Biology
Q24
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. That's why the answer is A (Choice B) Provided that a sample is representative of the population in question (ie, that the sample is randomly selected), a large sample size increases the validity of experimental results. An experiment in which data (eg, peritoneal fluid VEGF concentration) is collected multiple times from a single individual may yield skewed and invalid results if that subject is not representative of the relevant population. 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.
Q10
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. Educational objective: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.
Q4
According to the passage, deficiency in the LuxS enzyme leads to decreased AI-3 expression. AI-3 generally acts on neighboring bacteria to induce expression of LEE genes. These genes encode the type III secretion proteins necessary to introduce EHEC-secreted toxins into the host cells to cause disease. Because LuxS− mutants lack the LuxS enzyme and have insufficient AI-3 levels, these mutants lack type III secretion proteins, as shown in Figure 2. However, Figure 2 also shows that growing LuxS− cells in PCM resulted in the expression of type III secretion proteins. Based on Figure 1, which shows that LEE genes can be expressed via a LuxS-independent method, the banding patterns in Figure 2 suggest that compounds (eg, epinephrine, norepinephrine) released by epithelial cells into the PCM activated the transcription of LEE genes despite the deficiency of AI-3. Educational objective: Inducers are chemical substances (signals) that regulate gene expression. Specifically, secreted inducers can influence gene expression in other cells.
Q17
According to the passage, scientists proposed that a prolonged symbiosis period could induce gene transfer from eukaryotes to prokaryotes. In the given example, P. leiognathi (symbiont) was believed to have acquired the gene coding for Cu/Zn SOD from its eukaryotic host (ponyfish) after an extended period of symbiosis. 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. Educational objective: The endosymbiotic theory explains how primitive eukaryotic anaerobes engulfed ancient aerobic prokaryotes, and consequently acquired the ability to produce energy through oxidative phosphorylation.
Q7
According to the passage, symptoms of congential adrenal hyperplasia (CAH) include hypoglycemia (low serum glucose). In healthy individuals, low serum glucose can be expected to increase the release of glucagon, epinephrine, and norepinephrine because these hormones would act to restore serum level of glucose to the normal range (Numbers II and III). epi - increases glucose just like glucagon norepi - inhibits insulin Educational objective: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.
Q13
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 × 24 = 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
Q18
According to the question, SOD mRNA in P. leiognathi encodes a signal sequence that directs the transport of the SOD protein to the extracellular environment. P. leiognathi is stated to be a bacterium (prokaryotes) from the passage and prokaryotes do not have membrane-bound organelles Educational objective: 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.
Q24
All you had to realize is that norepi and epi are sympathetic NS hormones so a drug that increases their secretion mimics the activity of sympathetic neurons. Also know that 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. Educational objective: 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.
Q2
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. Educational objective: 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.
Q10
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.
Binary Fission
Asexual reproduction in prokaryotes. Results in two identical daughter cells and does not involve spindle fibers.
Q19
Ask Landon
Q16
Ask Landon for easy method so when doing dihybrid crosses, if they say: "and" --> multiply (decreasing/less chance) "or" --> add (increasing/more chance) can do two monohybrid crosses and just realize that when they ask you "or" it gets trickier and you have to find the ans choice that is the closest to yours!!
Q16
Based on the passage, mitochondria originated as aerobic prokaryotes that were engulfed by primitive anaerobic eukaryotes. The prokaryotic organisms (mitochondria) then began living inside the larger eukaryotic host, establishing a mutually beneficial symbiotic relationship. This endosymbiotic theory of eukaryotic evolution is widely accepted because, like bacteria, mitochondria have their own genome (ie, mitochondrial DNA) and possess an independent system for transcription and translation distinct from that of the nuclear eukaryotic genome. Prior to gene transfer, mitochondrial genes were prokaryotic in structure and located in the nucleoid region of the cell. According to the passage, there was a partial transfer of mitochondrial genes to the eukaryotic nucleus. These transferred mitochondrial genes became identical in structure to nuclear genes, but the residual mitochondrial genes remained in prokaryotic form. This characteristic means that, like bacteria, mitochondrial genetic information is still contained within a circular chromosome of double-stranded DNA that has no telomeres and is not associated with histone proteins. In contrast, eukaryotic genes are located on double-stranded linear chromosomes capped by telomeres, which are sequences of repetitive nucleotides located at each end of a chromosome. These regions shorten at each round of replication, protect the DNA from deterioration, and prevent the tip of chromosomes from fusing with adjacent chromosomes. (Choice A) Introns are found in eukaryotic DNA, not prokaryotic DNA. Introns are DNA regions that do not code for proteins but are removed by the spliceosome during formation of a mature messenger RNA molecule. (Choice C) Prokaryotic and eukaryotic genomes are double-stranded, not single-stranded. (Choice D) Histone proteins are restricted to the nucleus of eukaryotic cells and associate with DNA to facilitate its organization and packaging into chromosomes. Educational objective: 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.
