperm motility. This section also contains a pair of central microtubules that are anchored to the cytoskeleton and extend down the length of the flagellum (tail region). The tail, or flagellum, is a singular elongated structure specialized for wavelike movements to propel sperm through a fluid environment. Flagellum-driven motility is derived from the action of ATP-dependent motor proteins that act on the central microtubules.

Q8

The initial filtration process of the nephron is the glomerular filtration. Q-stem is saying there is a buildup of iron in the liver which causes problems and it says that if holoTf is the reason why then what would be the rate of holoTf filtration in the glomeruluar capillaries. It is asking abt what's happening to its rate at THAT moment when there is iron buildup. It is not asking for a solution or a way to correct this iron buildup. If there is iron buildup then that means kidney is not doing a good job at filtering iron out so the rate would be decreased for holoTf (which means increased holoTf in the body). If hydrostatic pressure within the glomerulus decreases, the GFR (and consequently, the rate of holoTf filtration) will decline because the flow rate of filtered fluid through the kidney will decrease. A is incorrect bc increased Na+ reabsorption in the nephron, which is stimulated by the adrenal hormone aldosterone, also leads to increased reabsorption of water. This results in increased blood pressure without altering blood osmolarity (blood solute concentration). Increased blood pressure increases the GFR, decreasing holoTf retention. Educational objective: Glomerular filtration rate is a measure of the rate at which fluid is filtered into Bowman's capsule from the glomerulus.

q30

The left ventricle contains oxygenated blood, so the opening in the ventricular septum causes oxygenated blood to flow into the right ventricle. If left untreated, this cardiac defect may cause: 1) Increased blood flow into the pulmonary artery. Blood flow from the left to the right ventricle increases the volume of blood in the right ventricle and subsequently increases the amount of blood pumped through the pulmonary artery by right ventricular contraction 2) Increased blood volume in the left atrium. Increased blood flow through the pulmonary circuit from the right ventricle likely increases the volume of blood flowing through the pulmonary veins into the left atrium 3) Low blood oxygen levels in systemic arteries. Because oxygenated blood abnormally flows from the left ventricle into the right ventricle and the pulmonary circuit, the level of oxygenated blood pumped from the left ventricle through systemic arteries decreases concept: Deoxygenated blood returning to the heart enters the right atrium, is transferred to the right ventricle, and is pumped to the lungs via the pulmonary circuit. The left atrium receives oxygenated blood and transfers this blood to the left ventricle, which pumps oxygenated blood throughout the body.

Q32

The newly discovered hepatitis strain, HXV, was identified while analyzing chromosomal DNA of liver (host) cells. This indicates that the virus is a retrovirus because it integrated with the host genome (DNA). Retroviruses are unique in that they are enveloped and carry two identical +ssRNA molecules. Successful viral replication of HXV depends on enzymes that allow its viral genome (RNA) to enter the nucleus and integrate with the host genome (DNA). RNA viruses such as HXV must convert their genomes (+ssRNA) into double-stranded DNA (dsDNA) using reverse transcriptase (RNA-dependent DNA polymerase activity) before integration into the host genome can occur. Once integrated, the viral genome is replicated along with the host cell's own DNA as a lysogenic provirus. Consequently, the original cell containing the integrated viral genome will divide and produce descendants that are also infected. Concept: (Choice D) According to the passage, viruses are obligate intracellular parasites, meaning they cannot replicate outside a host cell. Independent replication cannot occur because viruses lack the ability (resources and enzymes) to produce their own proteins.

question 13

The passage says that the drug VX-680 arrests cells after they complete metaphase. Figure 2B shows that in the presence of this drug the T brucei cells are able to complete the M phase but can't complete it to go back to G1. So the drug causes T brucei cells to complete metaphase but fail to enter anaphase. Educational objective: Mitosis typically consists of four phases: Prophase, in which the nuclear envelope disintegrates and chromatin condenses; metaphase, in which chromosomes align on the metaphase plate; anaphase, in which sister chromatids migrate toward opposite poles of the cell; and telophase, in which nuclear envelopes reform and chromatin reverts to its uncondensed form.

question 5

The passage stated that during sleep waste products are removed from the brain such as harmful compounds like amyloid proteins so if you have sleep deprivation then that will lead to accumulation of amyloid proteins which leads to alzheimers. Educational objective: Alzheimer disease is a progressive neurodegenerative brain disease characterized by the presence of plaques composed largely of amyloid beta proteins and neurofibrillary tangles composed of tau proteins.

q18

The passage states that SGLTs transport glucose out of the proximal tubule so it can be returned to circulation through the peritubular capillaries. Administration of a SGLT inhibitor prevents glucose reabsorption, increasing solute concentration within the tubule and consequently increasing the osmotic pressure of the tubular filtrate. As a result, water is drawn back into the renal tubule, increasing urine output. Concept: The kidneys regulate blood osmolarity, and in turn blood pressure, by selectively modulating the reabsorption and secretion of water and solutes.

question 10

The passage states that T. brucei cells were arrested in S phase. Because S phase is a part of interphase, T. brucei cells undergo interphase. In addition. Figure 2A shows that all three stages of interphase (G1, S, and G2) occur in T. brucei. Figure 1 shows that in M phase, T. brucei cells undergo cytokinesis longitudinally, along the longest axis. concept: The cell cycle consists of interphase (G1, S, and G2) and the mitotic (M) phase. In M phase, chromosomes are segregated into two nuclei. During the first phase of M phase (ie, prophase), chromatin condenses into chromosomes and the nuclear envelope disintegrates, allowing the cell to progress into the remaining phases of M phase (ie, metaphase, anaphase, and telophase). Cytokinesis, or the division of the cell membrane, then yields two daughter cells.

question 11

The passage states that a DNA replication inhibitor which is hydroxyurea was used to arrest T brucei cells in S phase, after which these cells were washed to remove the inhibitor and cell cycle progression after the release was monitored. So the purpose of the protocol was to make sure that most cells stayed in the same phase at each time point measured. Educational objective: A random population of cells will likely contain some cells in each phase of the cell cycle. Study of their progress through the cell cycle requires that these cells first be brought to the same phase of the cell cycle (ie, synchronized). These synchronized cells can then be monitored when they are released and progress through the cycle together.

q18

The passage states that harvested muscle samples were placed in an oxygen-infused, electrolyte-rich bath that contained Ca2+ ions; it also states that these ions could enter the muscle fibers and enable maximal contractile force. Based on this information, the calcium ions in the bath were able to enter and aggregate inside the SR of the muscle fiber samples, enabling their contractile activity. concept: The sarcoplasmic reticulum (SR) is a muscle fiber organelle that tightly regulates intracellular calcium (Ca2+) concentration. The release of calcium from the SR promotes muscle fiber contraction whereas the transport of calcium into the SR promotes and maintains muscle fiber relaxation.

Question 21

The passage states that the hCaSR is a G-protein coupled receptor (GPCR) whose activity is marked by IP-1, a downstream metabolite of IP3. This indicates that hCaSR specifically activates the IP3-DAG pathway. In addition, AMG 416 is an agonist of hCaSR, meaning it functions to bind and activate the receptor. In GPCR, GTP binds to the alpha subunit and then breaks off promote downstream signaling Concept: When the alpha subunit of a G-protein is bound to GDP, the protein is inactive. On ligand binding, GTP binds in place of GDP, causing activation of phospholipase C (PLC). Consequently, PLC breaks down phosphatidylinositol bisphosphate into diacylglycerol (DAG) and inositol trisphosphate (IP3). IP3 then triggers calcium release from the endoplasmic reticulum, and protein kinase C is subsequently activated by both DAG and calcium.

q6

The psg says that after 20 hrs 2.3 mL of fluid was in eels gut and the concentration of ingestion was 12.3 mL of water/ that means the eel absorbed 10 mL of water thru its gut (12.3ml - 2.3 mL). And they say that all this water was lost. Of the 10 mL absorbed, the passage states that the eel had a 2.3-g weight loss due to urine output. Because the density of water is 1 g/mL, a 2.3-g weight loss would be equivalent to 2.3 mL being excreted as urine. so the total amount of water loss = renal water loss + extrarenal water loss: 10 mL - 2.3 mL = 7.7 mL that was lost extrarenally. concept: Intestinal fluid ingestion, weight loss as water, and renal and extrarenal mechanisms of fluid excretion must be taken into account when calculating an organism's fluid volume.