Q18
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 [Ca10(PO4)6(OH)2], 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. Educational objective: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.
Q11
Catecholamines are also hormones released during the sympathetic NS so increased dilation of airways (bronchioles) enables increased respiratory function and oxygen delivery to tissues. Educational objective: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.
conjugation
Conjugation is the transfer of genetic information from one bacterial cell to another via direct contact. The donor cell contains the F (fertility) factor plasmid, a circular piece of DNA containing genes that direct the formation of the sex pilus. During conjugation, the sex pilus from the donor cell attaches to a recipient cell (one that does not contain F factor) and facilitates the transfer of a single strand of the F factor plasmid DNA to the recipient cell. The recipient then synthesizes a complementary strand and becomes capable of passing genetic information to another bacterium. The F factor plasmid is typically found outside the bacterium's genome, but it can integrate into the bacterial chromosome. When integration occurs, bacterial genes can be transferred with the F factor.
Q30
Did not know fungus and protozoa are eukaryotic Q is asking if pathogen lacks phospholipid bilayer, what is the pathogen? It would be a virus bc bacteria, fungus, and protozoa all have phospholipid bilayers cuz of PM. Unlike prokaryotic and eukaryotic cells, viruses are obligate intracellular parasites that cannot reproduce outside of a host cell. The genetic material of viruses can be RNA or DNA that is double-stranded or single-stranded and is surrounded by a protective protein coat known as the capsid. In addition, viruses can be either enveloped or nonenveloped. 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, nonenveloped (naked) viruses lack a phospholipid bilayer. (more resistant to heat and denaturing) In this scenario, the pathogen identified from the patient's blood sample was found to lack a phospholipid bilayer. Because all prokaryotic (eg, bacteria) and eukaryotic (eg, protozoa, fungi) cells are enclosed by a phospholipid bilayer (ie, cell membrane), the only possibility is that the pathogen lacking this bilayer is a virus (nonenveloped); therefore, it should be treated with antiviral drugs. Educational objective: 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.
Q28
Did not understand what the Q was asking but basically its asking what would you not find in eukaryotes. The clue here is that genome means an organism's DNA. Single-stranded genomes can be found in certain viruses. However, eukaryotic cells do not possess this kind of genomic material and instead have double-stranded genomes. Peroxisomes are small organelles that help maintain the oxidation state of the cell by facilitating oxidation-reduction reactions to remove peroxides (chemicals harmful to the cell) (Choice A). The nucleus is the storage site for DNA (ie, linear chromosomes) and where replication and transcription (but not translation) occur. Because translation takes place in the cytoplasm, the mRNA molecule resulting from transcription must exit the nucleus. Consequently, nuclear membrane channels known as nuclear pores facilitate the passage of mRNA out of the nucleus and into the cytoplasm (Choice B). Nuclear pores, which also regulate the passage of certain proteins in and out of the nucleus, allow the compartmentalization of molecular processes. Educational objective: 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.
Q23
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.
Q9
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.
phospholipid catalyzed movements
Flippase brings a phospholipid from the outer leaflet to the inner leaflet, and floppase brings a phospholipid from the inner leaflet to the outer leaflet. Both flippase and floppase require ATP. Scramblase brings a phospholipid from the outer leaflet to the inner leaflet AND a phospholipid from the inner leaflet to the outer leaflet. Scramblase does not require ATP.
Q12
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.
Q6
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).
Q6
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.
Q19
Had to look at the graph in figure 3 and realize that as AMG 416 dosage increases, PTH levels decrease. And then in the passage it states that PTH functions to increase plasma Ca2+ conc and reduce phosphate levels. The Q states that AMG 416 suppresses PTH release. So had to look for the graph that showed increase in phosphate levels from low AMG to high Educational objective: 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.
Q7
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.
Q14
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.
Q8
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.
Q15
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).
Q15
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.
Q13:
Incorrect the tRNA has a anticodon to the mRNA codon and the 3rd nucleotide is the one that changes to allow for "Degeneracy" which is where more than codon can code for the same amino acid. This redundancy of the genetic code can be explained in part by the wobble hypothesis. This hypothesis states that the first two nucleotides on the mRNA codon require traditional (Watson-Crick) base pairing with their complementary nucleotides on the tRNA anticodon, but the third nucleotide on the codon may undergo less stringent ("wobble") base pairing in a non-Watson-Crick manner. Codon degeneracy is therefore explained by nontraditional basepairing at the third position of the codon and anticodon.