q27

The psg states that the alimentary limb has a smaller gastric pouch has the same function as a normal sized stomach and would secrete - The zymogen (inactive) pepsinogen is released by gastric chief cells and is converted into pepsin (active form) on mixing with hydrochloric acid (HCl) in the gastric juice. Pepsin then digests proteins in the stomach (Choice A). - Gastrin, a peptide hormone released by gastric G cells, stimulates HCl secretion from parietal cells in the stomach and promotes stomach churning to assist gastric motility (Choice C). - The stomach normally protects itself from autodigestion by HCl and proteolytic enzymes in the gastric juice. Mucus is released by mucous cells in the stomach to form a protective physical barrier against autodigestion (Choice D). Bicarbonate ions are released from stomach epithelial cells form a chemical buffer barrier under the mucus barrier. The pancreas generally secretes proteolytic enzymes (eg, trypsinogen, chymotrypsinogen) into the pancreatic duct, which empties into the duodenum to aid digestion of chyme in the lumen. However, in RYGB patients, the duodenum is located in the biliopancreatic limb. Accordingly, pancreatic trypsinogen is released from the pancreas and secreted into the duodenum within the biliopancreatic limb, not the alimentary limb. Educational objective: The cells in the stomach secrete the following: gastrin (from G cells) signals parietal cells to secrete HCl (from parietal cells); pepsinogen (from chief cells) cleaves polypeptides when activated by HCl; and mucus (from mucous cells) and bicarbonate (from epithelial cells) protect the stomach lining against autodigestion by gastric juice

q25

The pyloric sphincter is a muscular ring located at the junction of the stomach and the duodenum. This sphincter controls the flow of chyme, or partially digested food, from the stomach into the duodenum. In RYGB patients, ingested food passes through the alimentary limb. Nutrients flow from the small gastric pouch into the surgically attached jejunum, bypassing the pyloric sphincter in the biliopancreatic limb. Concept: At certain sections along the gastrointestinal tract, sphincters, or rings of muscle, divide the tract into segments with distinct functions. Ingested food passes from the esophagus into the stomach through the cardiac sphincter, and from the stomach into the duodenum through the pyloric sphincter.

q1

The q stem is asking for the cells important for the differentiation of OPCs which are the precursor of RANK. In the psg they say that osteoclast differentiation is related to the RANK/RANKL. So, RANKL is important for the differentiation. RANKL is the receptor and its expressed by osteoblasts which make bone aka they would put calcium from the bloodstream into the bone tissue concept: Bone remodeling is a continuous process in which osteoclasts (bone-resorbing cells) break down old bone and osteoblasts (bone-depositing cells) secrete new bone matrix. This process functions to maintain the strength and integrity of bone over time. Osteoblasts promote the precipitation of calcium and phosphate from the bloodstream and their transfer and incorporation into the bone matrix. In contrast, osteoclasts secrete acids that break down the mineral components of bone, releasing calcium and phosphate into the bloodstream.

Q19

The question is asking about the survival rate so we have to individually look at the relationships between the genotypes and diet with survival rate. In the passage it says that lac operon reduced virulence so that means that the WT is going to be more virulent aka less % survival compared to the lac+. Comparing the diets, in the passage it says that GR is more virulent than LR so for both WT and lac+, GR will have less % survival than LR. Answer choice B is correct because its showing the virulence in the correct order --> WT GR has the least % survival, then WT LR and then lac+ GR and then lac+ LR concept: Bacterial virulence can be induced or repressed by the surrounding conditions. The lac operon makes lac+ bacteria less virulent than the wild type. In the presence of lactose, both strains become less virulent because they either cannot survive (wild-type) or cannot infect cells (lac+).

Q22

The question is asking how can you stop the entry of virus into the cell. The image in the passage shows that the virus is entering thru receptor mediated endocytosis where an external ligand (the virus here) binds a receptor on the cell surface. This binding causes the plasma membrane to bud inward toward the cytosol before pinching off as a vesicle that contains both the ligand and its receptor. Thats why a drug that inhibits this inward budding would most likely prevent the virus from entering cells. Phagocytosis. = cell eating Pinocytosis = cell drinking concept: Cells take up their surrounding environment via endocytosis, which includes mechanisms of phagocytosis, pinocytosis, and receptor-mediated endocytosis. Many viruses use receptor-mediated endocytosis to enter cells. Enveloped viruses can also enter cells by fusing their membrane with the membrane of the cell.

Q23

The question is asking how does the receptor move across the plasma membrane Since this is a transmembrane protein, the receptor can move laterally through the cell membrane and thats why the ans is D The cell membrane is said to be fluid as its various nonphospholipid components are able to migrate laterally through the entire phospholipid-rich surface of the cell in any direction. These other important components include cholesterol, glycoproteins, and glycolipids (proteins and lipids that have been modified with carbohydrates). Content: Cell membranes are composed largely of phospholipids, which act as a fluid that allows other membrane components such as transmembrane proteins, glycoproteins, cholesterol, and glycolipids to migrate through this environment laterally. Accordingly, the structure of the animal cell membrane is known as the fluid mosaic model.

Q28

The question is asking how would you reduce the number of chromosomes. The two ways to do that is through: 1) end to end fusion of two chromosomes and inactivation of one of the centromeres - This fusion would initially generate a larger chromosome with two centromeres, and inactivation of one of these centromeres would produce a single new chromosome, reducing the chromosome number in the cell by one 2) breaking of a chromosome at the centromere and fusion of each chromosomal pathway to the ends of other chromosomes - This initial breakage would result in two individual chromosomal portions. Fusion of these portions to other chromosomes would cause the original chromosome to be lost, also reducing the overall chromosome number in the cell by one concept: Eukaryotic chromosomes have both protein-coding DNA and distinct noncoding regions. Noncoding regions include the centromere (facilitates spindle fiber attachment during cell division) and telomeres (protect chromosomal ends from degradation).

question 19

The question is asking what happens to LHA signaling at shorter time intervals (time intervals closer to 0). LHA is part of the gustatory salivary reflex arc and so stimulates salivation when theres more firing. At the 50 msec mark, LHA signaling, which increases the activity of the preganglionic fiber of the gustatory-salivary pathway, would be expected to increase the quantity of salivation due to this reflex pathway. Thats why B is the correct answer Concept: Input from higher areas in the central nervous system can modulate the activity of reflexes by either strengthening or weakening the magnitude of the response.

Question 8

The question is asking what will happen if ACTH is blocked. So the stress response system would be effected. The hypothalamic-pituitary-adrenal pathway controls the secretion of glucocorticoids from the adrenal cortex as follows: 1. The hypothalamus secretes corticotropin-releasing hormone (CRH) in response to low glucocorticoid levels and increased stress. 2. CRH acts on the anterior pituitary, which causes the release of adrenocorticotropic hormone (ACTH). 3. ACTH stimulates the adrenal cortex to synthesize and release cortisol. 4. Cortisol targets tissues (eg, muscle and liver) to increase the ability to cope with stressors. In addition, cortisol functions as a negative feedback signal by inhibiting the secretion of CRH and ACTH. Concept: Corticosteroids (glucocorticoids and mineralocorticoids) are steroid hormones released by the adrenal cortex. The release of cortisol from the adrenal cortex is mediated by the secretion of adrenocorticotropic hormone from the anterior pituitary.

Q17

The question is asking which lac gene would be useful when the bacteria is in a glucose rich media. When there is a lot of glucose, metabolism of lactose does not take place. So we had to look for the lac gene that would inhibit the interaction of lac with its operon to inhibit transcription. According to the passage, lacl is a lac repressor and so that would be useful in glucose rich media concept: The lacI gene codes for the lac repressor and prevents transcription of the lac genes. When sufficient glucose is available, the repressor helps conserve ATP by inhibiting expression of unnecessary genes. When glucose is depleted and lactose is available, lac genes are again expressed to make use of available energy sources.

question 23

The question is asking which one is not consistent Looking at the IP-1 (downstream signal of IP3) concentration from figure 1, you can see that IP-1 level is high only when hCaSR is present and not when its lacking. Thats why C is wrong bc its saying IP3 levels are high whether hCaSR is present or not and this contradicts the results from figure 1 concept: An agonist enhances or duplicates the effect of a ligand binding to a receptor. Agonists are able to bind a receptor and induce a biological response that is similar to that of normal ligand binding.

Q6

The question is basically asking what doesn't eukaryotic cells have. The answer is single stranded genomic material bc genome means the organisms DNA. I thought that was referring to RNA and thats why I got it wrong. Single stranded genomes can be found in certain viruses. Peroxisome - small organelles in eukaryotes that maintain the oxidation state of the cell by facilitating oxidation reduction rxns to remove harmful chemicals from the cell Nuclear pore proteins facilitate the passage of mRNA out of the nuclues and into the cyto concept: Eukaryotic cells can be distinguished by their unique characteristics. For example, they have a genome with multiple linear chromosomes (double-stranded DNA), they use meiosis as one mechanism of cell division (sexual reproduction), and they have membrane-bound organelles.

Q13

The question is saying that this bacteria is exposed to a spindle fiber toxin. Spindle fibers aid the separation of chromosomes in meiosis and mitosis and so if theres a toxin stopping the spindle fiber from doing its job, you will get nondisjunction (no separation) concept: Most eukaryotic cells (except germ cells) undergo cell division via mitosis. In contrast, prokaryotic cells duplicate via binary fission, a simple form of reproduction that does not involve the separation of chromosomes by spindle fibers.

q1

The question states that the ability of a resting sarcomere to contract is dependent on the existence of overlap between actin and myosin filaments and the presence of a gap between the ends of the actin filaments and the M-line. At the ideal sarcomere resting length, actin filaments must overlap with myosin filaments to allow maximum binding of myosin heads to actin on contraction Also, the distance between the ends of the actin filaments and the M-line should be maximized to provide the greatest distance for actin filaments to slide during contraction. Sarcomeres exhibiting these characteristics will be able to generate the greatest amount of contractile force. concept: Skeletal muscle fibers consist of thousands of individual functional units called sarcomeres, and each sarcomere is comprised of both thin actin and thick myosin filaments. On muscle contraction, myosin heads bind the actin filaments and slide them toward the center of the sarcomere.

q22

The research objective was to investigate how varying concentrations of anti-nAChR antibodies affected muscle tension, so the ACh concentration of infusions B and C should have remained unchanged. This would have allowed researchers to assess how muscle tension was affected by varying concentrations of anti-nAChR antibodies only. However, Table 1 shows that the ACh concentration of the infusions was inconsistent. Infusion B contained 7.5 mL of a 2.5 mg/mL ACh solution (ie, 18.75 mg of ACh). In contrast, Infusion C contained 3.75 mL of the same 2.5 mg/mL ACh solution (ie, 9.36 mg of ACh). Muscle samples in the electrolyte bath exposed to Infusion B generated more muscle tension than those exposed to Infusion C. As a result, muscle tension measurements were invalid as they were affected by varying concentrations of both ACh and anti-nAChR antibodies, instead of being affected by anti-nAChR antibodies alone. Educational objective: When the research goal is to assess how changes in a specific independent variable affect a dependent variable, researchers should ensure that any other variables that are not under study remain controlled or unchanged.

q10

The researchers used Caco-2 monolayers to carry out their experiments, and these Caco-2 cells share the same morphology (shape) and function as cells lining the small intestine. In addition, the question states that these cells are highly proliferative. Caco-2 cells are nongametic (somatic) cells; therefore, each parental Caco-2 cell would only be able to proliferate (divide and multiply in number) in the manner that a somatic cell can, by completing the M phase (mitosis) of the cell cycle to create two genetically identical daughter cells. concept: In multicellular organisms (eg, eukaryotes), nongametic (somatic) cells can divide and multiply via mitosis whereas gametic (reproductive) cells divide and multiply via meiosis. Biological homeostatic mechanisms function to maintain a balanced ratio of proliferating to dying cells in human tissues and organs. In contrast, cell division in single-celled organisms (eg, prokaryotes) occurs via binary fission.

question 4

The serum level of glucose is also subject to homeostatic control primarily through the actions of hormones. When blood glucose levels are low (eg, during periods of fasting), the hormone glucagon is secreted from the alpha cells of the pancreas, and the autonomic nervous system promotes the release of epinephrine and norepinephrine by the adrenal medulla. Glucagon is a peptide hormone that promotes gluceoneogenesis (the synthesis of glucose from other molecules) and glycogenolysis (the breakdown of glycogen into glucose). Epinephrine and norepinephrine also promote glycogenolysis. Therefore, the overall effect of glucagon, epinephrine, and norepinephrine is to promote the synthesis of glucose and its release into the systemic circulation, thereby correcting the low serum level of glucose. Concept: Glucose is a key energy source for many tissues in the human body. The serum level of glucose is tightly regulated through the actions of the hormones insulin, glucagon, epinephrine, norepinephrine, and the glucocorticoids.

q14

The sharp rise in the graph shows higher affinity for O2 I is wrong bc hb T state is low oxygen and hb R state is high oxygen II is right bc positive cooperativity increases hb affinity for oxygen III is right bc again positive cooperativity IV is wrong bc negative cooperativity doesn't take place concept: Hemoglobin's affinity for oxygen increases as successive oxygen molecules bind to individual subunits and change neighboring subunits from a tense (T) to a relaxed (R) state. Positive cooperativity between binding sites in proteins and enzymes is indicated by the sigmoidal (S) shape of kinetic graphs.

question 22

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. concept: The female reproductive tract consists of the ovaries, fallopian tubes, uterus, cervix, and vagina. Each of these tissues performs specialized functions that enable pregnancy and delivery.

question 3

This question is saying that the turtles experience a temperature dependent sex determination. The enzyme's expression level (and corresponding amounts of male or female sex hormones) determines whether an embryo develops as male or female. High expression of the enzyme in embryos increases the conversion of male to female sex hormones, increasing the presence of female hormones and inducing female development. In contrast, low expression of this enzyme decreases the conversion of male to female sex hormones, increasing the presence of male hormones and inducing male development. The graph shows that most embryos in eggs incubated at 26°C developed as male, signifying that enzyme levels would be low at 26°C because reduced expression of this enzyme promotes male development. In contrast, the graph shows that most embryos in eggs incubated at 30°C developed as female, meaning that enzyme levels would be high at 30°C because elevated enzyme levels promote female development Concept: In general, the expression of genes produces proteins, and the properties and activities of these proteins lead to the expression of certain traits (phenotypes) in organisms. However, in a gene-environment interaction, certain environmental conditions (eg, temperature, diet, oxygen level) can influence gene expression or protein activity and alter an organism's phenotype.

question 12

To complete one cycle, a cell starting in any phase must pass through all other phases and return to the phase in which it began. For example, a cell in G1 must pass through S, G2, and M, then return to G1 to complete the cycle. Looking at figure 2, at 0 hrs the cell is in S phase. At 8 hrs, the cell is again in S phase meaning it takes 8 hrs to complete a whole cycle concept: To complete one full cell cycle, a cell starting in any given phase must pass through all the other phases and return to the initial phase.

Q7

Transferrin in blood is 679 AA but is actually composed of 698 AA. So, we can assume that those 19 AA is the signal sequence that is cleaved in the secretory pathway. Signal sequence is a short AA sequence present at the N-terminus of immature proteins. This sequence directs proteins to the rough ER for entry into the secretory pathway before its cleaved from the protein. After entering the secretory pathway, the protein and in this case the transferrin will go through the whole pathway and then be secreted into the ECM aka the blood and thats why you would see the 679 form of transferrin in the plasma portion of the blood concept: Plasma is the liquid portion of blood, and is made up of water, electrolytes, gases, hormones, nutrients, metabolic waste, and proteins. Although derived from blood, lymph is not a component of blood.

Q9

Viruses are obligate intracellular parasites with DNA or RNA genomes that can be double-stranded or single-stranded. Specifically, some viruses with single-stranded RNA genomes can directlytranslate their genetic material into viral proteins upon infection of a host cell. The virus shown in the diagram possesses a genome that can be immediately translated into protein following infection of the Escherichia coli cell. Therefore, this virus must have a single-stranded RNA genome, which can be detected with northern blot analysis. concept: Northern blotting is a technique used to detect and measure the concentration of a specific RNA sequence in a cell or tissue sample.

q22

Water generally flows from low to high osmotic pressure via osmosis, increased osmotic pressure within interstitial fluid and blood plasma increases intestinal water absorption. In contrast, some epithelial cells secrete ions into the intestinal lumen, increasing osmotic pressure within the intestine to draw water in and lubricate luminal surfaces. In this scenario, Chloride (Cl−) secretion from intestinal epithelial cells would increase the osmotic pressure of intestinal contents, drawing more water into the intestine and increasing the water content of stool. Therefore, drugs inhibiting luminal secretion of Cl− would likely decrease water secretion into the intestine and make the stools less watery. concept: Most nutrient and water absorption occurs as food moves through the small intestine. The large intestine absorbs any remaining water and electrolytes prior to waste excretion.

question 7

You had to look at the results in table one. Choice A: ICSI is more effective than IVF because it has higher percentages of fertilization success and blastocyst development in both BCB+ and BCB− groups. Choice B: Oocytes (not blastocysts) were supplemented with mitochondria prior to ICSI, not IVF. The passage makes no mention of mitochondrial incorporation in combination with IVF treatment. However, this should have been done to ensure the experiment had improved validity when comparing fertilization methods. Choice C: The passage provides no information on the fate of blastocysts that continue development. Therefore, the data do not support any statement about the viability of blastocysts produced after fertilization. Choice D: (CORRECT) looking at the table mlCSl gave the highest percent in the blastocyst stage in the BCB- compared to IVF and ICSl concept: Following fertilization, the mammalian zygote undergoes cleavage (successive mitotic cell divisions) and forms the blastocyst, a hollow ball of cells that implants into the uterine wall.

q14

You have to know how ELISA works A specific primary antibody linked to a "reporter" enzyme is then added to bind the antigen, as stated in the passage. The samples are washed to remove unbound antibodies, and the substrate of the reporter enzyme is added. The enzyme reacts with its substrate to generate ("report") a colored product, and if a color change is detected, its intensity is proportional to the amount of bound protein. Protein expression levels are ultimately quantified by comparing the color change in the well plate to the color change observed from a series of known concentration standards. Identifying a color change (minimal, medium, or intense) proportional to protein concentration would be a normal and accurate finding that indicates the presence and quantity of the protein. concept: Enzyme-linked immunosorbent assay (ELISA) can detect and quantify proteins. Initially, a primary antibody (linked to a "reporter" enzyme) is added, which binds the antigen (protein). The samples are washed to remove unbound proteins, and the reporter enzyme substrate is added. The enzyme-substrate reaction creates a product that results in a quantifiable/detectable signal.

q21

concept: The right heart receives deoxygenated blood from the body and pumps it out toward the lungs. The left heart receives oxygenated blood from the lungs and pumps it out toward the rest of the body. Atria receive blood; ventricles release blood.