Q6
Just had to find the same sequence in the gel. They will state complementary if they wanted us to find complementary. Concept: 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.
Q16
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.
Q10
Microtubules (composed of alternating alpha and beta tubulin subunits) 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.
Q31
Northern blotting is a biomolecular assay used to detect and measure the concentration of specific RNA molecules within a cell or tissue sample as follows: RNA is isolated from other cellular components and denatured. 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 directly translate 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. All other choices had to do with DNA btw (Choice A) Southern blotting is a technique used to detect and measure the concentration of DNA in a cell. Accordingly, a southern blot could detect the presence of a viral DNA genome but not an RNA genome. (Choice B) Researchers use DNA sequencing to determine the specific nucleotide order of a DNA molecule. This technique could be used to analyze viral DNA but not viral RNA genomes. (Choice D) Gene cloning, a technique used to generate many copies of a previously identified gene of interest, is not useful for the detection of specific DNA or RNA sequences. Educational objective: Northern blotting is a technique used to detect and measure the concentration of a specific RNA sequence in a cell or tissue sample.
Q19
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.
Q35
Passage states that AD is caused by hyperphosphorylated tau proteins so a therapeutic strategy would be phosphatase upregulation to concept: 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.
Q2
Passage states that EHEC is a bacterium and bacteria are classified by morphology or shape Educational objective: 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).
Q32
Passage states that SMA mice produce insufficient FL-SMN and exhibit progressive tail necrosis, loss of tail length and helps determine SMA severity. WT mice have normal levels of FL-SMN transcripts and show no tail necrosis. ASOs bind to pre-mRNA and help increase tail length when added to SMA mice so means exon 7 is not being spliced out so processing is the correct answer. concept: Processing (which includes splicing) of the pre-mRNA transcript into mature mRNA occurs in the nucleus.
Q18
Passage states that protein HMGN2 contains a nucleosomal binding domain which means that it most likely interacts with a nucleosome. A nucleosome is DNA double helix wrapped twice around a histone octamer (eight-protein complex). H1, H2A, H2B, H3, and H4 are the five major histone types, but the DNA-wrapped octamer core is composed of eight subunits (ie, two molecules each of H2A, H2B, H3, and H4). These histone proteins have a net positive charge at physiological pH because they are rich in arginine and lysine (positively charged, basic amino acids). This net positive charge facilitates histone binding to negatively charged DNA. Concept: 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.
Q29
Q basically asking what is only found in eukaryotes? And thats intron splicing. In addition, eukaryotic cells have a unique mechanism for processing messenger RNA (mRNA). Eukaryotic mRNA is initially transcribed as pre-mRNA, containing both introns (noncoding sequences) and exons (coding sequences). Following transcription, the spliceosome (an RNA and protein complex) removes introns from the pre-mRNA molecule to form mature mRNA in a process known as RNA splicing. This mature mRNA is transported out of the nucleus and into the cytoplasm, where it is translated into protein. Protozoa (eg, P. falciparum) are single-celled eukaryotic organisms, and therefore possess all of the unique characteristics of eukaryotic cells, including a mechanism for splicing introns from mRNA in the nucleus. Educational objective: 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.
Q37
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.
Q8
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. **signal sequences are a part of the AA and are typically mentioned for the ER or for the nucleus (but called nuclear localization signal), helps localize proteins to those areas*** 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.
Q46
Q is asking what enzymes are present when making a vector that will be inserted into PSNF5 cells? 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 the 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. PSNF5 cells were transfected with a vector containing BLM complementary DNA (cDNA) whereas PSNG13 cells were transfected with an empty vector (without BLM cDNA). Only PSNF5 cells were created via cDNA cloning, which involves the isolation of a single, self-replicating organism that contains a cDNA of interest. In this technique, target messenger RNA (mRNA) sequences are mixed with complementary primers known as oligo(dT) primers, which are composed of thymine nucleotides. These primers bind the mRNA 3′ poly-A tail, and the enzyme reverse transcriptase uses dNTPs to generate a single strand of cDNA from the mRNA (Choice A). The mRNA strand is degraded and the resulting cDNA sequence (eg, BLM cDNA) is amplified using DNA polymerase in PCR (Choice C). The cDNA sequence containing the gene of interest can then be inserted into a cloning vector, a circular DNA molecule found in bacteria and viruses. The foreign cDNA and vector are cut by a restriction enzyme (Choice D), generating complementary sticky ends that anneal when both molecules are mixed together. DNA ligase joins the cDNA to the vector, resulting in the formation of a recombinant vector that can be transfected (inserted) into cells, as was done with BLM cDNA in PSNF5 cells. (Choice B) RNA polymerases synthesize various types of RNAs but are not used as standard reagents in cDNA cloning because the target mRNA is already present in the cell (ie, does not need to be synthesized). concept: 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).