Q10

didn't understand the passage or the question properly The passage is saying that mitochondria was engulfed by an eukaryotic cell and only eukaryotic cells have telomeres on their chromosome. So the mitochondria would have their genes located on chromosomes that would not have telomeres concept: In the nucleoid region of prokaryotic cells, double-stranded DNA is condensed into a circular chromosome that has no telomeres or associated histones. By contrast, eukaryotic cells package their histone-wrapped, double-stranded DNA into linear chromosomes with ends capped by telomeres to prevent DNA from unraveling.

q6

had to know that macrophages would phagocytize organisms and basophils would release histamines Educational objective: Macrophages are one of the cells responsible for phagocytosis, the process of engulfing solid particles (eg, bacteria) designated for destruction into phagocytic vesicles called phagosomes. In contrast, basophils function to release chemical mediators such as histamine that stimulate an inflammatory response

question 13

insulin is a peptide hormone so it uses GPCR concept: Peptide hormones are amino-acid based, water-soluble molecules that travel freely through the bloodstream and act as first messengers by binding to an extracellular receptor on the target cell membrane, which leads to activation of intracellular second messengers.

q17

q says they are looking for OXIDATIVE enzyme on the inner mitochondrial membrane and they measure that by dark colored staining. So, cross section 1 would be non-oxidative bc it has very little staining and 2 would be oxidative bc it has a lot of staining. Hence, 1 would have less o2 storage proteins than 2 concept: Oxidative muscle fibers use aerobic respiration for ATP synthesis, requiring large amounts of oxygen supplied by high levels of myoglobin and extensive capillary networks. In contrast, nonoxidative fibers generate ATP through anaerobic glycolysis and require less oxygen than oxidative fibers.

q24

q stem says, NSAIDs nonionized in the acidic stomach and ionized in the less acidic small intestine. But RYGB creates a smaller stomach with a decreased internal surface area compared to the pre-surgery stomach. As a result, the small gastric pouch formed in RYGB contains significantly fewer parietal cells than a normal-sized stomach = less GCL = increase in pH in the lumen. Due to this increased pH (decreased acidity) of the gastric juice, NSAIDs ionize more readily and exhibit increased solubility in the gastric pouch. Greater amounts of soluble ionized drug can then be passively absorbed in the small intestine. concept: The acidic pH of the stomach is maintained by gastric juice, which is primarily composed of hydrochloric acid (HCl). This acidic environment is required for protein digestion and to kill harmful bacteria. When gastric juice mixed with food (chyme) enters the duodenum, it is neutralized by bicarbonate ions (from the pancreas) and bile (released from storage in the gallbladder).

question 17

the cilia in the fallopian tube guide the egg toward the uterus for implantation. But extrauterine implantations are caused by reduced fallopian cilia concept: After fertilization, fallopian cilia help propel the fertilized oocyte toward the uterus for implantation. An inadequate number of cilia in the fallopian tube can cause implantation of the fertilized egg outside the uterus.

question 24

the question is asking if neural crest cells would be involved in the pathology of myelomeningocele. In the passage they say that NTDs arise from abnormal development of the neural tube and myelomeningocele is spina bifida which is a form of NTD. Based on this, because the neural crest cells derive from the residual portions of the neural folds that do not contribute to neural tube formation, they are likely not involved in myelomeningocele pathology and continue to act normally as temporary migratory cells that give rise to a diverse lineage of cells. Concept: Cell migration in embryogenesis is the movement of cells into their final positions within the embryo. The migratory action of neural crest cells during neurulation (the formation of the nervous system) gives rise to many peripheral nervous system structures. In contrast, the central nervous system is derived from the neural tube.

Q15

the question is asking what will not be present during anaerobic conditions. Its ATP synthase bc thats used in ETC to make ATP during aerobic conditions concept: The Krebs cycle and the electron transport chain are only active in the presence of a final electron acceptor, such as oxygen (aerobic respiration) or inorganic ions (anaerobic respiration).

question 17

the question is asking what will you see a decrease in during period of stress aka during fight or flight response. Ans is D bc peristalsis is only increased during parasympathetic response concept: The parasympathetic and sympathetic divisions of the nervous system are broadly antagonistic in their effects. The parasympathetic division promotes digestion and other tasks related to an organism's long-term needs. Alternatively, the sympathetic division enables the organism to meet more immediate needs.

q1

the question is asking which one is not a function of B lymphocytes. Ans is C bc natural killer (innate) and cytotoxic T cells (adaptive) respond to antigens by releasing toxins that induce apoptosis (cell death) in nearby infected cells concept: Humoral immunity is an antibody-mediated immune response. During an infection, B lymphocytes bind foreign antigens and become activated. Activated lymphocytes divide into many identical cells, some of which differentiate into short-lived plasma cells that secrete antibodies during the immediate immune response. The remaining cells differentiate into long-lived memory cells, which can respond more rapidly to future infections.

question 22

the question is saying that there is a low serum calcium level so as a result calcitonin from the thyroid gland will be released to fix the calcium level concept: Calcitonin, a hormone synthesized by the thyroid gland, decreases calcium concentrations in the blood by inhibiting osteoclast activity (bone resorption) and promoting calcium excretion in the kidneys.

question 1

the question says that Na+ ions are not let in thru voltage gated channels meaning depolarization will not occur and thats why you have to pick the graph that shows that even when the threshold is reached the channels cannot open and the rising phase is not initiated = no AP concept: An action potential is fired in an all-or-none manner based on the cell's membrane potential. If the threshold is reached, voltage-gated ion channels open and the membrane rapidly depolarizes. The resting membrane potential is restored by the Na+/K+ pump.

q19

vitamins in the psg cause hemolysis according to the table vitamin B12 has the highest absorbance so it has the greatest rate of hemolysis. Hematocrit = measurement of RBC as a percentage of total blood volume. So higher the absorbance = higher the rate of hemolysis = greater reduction in hematocrit (RBC) levels. According to the table that would be vitamin B12 concept: Hematocrit, measured as a percentage of the total blood volume, indicates the number of red blood cells in a blood sample. Hemolysis induced by oxidative stress can reduce hematocrit.

question 6

white matter (myelinated) is involved in communication and gray matter is unmyelinated White matter consists of afferent and efferent axons: Afferent (ascending) axonal tracts carry sensory information from the body to the brain in the dorsal and lateral columns. Efferent (descending) axonal tracts carry motor commands from the brain to the body in the ventral and lateral columns. SAME DAVE - mnemonic concept: In the CNS, gray matter is composed of neuronal cell bodies and white matter is composed of axons that allow long-distance communication between neurons. In the white matter of the spinal cord, afferent axons carry sensory information to the brain and efferent axons carry motor commands to the body.

What is the difference between single cope DNA, repetitive DNA and highly repetitive DNA?

1. Single copy DNA - holds most of the humans genetic info; transcribed and translated and the mutation rate is low 2. Repetitive - found near centromeres, many contains genes that are transcribed and translated, higher rate of mutation than single copy DNA 3. highly repetitive - no genes and not transcribed or translated ex: telomeres (5'-GGTTAG-3')

Question 30

A. Intrinsic charges of the proteins are masked by SDS not by reducing agents B. Disulfide bonds are broken by reducing agents not by SDS C. Figure 1 shows the chromatin binding motifs of HMG proteins, not DNA. In agarose gel electrophoresis, ethidium bromide is used as a stain to make DNA bands (not proteins) fluoresce; this effect occurs as a result of ethidium bromide's ability to intercalate ("slip in") between DNA base pairs D. Only one thats correct Content: SDS-PAGE is used to separate proteins by molecular weight. During the procedure, SDS coats proteins with a negative charge. An electric current is then applied, and smaller proteins travel through a polyacrylamide gel and toward the positive anode faster than larger ones, creating lanes of size-separated protein bands.

Question 21

AD is caused by hyperphosphorylated proteins so a phosphatase to get rid of the phosphates would be a potential therapeutic agent Content -> Phosphorylation and dephosphorylation are post-translational modifications that alter the function or activity of proteins by changing their conformation. Kinases catalyze the transfer of phosphate groups from ATP or GTP to proteins, and phosphatases catalyze the removal of phosphate groups via hydrolysis.

Question 26 ** understand a little better**

ALSO MISSED THE FACT THAT IT SAID EXCEPT **** In these experiments, antibody and interferon levels were used as markers of immune response. Although antibody production is specific to the adaptive immune response, interferons represent a large class of signaling molecules that act in both the innate and adaptive immune systems. Because none of the experiments are specific to innate immunity, researchers would not be able to differentiate the influence of genotype on innate versus adaptive immune response. Educational objective: The innate immune system is a nonspecific system that responds within minutes to hours to foreign antigens. The adaptive immune system is based on learned recognition of specific antigens and can be subdivided into cell-mediated (T cells) and humoral (B cells) immunity. Antibodies are specific to humoral immunity whereas interferons play roles in both innate and adaptive immunity.

Question 16

According to Table 1, the mean age of ADPKD onset is ~58 for patients with PKD1 mutations and ~80 for patients with PKD2 mutations. This finding signifies that the observable characteristics of the disease do not emerge until after child-bearing age. Although ADPKD-causing dominant alleles are deleterious, they are able to remain in the gene pool and escape eradication by natural selection because they can be passed to an affected individual's offspring before the individual's reproductive capabilities are affected. concept: Deleterious dominant alleles can remain in the gene pool if the organism's fitness remains unaffected, even when the deleterious alleles reduce survival after the reproductive years. Deleterious recessive alleles evade elimination by natural selection through phenotypic masking in heterozygotes.

Question 39

According to the passage, c-Myc overexpression leads to early cancer development (ie, B-cell lymphoma) in Eμ-Mycmice. Therefore, c-Myc is most likely an oncogene, a mutated or overexpressed gene that induces uncontrolled cell growth by promoting cell cycle progression or inhibition of apoptosis. As a result, c-Myc levels would be higher in Eμ-Myc (Eμ) mice than in wild-type mice to promote cancerous cell growth. The passage states that p21 inhibits cell cycle progression, a key feature of a tumor suppressor gene. As a result, p21 levels would be lower in Eμ-Myc mice compared to wild-type mice because p21 was likely inactivated by loss of function mutations and lost the ability to prevent abnormal growth/division of cancerous cells. Content: In the setting of cancer, oncogenes are mutated or expressed at abnormally high levels and contribute to cancer development by promoting cell growth/proliferation or suppressing apoptosis. In contrast, tumor suppressor genes, which induce programmed cell death, are inhibited in cancerous cells.

Question 35

According to the passage, the SRY gene is a mammalian sex-determining gene. SRY is found on the Y chromosome, which is present only in males, and its expression induces the development of the male gonads (testes). When the SRY protein is produced, fetuses will develop as male, but in the absence of SRY, fetuses will develop as female. In this scenario, researchers expressed the SRY gene in female rabbit embryos. Despite being genetically female (XX), the presence and expression of the SRY protein would promote testis development, causing these embryos to display male sexual characteristics. Educational objective:Mammalian sex is determined by the expression of the SRY gene on the Y chromosome. In males (XY), SRY expression induces the development of male sexual characteristics. In females (XX), the lack of SRY expression leads to the development of female sexual characteristics.

Question 36

According to the passage, the i2 isoform of the TERT gene contains a portion of intron 2 and is believed to downregulate the active isoform. The passage also states that telomerase activity in somatic cells is associated with tumorigenesis, or the formation of cancerous tumors. Therefore, i2-mediated downregulation of telomerase activity in somatic cells would be expected to reduce the risk of cancer development Content: DNA is initially transcribed as precursor mRNA (pre-mRNA), consisting of introns and exons. The pre-mRNA can then be spliced into one of several possible isoforms, each containing multiple exons, and some containing portions of introns. Many inactive isoforms are believed to help regulate the activity of active isoforms

Question 39

Adaptive radiation is the diversifying characteristics in a subgroup of individuals from a single species so there's less competition for resources and the fitness is improved for the entire species A. In stabilizing selection, phenotypes are narrowed toward an average, homogeneous phenotype by selecting against extreme phenotypes; as a consequence, diversity is decreased within the population. B. Disruptive selection results in the selection of 2 extreme phenotypes that differ from the average. In disruptive selection, the average phenotype is selected against while 2 extreme phenotypes are selected for. Although the subgroup of cheetahs has larger claws and teeth, the average cheetah phenotype is not being selected against; therefore, this scenario does not reflect disruptive selection C. Speciation is the evolutionary process of forming a new species from a previously existing species. This process occurs over many generations and results in organisms that cannot interbreed with the ancestral species. Concept: Adaptive radiation is the process of diversifying characteristics (eg, claw and teeth size) to better fill an ecological niche. Adaptive radiation can eventually lead to speciation if the subgroup continues to diverge and loses the ability to interbreed with individuals from the original species.

Question 36

All cells in our body have the same genome but only particular ones are expressed for particular thing. For ex. Cardiac cells only express cardiac cells - CALLED DIFFERENTIAL GENE EXPRESSION. Educational objective:Heterochromatin is tightly packed DNA that is inaccessible to transcriptional machinery, preventing gene expression. Euchromatin is loosely packed DNA that is accessible to transcriptional machinery, allowing gene expression.

Question 28

Ask landon Educational objective:In a monohybrid cross, members of the P generation are homozygous for different alleles at one locus. The first cross produces heterozygous offspring in the F1 generation. The second cross produces a mix of genotypes and phenotypes observed in members of the F2 generation.

Question 26

Asking about the translation process 1. Initiation: The small 40S ribosomal subunit binds the 5′ cap, an initiator transfer RNA (tRNA) is recruited to the start codon, and the large 60S subunit binds the initiator tRNA. 2. Elongation: The ribosome continues to elongate the polypeptide chain by reading each mRNA codon in a 5′ to 3′ direction. 3. Termination: A stop codon is read, and release factors induce translation complex dissociation

Q2

Bacilli is talking about the shape which is morphology Bacilli are rod-shaped bacteria. Cocci are spherical bacteria. Spirilli are spiral bacteria. Concept: Bacteria can be classified by morphology, or shape. In morphological classification, three basic shapes are used to classify bacteria: rod-shaped (bacilli), spherical-shaped (cocci), and spiral (spirilli).

Question 34

Because the ovaries and ovarian follicles facilitate oogenesis, which is required for fertilization and pregnancy, femaleswith underdeveloped ovaries or reduced follicle count may have fertility issues. The passage states that Sp1 KO,XY rabbits develop as female but have smaller ovaries and fewer follicles than WT,XX female rabbits, indicating that the second X chromosome may contribute to fertility by promoting normal ovarian and follicular development. Educational objective:The ovaries and ovarian follicles play a role in reproduction by facilitating the maturation and release of female gametes

Question 9

DNA grows from 5' to 3'. The answer is explaining how that happens. The 3' OH of the growing strand attacks the 5' phosphate of the incoming nucleotide. Choices A and B are wrong bc the question states after the 85 min incubation time meaning that CldU is not being added to the solution anymore so only the nucleotides would incorporate into the growing strand. When you are joining two nucleotides and forming covalent phosphodiester bond, its endergonic and cleaving it would be exergonic. Content -> During replication, DNA polymerase joins uncoupled deoxyribonucleoside triphosphates to the new DNA strand. As a result, the exergonic release of a pyrophosphate molecule leads to the formation of a covalent phosphodiester bond between the last nucleotide of the growing strand and the incoming nucleotide

Question 2

DNA mismatch repair takes out the wrong nucleotide so that DNA polymerase can incorporate the right nucleotide A-T G-C Content -> base pair mismatches happen when DNA polymerase puts in the wrong nucleotide in the new strand

question 33

Educational objective: The Hardy-Weinberg equations can be used to relate allele frequencies to genotype frequencies. Allele frequencies are used to calculate the expected genotype frequencies of populations in Hardy-Weinberg equilibrium

Question 48

For a cell that expresses a gene of interest, one way to determine the cell type is to identify the presence of another mRNA that is known to be specific to the cell type of interest. The passage gives the names of three genes that are specific to oligodendrocytes: plp1a, Sox10, and mbp. If cldnk mRNA is found in the same cells as the mRNA of any of those genes, then cldnk must be expressed in oligodendrocytes. Therefore, if a probe that is complementary to cldnk mRNA hybridizes to cells that also express plp1A, then cldnk is expressed in oligodendrocytes Content: In genetic studies, hybridization is the annealing of two complementary nucleic acids. The presence of a particular mRNA within a specific cell type can be assessed by hybridization to a complementary probe.

Question 17

Forgot the structure of guanine Content -> Deoxyribonucleotides are composed of a phosphate, a deoxyribose sugar, and a nitrogenous base (A, T, C, or G). Cytosine and thymine are pyrimidines (single-ringed bases) whereas guanine and adenine are purines (double-ringed bases). In double-stranded DNA, cytosine always aligns with guanine (and vice versa), and adenine always aligns with thymine (and vice versa) via complementary base pairing

Question 37

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) According to the passage, cheetahs experienced a catastrophic event that drastically reduced their population size; this is known as a bottleneck event. Bottlenecks (due to environmental events or human action) greatly reduce the genetic diversity of a population. Consequently, the smaller population of cheetahs has a reduced ability to buffer the negativeimpacts of random changes in allele frequencies (genetic drift) that may result in extinction All the other choices are beneficial A. Natural selection is when beneficial alleles are selected for so they survive rather than go extinct B. Gene flow changes in allele frequency due to migration which would increase genetic diversity by introducing new alleles to the gene pool C. Random mating increases a population's genetic diversity bc of the inherent variations that arise during meiosis (recombination and sex reproduction (fusion of gametes) Concept: Genetic drift is a mechanism of evolution; however, unlike natural selection, the variations in allele frequencies occur randomly by chance. Because smaller populations have a smaller gene pool, the random (good or bad) variations due to genetic drift cannot be buffered.

question 32

Genetic drift is the fluctuation of allele frequencies within a population due to chance events. This can lead to random loss of alleles within a population. In general, low-frequency alleles have an increased probability of being eliminated by random events than do high-frequency alleles. Because alleles on the Y chromosome exist at the lowest frequencies, they are most susceptible to loss by random chance alone. And in the passage there are only 5 Y chromosomes Concept: Genetic drift refers to random genetic changes in allele frequency that are due to chance events (not natural selection). Low-frequency alleles have an increased probability of being lost by genetic drift compared to those present at higher frequencies.

Question 15

Histone acetylation is when the DNA is silenced and so when you deacetylase you make the DNA more accessible. But the question is saying deacetyl inhibitor so the DNA will not be silenced Content -> Heterochromatin, or tightly packed chromatin, is composed of deacetylated histones (due to histone deacetylase activity) and is transcriptionally repressed. In contrast, euchromatin or relaxed chromatin is highly acetylated (due to histone acetylase activity) and transcriptionally active.