Q11
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.
Q29
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.
Q27
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.
Q21
Q is asking what will NOT be present during anaerobic conditions in eukaryotes. the answer is ATP synthase cuz ETC is aerobic. The enzyme ATP synthase harnesses the potential energy of the ETC proton gradient to generate ATP from ADP and inorganic phosphate. It is stated in the question that P. leiognathi can utilize the ETC in anaerobic environments because it can use compounds other than O2 as the final electron acceptor. Although ATP synthase does not directly need O2 to generate ATP, the formation of the proton gradient by which it functions is oxygen-dependent in ponyfish cells but not in P. leiognathi. Consequently, under anaerobic conditions, ATP synthase is not active in ponyfish cells but is active in P. leiognathi. Educational objective: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).
Q34
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.
Q32
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.
Q1
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.
Q36
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.
Q13
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.
Q25
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.
Q3
Q states in WT cells, so only had to look at the WT lane dummy, also even if you didnt know the size direction of amino acids you could've figured it out through the "greatest concentration" which is also what they are asking. Western blot, smaller MW molecules travel further Educational objective: 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.
Q11
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.
Q26
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.
Q19
Q states that there is a "spindle fiber toxin" so that will obviously inhibit chromosomes properly splitting apart during nuclear division. Also Q asks abt ponyfish cells which are eukaryotic and P.leiognathi are prokaryotes and don't use microtubules for division bc they divide by binary fission. Educational objective: 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.
Q41
Seems more like a stand-alone Q. The answer choices give hints. The choices give you different numbers of origins of replication and that should clue you into the fact that yeast and human are eukaryotic and have multiple origins of replication whereas bacteria have one origin of replication. concept: Prokaryotes typically have circular DNA with a single origin of replication in the cytoplasm whereas eukaryotes have linear DNA with multiple origins of replication in the nucleus. An origin of replication expands to form a replication bubble, which contains two replication forks that move apart in opposite directions during DNA synthesis.
Q26
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. (Choice D) A constitutively active ACTH receptor would mimic the effects of high serum ACTH levels, stimulating excessive glucocorticoid secretion by the adrenal cortex. This would result in adrenal cortex hyperactivity, not adrenal insufficiency (adrenal cortex hypoactivity). ACTH administration, which promotes adrenal cortex function, would not be used as a treatment for these patients. Educational objective: 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. Peptide hormones are synthesized in the RER
Q5
The cell membrane is a phospholipid bilayer that separates the interior of the cell (the intracellular space) from the external environment (the extracellular space). The two phospholipid layers are oriented such that the hydrophobic tails of the phospholipids in the inner and outer layers are in contact with each other, leaving the hydrophilic head groups in contact with either the intracellular or the extracellular fluid environment. The cell membrane is described by the fluid mosaic model as a nonrigid (fluid) phospholipid bilayer through which diverse proteins are scattered (the mosaic) and are able to migrate laterally. 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. Educational objective: 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.
Q9
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 Educational objective: 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 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.
Q9
The hormones testosterone, estrogen, and progesterone are referred to as sex hormones because of their role in sexual development and reproduction. Although testosterone can be detected in the serum of females, it is typically considered a male sex hormone because of its role in the development of male traits and behaviors. 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. (Choice D) Glucocorticoids (eg, cortisol) do not play a role in the sexual development of males or females, and therefore cannot explain the genital abnormalities seen in patients with 21-hydroxylase deficiency. Educational objective:Testosterone is a sex hormone that facilitates the development of male sex organs and secondary sexual characteristics.
Q1
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. adrenal cortex - secretes glucocorticoids (cortisol) and mineralcorticoids (aldosterone) and androgen (testosterone) and estrogen Adrenal medulla - secretes norepi, epi, dopamine Educational objective: 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.
Q26
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). Retroviruses enter host cells through endocytosis, which allows the capsid and envelope to uncoat (disassemble). Uncoating releases the viral genetic material inside the host cell. 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. Educational objective: Retroviruses are enveloped, positive-sense, single-stranded RNA viruses that convert their RNA genomes into double-stranded DNA using the enzyme reverse transcriptase. During their lysogenic cycles, retroviruses enter the nucleus and integrate their reversed transcribed DNA with the host genome.
Q20
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 (Number III). (Number I) Increased intestinal absorption of calcium will raise blood calcium levels, which will serve as a negative feedback signal to end PTH secretion. As a result, increased intestinal calcium absorption would not cause the increased PTH release characteristic of secondary hyperparathyroidism. (Number II) Phosphate normally binds to and forms an insoluble complex with calcium in the blood. If phosphate cannot be properly reabsorbed in the kidney, it will be excreted in urine, and free calcium ions in the blood will increase. However, decreased circulating phosphate effectively increases the concentration of detectable free (ionized) calcium, indirectly suppressing the release of PTH, which is not a hallmark of secondary hyperparathyroidism. Educational objective: 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.
Q23
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.
Q20
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.
Q21
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 (also mentioned in passage). And then you just had to find something that would occur in the GPCR pathway. As a result, the following processes would occur due to hCaSR activation by AMG 416: 1. The G-protein heterotrimer (alpha, beta, and gamma subunits) attaches to the intracellular portion of the GPCR, a transmembrane cell surface receptor. When the ligand binds to the extracellular domain of the receptor, the GDP molecule initially bound to the G-protein alpha subunit is released, and a GTP molecule binds the G-protein in its place. 2. The GTP-bound alpha subunit dissociates from the beta and gamma subunits, and activates the membrane-bound enzyme phospholipase C (PLC). 3. PLC hydrolyzes the membrane phospholipid phosphatidylinositol bisphosphate (PIP2) into inositol trisphosphate (IP3) and diacylglycerol (DAG). 4. Water-soluble IP3 diffuses from the cell membrane through the cytoplasm to bind IP3 receptors located on the endoplasmic reticulum and mitochondria. Binding of the IP3 receptor opens calcium channels, allowing the release of stored calcium into the cytoplasm, which activates protein kinase C (PKC). 5. DAG diffuses within the membrane and also activates PKC. 6. PKC phosphorylates downstream intracellular proteins to produce its physiologic effects. Educational objective: 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.
Q35
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+).
Q6
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.
Q7
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.
Q22
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).
Q19
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 there's 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. That's why B is the correct answer. In this scenario, researchers are trying to determine whether the lateral hypothalamic area (LHA) of the brain modulates the gustatory-salivary reflex response and, if it does, whether that modulation results in a diminished or enhanced reflex. First, the researchers measured baseline activity in the preganglionic fiber of the gustatory-salivary parasympathetic reflex pathway, which was represented by the firing rate of the preganglionic fiber in response to tongue stimulation alone. The researchers then measured the activity of the fiber in response to sequential stimulation of the LHA and the tongue such that the LHA was stimulated at various time intervals before the tongue. According to the data, the preganglionic fiber responded to this sequential stimulation by exhibiting greater activity at shorter time intervals. This effect was most pronounced when the LHA was stimulated shortly prior to the tongue (<100 msec). The effect weakened as more time elapsed between LHA stimulation and tongue stimulation, and the firing rate of the fiber returned to baseline at a time difference of approximately 200 msec. Therefore, properly timed 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. 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.
Q33
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.
Q1
The unequal concentration of charged ions between the extracellular and intracellular fluid of all living cells results in an electrochemical gradient across the membrane that determines the membrane potential (voltage difference). The resting membrane potential of neurons is primarily due to the high concentration of potassium ions (K+) and the low concentration of sodium ions (Na+) inside the cell as compared to the outside. The following mechanisms contribute to maintenance of the resting membrane potential: 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+ (Numbers I and III). 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 (Number II). Educational objective: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.
Q13
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.
Q22
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.
kinesin vs dynein
both are motor proteins on microtubules kinesin - anterograde transport (away from cell body) dynein - retrograde transport; (-ve) end-directed --> nucleus (towards cell body)
Q22
calcitonin is secreted by the thyroid glands. and it acts to reduce plasma calcium by decreasing osteoclast activity (bone resorption) and increasing renal excretion of calcium. According to the passage, 5/6 Nx mice are animal models of secondary hyperparathyroidism in ESRD, and a hallmark of this disease is elevated PTH release due to insufficient circulating calcium. However, if increased serum calcium levels were sustained in these mice via regular CaCl2 infusions, this physiological condition would trigger increased calcitonin release by the thyroid. Therefore, as the site of calcitonin synthesis, the thyroid is most likely where a calcitonin-specific radiolabeled antibody would localize and bind. Educational objective: 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.
Q17
did not know that leptin is secreted from white adipose tissue and induces feelings of satiety and ghrelin is secreted from the gastric cells Educational objective: 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.
Q12
forgot that endocrine and exocrine mean diff things. We are studying the endocrine system first of all. endocrine - secrete hormones into the bloodstream to cause an effect in a different part of the body. exocrine - secrete substances (eg, saliva, sweat, enzymes) through a duct and onto an epithelial surface The pancreas has endocrine, paracrine, and exocrine functions. The functional units of the endocrine pancreas are the islets of Langerhans, which are made up of alpha, beta, and delta cells that secrete different hormones into the blood. - 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. Given that T1DM is caused by a failure in insulin production, affected patients have impaired endocrine function of pancreatic beta cells (Number II). Educational objective: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.