Question 33

I knew it had to do with RNA polymerase because we are taking about transcription factors here. In the passage they say that when WT1 is present in cells, TERT is not and thats a healthy cell but when c-Myc is present in cells so is TERT and thats a cancerous cell. That means WT1 inhibits the gene promoter region for TERT while c-Myc facilitates it. Also, if TERT is cancerous its fair to say that c-myc would facilitate it bc c-myc is an oncogene Content: Transcription factors can upregulate or downregulate transcription by influencing the ability of RNA polymerase to bind a promoter. Transcription factors that increase transcription are called activators and facilitate RNA polymerase binding whereas those that decrease transcription are called repressors and inhibit binding

Question 7

I looked at figure 3 and in the graph the new sites of replication was high for the mutation that were treated with Aph. Since there are more replication sites, the following enzymes would be more active in them: 1. Topoisomerase - introduces negative supercoiling in the DNA double helix ahead of the replication fork to reduce the strain produced by unwinding, which causes positive supercoiling. 2. DNA helicase - unwinds the DNA double helix and separates the parent strands at the origin of replication. 3. Single-strand DNA-binding protein binds to each strand to prevent spontaneous reannealing of unwound single-stranded DNA. 4. Primase - synthesizes RNA primers and positions them at the beginning of each DNA strand. Only one primer is needed for leading strand synthesis, but lagging strand synthesis requires many RNA primers. 5. DNA polymerase synthesizes daughter strands in a 5′ → 3′ direction only. One daughter strand is synthesized continuously toward the replication fork (leading strand); the other strand is synthesized discontinuously in a direction away from the replication fork (lagging strand), with more and more segments added as the replication fork progresses. This process results in the formation of Okazaki fragments, short stretches of newly synthesized DNA separated by RNA primers. 6. DNA ligase joins Okazaki fragments. Content -> DNA replication occurs in the 5′ → 3′ direction; DNA helicase first unwinds the double helix and separates the parent strands. Single-strand DNA-binding proteins prevent spontaneous reannealing of unwound single-stranded DNA. Leading strand synthesis is continuous, whereas synthesis of the lagging strand (composed of RNA primers and Okazaki fragments) is discontinuous.

Question 29

I was correct because the table showed euchromatin and heterochromatin which is structural chromatin organization and III was correct bc in the passage they stated that the two proteins compete for the same regions Content: Chromatin compaction can influence accessibility of regulatory factors to DNA. The known H1 function is to stabilize chromatin compaction by securing nucleosome packaging. However, competing proteins would have the opposite effect, which would be to destabilize chromatin compactness and enhance accessibility to DNA

Question 8

In eukaryotes, genetic recombination occurs via crossover events (exchange of DNA segments between homologous chromosomes). Synapsis or the joining of homologous chromosomes into tetrads occurs during prophase I of meiosis and is required for crossing over to occur. Crossovers increase genetic diversity by mixing maternal and paternal alleles into a single chromosome that is then inherited by the offspring.

Q1

In the passage it said that E. coli is a nonvirulent bacterial strain and the question says that it contains F factor plasmid. The only only process that transfers gene based on plasmids is conjugation. Transformation is the cellular uptake of foreign DNA from the environment and this is enhanced with increased cell membrane permeability that occurs bc of physical manipulation Transduction involves DNA transfer from one bacterial cell to another by a bacteriophage. Subsequent infection of other cells with these new bacteriophages results in the transfer of bacterial DNA into a new host. Think about Type III secretion Images of these in your notes concept: Conjugation is the transfer of genetic information from one bacterial cell to another via direct contact facilitated by the sex pilus. The sex pilus is encoded by genes contained in the F factor plasmid.

Question 6

In the passage they say that in BS there is excess somatic recombination which leads to somatic instability. So, that means the cells that don't have the BLM will show increase somatic recombination aka instability and lower replication recovery. The choice is basically just saying that there is a difference between recovery and instability in the two types of cells which is true. The ans that I picked is wrong bc the cell lacking the BLM would be more sensitive and damaged then the one that has BLM Content -> DNA synthesis can be arrested by replication inhibitors. Cells lacking the genes necessary to promote replication stability are more sensitive (damaged or affected) to the replication blockade induced by these agents, and recovery of replication may be impaired.

question 27

In the passage, in vitro studies of F1 spleen cells suggest that both wild-type and mutant TLRs are observable by protein electrophoresis. This co-expression of the normally functioning receptor and the mutated receptor can be attributed to the codominance of tlr4(+) and tlr4(d) alleles. F1 cells [tlr4(+)/tlr4(d)] display intermediate production of anti-SRC antibodies and interferon because half of their TLR receptors are functional while the other half are mutant. As a result, the quantitative data for their immune response were lower than He-N mice [tlr4(+)/tlr4(+)] mice but higher than He-J mice [tlr4(d)/tlr4(d)]. Educational objective:Codominance refers to the co-expression of two alleles at the same locus, where both allelic products are observable in the phenotype of heterozygous individuals.

question 29

In this example, p is the frequency of the normal-functioning Tlr4 allele and q is the frequency of the mutated Tlr4 allele. The frequency of mice homozygous for the mutated allele (q2) is 1,600/10,000, or 0.16. Therefore, q is the square root of 0.16, or 0.40, and this value can be used to calculate p: p = 1 − 0.40 = 0.60 The p and q allele frequencies can then be used to calculate the frequency of the heterozygous genotype and the total number of heterozygous mice in the population of 10,000: 2pq = 2(0.60)(0.40) = 0.48 (0.48)(10,000) = 4,800 mice Educational objective:Hardy-Weinberg equations can be used to relate allele frequencies and genotype frequencies. Genotype frequencies are equal to the probability of inheriting each genotype, with the frequency of homozygotes equal to p2 or q2 and the frequency of heterozygotes equal to 2pq.

Question 19

It was based on the passage figure content -> The p-value is the probability of observing a result due to chance alone, assuming that the null hypothesis is true. A value of p ≤ 0.05 is generally considered statistically significant whereas p > 0.05 is not considered statistically significant.

Q3

Looking at the western blot, proteins with the shortest amino acid sequences have lower molecular weights and migrate quickly through the gel. Band intensity (thickness) demonstrates the relative levels of expression of the target proteins. In the question, tir moved the least and so it has the lowest molecular weight and EspB was at the top so it has the higher molecular weight concept: Western blot involves protein separation by gel electrophoresis, transfer onto a blotting membrane, and detection with protein-specific antibodies. Proteins with low molecular weight migrate farthest, and protein concentration is generally proportional to band intensity.

Question 13

Loss of exon 7 is the SMA pathology and B is the only one that shows that Content -> The spliceosome removes introns from pre-mRNA by locating specific sequences within introns called 5′ splice donor sites and 3′ acceptor sites. Splice donor sites are located in the 5′ end of the intron next to the exon, and splice acceptor sites are found at the 3′ end of the intron adjacent to the exon

Question 1

Meiosis II then proceeds similarly to mitosis but differs in that each cell begins with a haploid, rather than a diploid, genome. During anaphase II of meiosis, sister chromatids are separated to opposite poles of the cell. Assuming no recombination, the identical wild-type Ena alleles on one set of sister chromatids will be pulled to opposite poles by the mitotic spindle during anaphase II. Anaphase I is wrong bc sister chromatids are copies of each other, they have identical Ena alleles. Therefore, in anaphase I in an Enaheterozygote, wild-type Ena alleles remain together on one set of sister chromatids, as do mutant alleles on the other set. Educational Objective: Gamete formation requires cellular division by meiosis I and meiosis II, in which homologous chromosomes and sister chromatids, respectively, are separated. In the absence of recombination, maternal and paternal alleles are separated from each other during anaphase I, and identical alleles on each sister chromatid are separated from each other during anaphase II.

Question 38

MicroRNAs are small noncoding eukaryotic or viral RNA molecules that bind complementary sequences on target messenger RNA molecules, inhibiting their expression. So, the miR17-92 gene cluster products would also be miRNAs that would interfere with the expression of specific genes by binding target transcripts containing complementary nucleotide sequences Content: MicroRNAs (miRNAs), an example of noncoding RNA, silence gene expression at the translational level. miRNAs bind complementary sequences on target messenger RNA (mRNA) molecules, consequently inhibiting expression of the target mRNA by either blocking its translation or marking it for degradation.

Question 25

Paragraph 4 states that eIF4E facilitates ribosome binding to the m7Gpp cap of mRNA, the first step of cap-dependent translation. If elF4E is absent, a transcript requiring cap-dependent translation cannot be made into a protein. This depletion of eIF4E protein induced by siRNA leads to noticeably reduced expression of elongation factor eEF2K because the eIF4E protein is mostly absent and unable to help the ribosome bind to the m7Gpp cap of eEF2K mRNA. Therefore, eEF2K mRNA is translated in a cap-dependent manner. content -> Small interfering RNA (siRNA) molecules are short, double-stranded RNA sequences that decrease the translation of target proteins. siRNAs contain complementary sequences that bind to the mRNA of the target protein and signal for its degradation.

question 31

Question says brown = dominant and white = recessive so A. brown and white ration would not be equal in F1 C. F1 would not majority express recessive trait D. F1 would produce more of the dominant than the recessive Concept: Genetic linkage leads to an unequal combination of alleles (haplotypes) in haploid cells (gametes) formed by meiosis. If genes are linked, a greater number of nonrecombinant cells (original allele pairs) will be found than recombinant cells (altered allele pairs), resulting in a greater number of nonrecombinant offspring, or offspring that retain the original linked alleles of the parents. Genetic linkage refers to the tendency of alleles in close proximity to remain on the same chromosome and be inherited together by offspring. This tendency occurs because of fewer crossover events between these loci during meiosis, resulting in a greater number of haploid gametes with nonrecombinant genotypes.