Q8
know function of aldosterone - increases blood volume and increases sodium reabsorption Educational objective: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.
Q16
miR-26a overexpression is hypothesized to help rescue obesity-induced insulin resistance, so which of the following does not support their hypothesis? Knew that you first had to compare against the same independent variables, you cannot compare against two diff ind variables like HFD vs SFD and Alb Tg mice and WT mice. 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 overexpression in Alb Tg mice does not improve obesity-induced insulin resistance. If Alb Tg mice fed a HFD had similar levels of serum insulin and glucose as WT mice fed a SHD, this would mean that three-fold overexpression of miR-26a in Alb Tg mice was able to decrease insulin and glucose levels to the same levels as the control mice on a normal diet. As a result, Alb Tg mice fed a HFD would show improved insulin resistance (higher insulin sensitivity) (Choices A and B). Glucose tolerance refers to the speed at which glucose is cleared from the blood. Thus, if WT mice fed a HFD had lower glucose tolerance (clearance) than Alb Tg mice fed a HFD, this would support the claim that miR-26a overexpression is protective against insulin resistance (Choice D). Educational objective: 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.
Q5
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. Educational objective: 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.
Q17
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.
Q10
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.
Q9
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.
Q6
Adrenal hyperplasia is excessive secretion of hormones from the adrenal gland. The pathway to activate the adrenal gland stays the same so it goes from hypothalamus --> pituitary gland --> adrenal glands Educational objective: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.
Q3
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. Educational objective: 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 hyperpolarization of the membrane potential and inhibits action potential initiation in the post-synaptic neuron.
Q7
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.
Q18
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.
Q11
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.
Q37 COME BACK TOO!!!
Concept: Eukaryotic translation initiation begins when ribosomes bind the m7Gpp cap at the 5′ end of the mRNA sequence; however, cap-independent processes do exist.
Q9
Educational objective: Incomplete dominance between alleles results in the expression of phenotypes in heterozygous offspring that are intermediate to the phenotypes of homozygous parents.
Q2
Educational objective: 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.
Q14
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). Educational objective: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.
Q14
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.
Q27
High cortisol levels means decrease in endogenous cortisol through negative feedback inhibition. Educational objective: Negative feedback is the process by which the product of a metabolic or signaling pathway inhibits one or more of the steps that promote the synthesis or secretion of that product.
Q14
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.
Q15
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
Prokaryotic vs 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
Q20
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.
Q3
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.
Q5
Q states that glucocorticoid deficiency acts on a variety of different cells so thats why i picked my answer that receptors for glucocorticoid hormones are expressed in diverse tissue types Educational objective: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.
Q27
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.
Sense strand vs antisense strand
Sense strand - coding strand antisense strand - template strand
Q2
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).
Q8
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.
Q4
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.
Q21
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.
Q23
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.
Q2
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. If a diploid cell with 46 chromosomes undergoes meiosis, four haploid daughter cells should be produced with each containing 23 chromosomes. However, in the given scenario two daughter cells have an extra chromosome and two daughter cells are missing a chromosome. Such a finding is most likely due to chromosomal nondisjunction, which occurs when either homologous chromosomes (during anaphase of meiosis I) or sister chromatids (during anaphase of meiosis II) fail to separate to opposite poles of the cell. Chromosomal nondisjunction during anaphase I may produce one daughter cell with an extra homologous chromosome (24 chromosomes) and one daughter cell missing a chromosome (22 chromosomes) (Choices A, B, and C). Division of these cells in meiosis II would yield two cells with 22 chromosomes and two cells with 24 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.
Q13
The solubility of a hormone in water affects how it is transported in the blood and how it exerts its physiological effect on target cells. Hormones are primarily divided into three different classes and their chemical structure determines their water solubility: Peptide hormones (eg, insulin, glucagon) are composed of amino acids linked by peptide bonds and are made in the rough endoplasmic reticulum (RER). Peptide hormones range from small to large depending on their number of (charged) amino acids. The overall charge makes peptide hormones water-soluble (hydrophilic), which means they can dissolve in the bloodstream but cannot cross the hydrophobic lipid bilayer of the plasma membrane. Accordingly, peptide hormones act as first messengers and must bind extracellular receptors to induce a signaling cascade via activation of intracellular second messengers (Choice A). Steroid hormones are lipid hormones derived from cholesterol and generated in the smooth ER. These hormones are fat-soluble (hydrophobic) and must therefore be bound to a carrier protein in the bloodstream. However, their hydrophobicity allows them to diffuse through the plasma membrane (lipid bilayer) and bind their receptor in the cytoplasm or nucleus. Steroid hormones are labeled first messengers as they perform the initial signaling that influences the nuclear transcription and cytoplasmic translation of physiologically required proteins. Tyrosine derivatives are derived from the amino acid tyrosine. These hormones function similarly to either steroid hormones (eg, thyroid hormones T3 and T4) or peptide hormones (eg, catecholamines epinephrine and norepinephrine). (Choices B and D) Steroid hormones, not peptide hormones, are able to dissociate from their carrier proteins in the blood and diffuse through the plasma membrane. Educational objective: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.
Q28
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.
Transduction
Transduction involves DNA transfer from one bacterial cell to another by a bacteriophage (a virus that infects bacteria). During assembly of bacteriophages inside an infected cell, bacterial DNA can become trapped within the capsid of newly created bacteriophages. Subsequent infection of other cells with these new bacteriophages results in the transfer of bacterial DNA into a new host.
Transfection
Transfection is the process by which genetic material, usually in the form of a plasmid, is introduced into eukaryotic cells
Transformation
Transformation is the cellular uptake of foreign DNA from the environment. This process is enhanced with increased cell membrane permeability (competence) that occurs in response to physical manipulation.
Q24
Viral load is measured by the total quantity of viral genetic material inside an organism. Viral genomes consist of either DNA or RNA that is double-stranded or single-stranded. Once inside a host, viruses rely on cellular (host) and/or viral polymerases to replicate their genomes and produce mRNAs that can be used for viral protein synthesis. DNA viruses such as HBV have double-stranded genomes that are similar to the host cell's genome; as a result, this type of virus can directly utilize host enzymes. Their viral genes are transcribed into positive-sense, single-stranded mRNA (+ssRNA) in the nucleus by the eukaryotic DNA-dependent RNA polymerase, also known as Pol II. This polymerase produces positive-sense mRNA that codes for viral proteins, which are subsequently detected by the immune system. In this scenario, inhibition of Pol II in mice infected with HBV would prevent the virus from utilizing the host enzymes, therefore stopping production of viral mRNAs and proteins. Consequently, there is a decrease in viral load when mice are treated with the antiviral drugs. Because the numbers of cells presenting viral particles follow patterns similar to the amount of viral load in the blood, there will also be a decrease in HBV fragments displayed by antigen-presenting cells in HBV-infected mice (Numbers I and III). (Number II) +ssRNA viruses such as HEV replicate their genomes in the cytoplasm of a host cell using their own viral RNA-dependent RNA polymerase, which is translated from the original viral genome upon infection. The administered compounds inhibit only the host's DNA-dependent RNA polymerase (Pol II) activity. The viral polymerase translated from the HEV genome will continue to replicate its viral genome by creating a complementary negative-sense mRNA intermediate. The final +ssRNA viruses produced by the viral RNA polymerase are then translated into viral proteins and assembled into new virions; therefore, the HEV viral load would not decrease as a consequence of Pol II inhibition. Educational objective: Most double-stranded DNA viruses utilize host machinery and resources to transcribe positive-sense mRNA from viral DNA. Positive-sense ssRNA viruses (except retroviruses) replicate their genomes by encoding their own viral RNA-dependent RNA polymerase.
Q4
Was debating bw B and D but should've known that adrenal cortex secretes aldosterone (increases BP) and adrenal medulla secretes norepi and epi which also increase BP (Sympathetic NS also secretes them) Educational objective: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.
Q18
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.
Q17
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.
Q25
hormones are secreted by the adrenal medulla like epi help increase glucose conc in blood so you can't have glycogen synthesis (decreasing glucose levels) Educational objective: 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.
peptide hormones
hormones composed of short chains of amino acids
Q15
knew the GPCR pathway and that if there was deficient binding to its GPCR it will lead to reduced protein kinase A activity Educational objective: 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.
Q3
pancreas secretes insulin (beta-islet) and glucagon (Alpha-islet). Anterior pituitary gland secretes GH and LH Educational objective: 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.
post-translational modification
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.
Q4
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.
Q16
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.
Q25
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.
transposons
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
Q5
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.
Q13
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.