Question 4

Recombinant is basically the product of crossing over. New combinations of alleles are called recombinant, whereas combinations that already existed in the parent are called parental. The recombinant progeny can be identified by the fact that there are fewer of them. Educational objective:Homologous chromosomes can exchange genetic material by recombining. Genes that are located close together on a chromosome have a lower probability of being separated by recombination than those that are far apart. Fewer progeny from a cross will have recombinant genotypes than will have parental genotypes.

Question 6

Researchers hypothesized that the function of PTP in axonal guidance is to inactivate Abl. If this is correct, then when PTP is inactivated by a mutation, Abl will be overly active, leading to abnormal axonal guidance. Therefore, Abl must be suppressed to some extent for normal axonal guidance to occur, and its suppression in the presence of inactive PTP would help restore normal axonal guidance. In effect, suppression of Abl activity carries out the same function that active PTPdoes and therefore should lead to the same outcome. Educational objective:Upstream gene products in a biological pathway act on downstream gene products. Failure of an upstream product to act can be compensated for by alterations to downstream targets that mimic the effect of the upstream product.

Question 5

Since the sequence homology was basically the same for all the peptides, we can assume that all of them can repair the arrested replication forks. Look at the choices for hint that you need to compare Ori Ans is B bc the E. coli is a bacteria aka prokaryote which means it has a circular DNA with a single Ori Yeast is considered eukaryotic so eliminates A, C and D bc it can do it in multiple Ori Content -> Prokaryotes have circular DNA with one Ori in the cytoplasm but eukaryotes have linear DNA with multiple Ori in the nucleus. And Ori expands to form a replication bubble which contains 2 replication forks that move part in opposite directions during DNA synthesis

question 30

Survival curves plot the fraction or number of surviving members as a function of an independent variable, usually time after diagnosis (for humans) or drug dosage. In animal-based safety studies, survival curves are often used to assess lethal dose, which is the minimum dose causing death in 50% of group members. On survival curves, this data point can be observed as the dose (x-value) corresponding to 50% survival (y-value). In this case, survival is a function of LPS dose. Experiment 3 states that compared to the LD50 for He-N mice, 10-fold and 100-fold increases in LD50 were found for F1 and He-J mice, respectively. This result means that the LD50 for He-J mice was 10 times greater than the LD50 for F1 mice. The only survival curve that corresponds to this ratio is the one that shows that 50% of F1 mice died at a dose of 1 mg (LD50 = 1 mg) and 50% of He-J mice died at a dose of 10 mg (LD50 = 10 mg). Educational objective:Survival curves plot percent survival against an independent variable, usually time or dosage. Survival curves can be used to determine LD50, defined as the minimum dose causing death in 50% of test subjects.

Question 12

Talking about the post transcriptional modification. I is talking about removing introns, II is talking about 3' poly A tail thats added (5' methyl cap is also added). Removal of the cap from the mature mRNA would degrade the mRNA when it moves to the cytosol from the nucleus. DNA polymerase catalyzes the synthesis of DNA replication and is not involved in transcription Content -> Transcription is the process of synthesizing RNA from template DNA and begins with RNA polymerase II binding to the gene promoter region. RNA polymerase II reads the DNA strand in a 3′ to 5′ direction to generate a 5′ to 3′ pre-mRNA molecule. The pre-mRNA transcript undergoes 5′ capping, the addition of a 3′ poly-A tail, and excision of noncoding regions (introns) to be converted into mature mRNA

Question 1

The anticodon base pairs with different mRNA codons that code for the same AA. The redundancy is explained by the wobble hypothesis which says that the 1st two nucleotides on the mRNA require traditional base pairing but the 3rd one does not go strict base pairing. So, the codon degeneracy comes from the nontraditional base pairing at the 3rd position of the codon and anticodon Content takeaway -> genetic code is degenerate bc 1 codon codes for the same AA. This is bc the 3rd position of the mRNA codon and tRNA anticodon can undergo nontraditional base pairing, allowing a single tRNA molecule to bind different codons

question 7

The given scenario describes the genetic transmission of an X-linked recessive disorder characterized by muscle degeneration. The mother (XMXm) is a carrier, meaning she has the disease-causing allele but does not express the disorder, whereas the father (XMY) is normal (unaffected by the disorder). Genetically, the woman has one copy of the wild-type dominant allele on one X chromosome (XM) and one copy of the recessive allele on the other X chromosome (Xm). Women always pass one of their two X chromosomes to offspring; therefore, this woman would always have a 50% chance of passing the recessive allele to each child, regardless of gender. In contrast, the man would always pass on a normal X chromosome to female offspring. Accordingly, if they were to have children, their daughters would never be affected and would be either normal (XMXM) or carriers (XMXm). Their male offspring, however, would always inherit the Y chromosome from their father and inherit either the normal allele (and be unaffected) or the recessive allele (and express the disorder) from their mother. Given the parental genotypes, the overall probability (P) that a child will express an X-linked recessive trait can be calculated as follows: P(child inherits affected X-chromosome from mother) × P(child is male) = 0.5 × 0.5 = 0.25

Question 32

The main info that I should have focused on was that t80 is the average amount of time req to recover 80% of the pre-bleach fluorescence. The graph would have to show that HMGB1-GFP + HMGA1 has a greater t80 value (ie, takes longer) than HMGB1-GFP alone. Accordingly, the fourth graph (Choice D) shows that in the presence of HMGA1, it takes longer for HMGB1 to move back into the photobleached region and bind chromatin because HMGA1 is already binding to the same sites. Content: Green fluorescent protein (GFP) is frequently used to track the mobility of proteins in a cell. The protein of interest is tagged with GFP, and its movement, localization, and expression can be studied in cells based on the fluorescence emitted.

Question 22

The passage says that the normal undigested protein is 161 bp and also says that it was treated with a restriction enzyme. Thats why you see the two bands at the bottom, not bc it was run under reducing condition. Educational objective: Polyacrylamide gel electrophoresis (PAGE) can be used to separate DNA molecules by size. A native PAGE gel contains no denaturants or reducing agents, and allows double-stranded DNA molecules to traverse the gel in their native (unaltered) state

Question 3

The passage states that males in the first cross of Experiment 1 were heterozygous for Ena, suggesting that they had two different copies of the gene. Therefore, Ena cannot be on the X chromosome and must be autosomal (Number I). In Experiment 2, F2 males and females have the same ratios of normal-shaped eyes to elongated eyes. Based on the cross used to generate the F2 generation, if Abl were sex-linked, all females would have elongated eyes (dominant) and all males would have normal eyes (recessive). Because this is not the case, Abl must be autosomal (Number II). Because PTP is on the same chromosome as Abl, it must also be autosomal (Number III). Educational objective:Genetic traits may be autosomal or sex-linked (X chromosome). Because males have only one X chromosome, sex-linked recessive traits tend to be present at greater rates in males than in females, whereas autosomal traits are present in the same proportions in both.

Question 46

The passage states that the cldnk gene was digested by EcoRI and XhoI. Table 1 shows that EcoRI cuts between the G and the A in GAATTC, leaving an AATT overhang, and XhoI cuts between the C and the T of CTCGAG, leaving a TCGA overhang. Therefore, the digested gene should have an AATT overhang on one end and a TCGA overhang on the other, as shown in Choice B (Note: Because the Xhol overhang is shown as the bottom strand, it is read from right to left [5′ to 3′]). Content: Restriction enzymes cleave DNA molecules at specific sites, known as restriction sites. Each enzyme has a unique restriction site and may cut the DNA such that an overhang, or sticky end, is left on the end of the strand

Question 34

The question is asking that since TERT is expressed only at low levels, where would they be found. So, they're talking about heterochromatin which is D Content: Chromatin can be broadly classified as heterochromatin or euchromatin. Heterochromatin consists of DNA tightly coiled around histones and is not readily transcribed by RNA polymerase. Euchromatin is more loosely associated with histones and is more easily transcribed.

Question 28

The question is asking what is histone complex rich in for HMGN2 bind to. The passage also states that HMGN2 contains a nucleosomal binding domain which means that it interacts with a nucleosome (DNA wrapped around histone) and a nucleosome core has a positive net charge so the correct ans would be the positive amino acids aka arg and lys Content: The nucleosome is a structural subunit composed of DNA wrapped twice around a histone octamer, which contains two molecules of H2A, H2B, H3, and H4. H1 is located outside the octamer core and serves as the "linker" protein that secures the DNA wrapped around the nucleosome. Histone proteins are rich in positively charged arginine and lysine, which facilitate binding to negatively charged DNA

Question 23

The question is asking which one does not explain table 1. In both male and female offspring, mitochondrial DNA (and any associated mitochondrial disease) is inherited only from the mother, meaning mitochondrial genes do not follow Mendelian inheritance patterns, which occur when organisms inherit two copies of each gene, one from the father and one from the mother. Due to heteroplasmy, the phenotype of a given tissue, organ, or whole organism, is influenced by the relative amounts of wild-type and mutant mtDNA, not by Mendelian patterns of dominance or recessivity. Educational objective:Variable expressivity and incomplete penetrance are common features of mitochondrial diseases. Penetrance is defined as the proportion of individuals with a specific genotype who express the corresponding phenotype. Variable expressivity refers to the range of phenotypic outcomes shown by individuals who carry the same genotype

Question 10

The question is asking which will be used while making the cDNA vector in the PSNF5 cell. cDNA is formed from mRNA using reverse transcriptase. And then the mRNA strand is degraded and resulting cDNA sequence is amplified using DNA polymerase in PCR. To create the cloning vector in the PSNF5 cells, the cDNA sequence is inserted into a cloning vector which is cut by restriction enzymes and then DNA ligase joins the cDNA vector which results in the formation of a recombinant vector that is inserted into cells like PSNF5 cells Content -> In cDNA cloning, reverse transcriptase generates a single strand of cDNA from a target mRNA sequence. DNA polymerase synthesizes the second complementary DNA strand and amplifies the target cDNA sequence. The target cDNA can then be inserted into a plasmid vector via the actions of a restriction enzyme (cuts both plasmid and vector) and DNA ligase (joins the cDNA to the vector)

Question 41

The question is indirectly talking about positive controls. In Experiment 2, verifying that each mutant construct works properly and expresses only the desired families serves as a positive experimental control. In Figure 2, the independent variable (treatment) is the type of vector transduced into Eμ-Myc cells with miR-17∼92 knocked out (Δ/Δ), and the dependent variable is the percentage of apoptotic cells. The purpose of measuring apoptosis under these conditions was to determine precisely which miRNA families contribute the most to miR-17~92 function. To reliably draw such conclusions, it was necessary to first ensure that deletion of one or more miRNAs encoded by the miR-17∼92 cluster did not affect expression/processing of the remaining miRNAs encoded by the mutant constructs. Content: Positive controls are often employed to ascertain test validity and allow researchers to correctly interpret scientific data. This kind of experimental setup can verify that the dependent variable to be studied is present or that the treatment (independent variable) meant to manipulate the dependent variable functions properly.

Question 25

The question is talking about PCR. You run PCR on cDNA stupid. Choice C is literally explaining cDNA. Educational objective:Polymerase chain reaction (PCR) is a thermal cycling technique used to amplify DNA fragments. PCR reagents include a source DNA template, GC-rich primer pairs, a thermostable DNA polymerase, and a buffer solution with positively charged ions.

Question 15

The question mentioned advanced ADPKD. The whole passage is about this disease and its a kidney disease. So, you had to look for which function is NOT a kidney function and that would be leukocyte production. A. DUMP the HUNK B. Blood pH - When extracellular fluid (eg, blood plasma, interstitial fluid) becomes excessively alkaline (high pH), the kidneys excrete HCO3− and retain H+, causing blood pH to decrease. Conversely, when the extracellular fluid becomes excessively acidic (low pH), the kidneys excrete H+ and retain HCO3−, causing blood pH to rise. Via this process, the kidneys are able to maintain blood pH within the narrow physiologically required range C. erythropoietin - The adult kidneys normally produce erythropoietin, a hormone that signals the bone marrow to increase red blood cell (erythrocyte) production Concept: The kidneys' primary function is to maintain the salt and water balance of the blood. They also play a key role in regulating multiple aspects of physiological homeostasis (eg, blood pressure, waste removal, osmolarity, blood pH, erythrocyte production).

Question 37

The question states that some organisms have multiple TERT genes with high sequence identity. These similar genes most likely arose by gene duplication. A is not correct because it produces multiples proteins products from the same gene but not multiple similar gene. Conjugation is the exchange of genetic info between prokaryotes in the form of plasmas DNA. Transformation is when prokaryotes pick up foreign genetic material from their surroundings Content: Genes with high sequence identity are evolutionarily related, having a common origin. They generally arise by gene duplication and, over time, may mutate and fulfill distinct roles within an organism

Question 11

The question stem says truncated. Truncated proteins result from nonsense and frameshift mutations with a downstream stop codon in the new reading frame. Insertion of A makes the mRNA codon UGA so thats why thats the right answer A. produces a silent or a missense mutation B. deletion of the G at the 5h position changes the reading frame but doesn't generate a stop codon C. this doesn't generate a nonsense or a frameshift Concept: Various types of DNA mutations can alter the protein product of a gene. Truncated proteins result from nonsense mutations and frameshift mutations with a downstream stop codon (UAA, UAG, UGA) in the new reading frame

Question 31

The question was indirectly asking about telomeres and the only answer choice that made sense was B Content: Centromeres join two sister chromatids and are essential for proper chromosome division during mitosis. Telomeres are regions at the chromosome ends that are repeatedly truncated each time a cell divides. Both centromeres and telomeres are composed of heterochromatin, a transcriptionally inactive and tightly condensed complex of DNA wrapped around histones.

Question 35

The sequence was given in the passage so had to find the complementary sequence Content: New nucleotide strands are synthesized using complementary strands as templates, with the two strands aligned antiparallel to each other (ie, the 5′ end of one strand aligns with the 3′ end of the other). In DNA, guanine (G) always pairs with cytosine (C), and adenine (A) always pairs with thymine (T). In RNA, the only difference is that uracil (U) is present instead of thymine

question 12

The woman's mother is unaffected, and therefore has the genotype aabb. However, the woman's father has ADPKD as he is heterozygous for a dominant PKD1 mutation, so he has the genotype Aabb. For PDK2, each parent contributes a wild-type b allele to the woman, giving her the genotype bb. A heterozygous parent has a 50% chance of transmitting the mutation to his or her offspring. Therefore, for PDK1 the father (Aa) will either transmit the mutant A allele with a 0.5 probability or the wild-type a allele with a 0.5 probability. In contrast, the mother (aa) can only pass on the wild-type aallele to the woman. This woman's genotype will either be aabb (normal) or Aabb (ADPKD). .5x.5 = .25 or 25% Concept: In autosomal dominant inheritance, transmission of only one copy of a dominant allele is necessary to produce the phenotype. A heterozygous parent has a 50% chance of transmitting the mutation to their offspring.

Question 44

You just look for the start codon and the stop codon and look which one is the longest Content: An open reading frame (ORF) is a set of codons within a strand of mRNA that can be translated by a ribosome. An ORF begins with a start codon (AUG) and ends with a stop codon (UAA, UAG, or UGA). The start and stop codons must be in the same reading frame.

Question 24

You would turn the mRNA into cDNA and then measure the expression using PCR at varying time points content -> Researchers can assess the half-lifes of mRNA isoforms by converting mRNA to cDNA via RT-PCR at varying time points, allowing for comparison of isoform concentrations. The isoform with greater cDNA concentration at the final time point has the longer half-life

Question 51

cDNA is derived from mature mRNA through reverse transcription. Because the mature mRNA has already been spliced (ie, introns were removed), the cDNA will not contain any introns. The cDNA can be cloned into expression vectors such as plasmids, which can be introduced into embryonic cells and integrated into the genome. These cloned genes can then be expressed in the organism. However, because the cloned gene does not contain any introns, its expression will not include splicing Content: RNA polymerase transcribes both exons and introns to form pre-mRNA. During RNA processing, introns are removed by splicing to yield mature mRNA. cDNA is generated from mature mRNA and does not contain introns, so it is not spliced during expression.

Question 10

convergent bc the species are distantly related and they form similar patterns

Question 40

figure 1 is showing the MHC heterozygosity (diversity) where the captive cheetahs are increasing so thats why B Is wrong. So the decreased fitness is explained by environmental stress bc chronic stress negatively impacts immunologic responses, increases the susceptibility for diseases and affect fecundity. Figure 2 also shows that captive cheetahs have higher cortisol level = higher stress so they are more likely to have decreased fitness concept: MHC are cell surface receptors in vertebrates that present foreign molecules in order to elicit an immune response. MHC variations are evolutionarily important to an organism's fitness because they influence the organism's ability to combat a variety of infections and diseases. Cortisol, a hormone that negatively impacts immunocompetence, increases in response to stress.

Question 18

its natural selection bc of the gradual change. If it was population bottleneck, there would be a sudden dip in the graph Concept: In natural selection, beneficial traits that improve fitness are more likely to be passed to subsequent generations than less favorable traits. Beneficial traits should become more common with each generation, allowing the species to adapt to its environment.

Question 47

subtract 32 from 2958 and then add 915 Content: Plasmids—small, circular DNA molecules that can carry a small number of genes—are often used in laboratories to amplify and express genes of interest. A gene may be inserted into a plasmid by digesting the gene and the plasmid with the same restriction enzymes, followed by ligation.

question 4

telomeres are noncoding DNA sequences that protect the rest of the chromosomes from degradation by degrading themselves When telomerase extends telomeres, it does so by adding the sequence 5′-TTAGGG-3′ to the end of one of the strands on the chromosome multiple times in succession Side note: RNA (not DNA) can self-complement (complementary base pair w/ ITSELF ex tRNA - (picture attached)

question 9

the more genetically similar you are to the ancestor, the more likely that you diverged recently from the ancestor Concept: The molecular clock model uses the theory that most genetic mutations are neutral and occur at a fairly constant rate across organisms to estimate the amount of time elapsed since species diverged from their common ancestor

question 17

the number of mutation increases over time as you diverge more and more from your ancestor Concept: Most genetic mutations are neutral and accumulate at a fairly constant rate across organisms over evolutionary time. The overall number of mutations in a species will therefore increase linearly across time. By analyzing the rate of neutral mutations in the genome, this "molecular clock" of evolution can be used to measure evolutionary time and estimate the evolutionary relationships between species.

Question 41

the passage mentioned that they were inbreeding and inbreeding causes decrease in fitness concept: Inbreeding depression is characterized by decreased genetic diversity, reduced fecundity, and decreased fitness. As genetic diversity decreases within a population, the frequency of homozygous deleterious recessive genes increases within that population. Consequently, inbred populations have an increased probability of becoming extinct.