Q10
A dimer is a protein that is composed of two noncovalently bound polypeptide chains (also called subunits, or monomers). A homodimer is a protein in which the polypeptide chains of the two monomers have the same sequence (number, order, and type) of amino acids (AA). According to the passage, LPL is a homodimer that can dissociate into two inactive and identical 225-residue monomers, meaning the active LPL homodimer is made up of 450 total amino acid residues. In the given scenario, the scientist must calculate two things: The number of mRNA nucleotides that must be translated to form the LPL homodimer, which can be calculated as follows: # AA in protein × 3 nucleotides per codon = # of nucleotides in mRNA sequence 450 AA in LPL homodimer × 3 nucleotides = 1,350 nucleotides The molecular weight of the LPL homodimer. Given that the average molecular weight of an amino acid is about 110 daltons (Da), the molecular weight of the LPL homodimer would be approximated as: # AA in protein × average molecular weight of an AA = molecular weight of the protein 450 AA in LPL homodimer × 110 Da = 49,500 Da Because 1,000 Da = 1 kDa, the approximate molecular weight in kDa of the active LPL homodimer is 49.5 kDa, or ~50 kDa. concept: 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).
Q7
Educational objective: 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.
Q50
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. Concept: 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.
Q8
Educational objective: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.
Q5
As shown in Figure 2, out of all the transduced cell lines, the cell line expressing all miR-17~92 families except miR-19 (Δ/Δ + 17~92 (Δ19)) had the highest percentage of apoptotic cells. This result signifies that removal of only miR-19 from the miR-17~92 cluster promotes apoptosis. Accordingly, the miR-19 family is therefore the most essential to cancer progression as its expression suppresses apoptosis. Concept: 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.
Q33
Chose wrong answer bc did not read the word "inhibitor" meansing it would do the opp of HDACs and increase gene expression. concept: 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.
Q40
Concept: Eukaryotic translation involves three stages: 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. Elongation: The ribosome continues to elongate the polypeptide chain by reading each mRNA codon in a 5′ to 3′ direction. Termination: A stop codon is read, and release factors induce translation complex dissociation
Q30
D and E are correct bc passage states "PKD2 gene, which codes for the protein polycystin-2 (PC2). The PC2 protein functions as a mechanically activated calcium (Ca2+) channel in ciliated renal epithelial cells." Meaning you need negative amino acids to bind with positive Ca2+ Due to their charged nature, ions cannot freely cross the plasma membrane because the membrane contains a middle layer of uncharged fatty acid chains. Consequently, an ion must pass through the pore of an ion channel to enter the cell. Ion channels are transmembrane protein complexes that fluctuate between an open or closed conformation to facilitate ion transport. These channels exhibit selectivity as only specific ions may pass through the pore and down their concentration gradient (high to low). For a particular ion to go through a channel, the inside of the channel's pore must be lined by amino acid residues with charges opposite to the ion's charge. This is because oppositely charged species attract each other whereas identically charged species repel each other. In addition, the ion must be briefly pulled out of its hydration shell to pass from the aqueous environment outside the plasma membrane to the aqueous environment inside the cell. The opposite charges within the pore stabilize the charge of the newly dehydrated ion. Educational objective: 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.
Q3
Evaluation of Figure 1. First of all, I and II contradict one another Concept: 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.
Q24
Figure 1 shows that prior to insertion of cldnk, the pSKII plasmid contained 2,958 base pairs. Figure 2 shows that the distance between the EcoRI and XhoI restriction sites is 32 base pairs (17 + 15). These bases will be removed by the digest. The passage states that the cldnk gene produces a processed transcript that is 915 base pairs long, so insertion of cldnk cDNA increases the plasmid size by 915 base pairs. Therefore, the final size of the plasmid after cldnk is inserted can be calculated as 2958 − 32 + 915 = 3841. Concept: 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.
Q11
Hardy weinberg equation 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.
Q44
I think Q is asking which lane in Fig 1 would you find replication stress response factors (ex. BLM) during HU (causes replication arrest) treatment of PSNF5 (BLM+) cells? The question is basically asking where would you see the HU cells in the immunostaining. 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 bc replication arrest is still occuring and thats where you will see the repl stress response factors. Concept: DNA replication (synthesis) occurs in S phase of the eukaryotic cell cycle.
Q48
Q is saying that TERT gene is transcribed at low levels in somatic cells so where in the chromosomes would this gene be found? In heterochromatin, because it supresses transcription bc of tightly wound DNA Concept: 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.
Q4
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.
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. Educational objective: 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.
Q8
In mammals, sex is determined by an inherited combination of sex chromosomes. Females (XX) inherit one X chromosome from their mother and one from their father. Males (XY) inherit an X chromosome from their mother and a Y chromosome from their father. Sex-linked genes, located on either the X or Y chromosome, exhibit different inheritance and expression patterns between males and females. Specifically, females will inherit two copies of X-linked genes, whereas males will inherit only one. The question describes a recessive X-linked mutation that prevents sweat gland formation within the skin. Heterozygous individuals (ie, having one mutant and one wild-type gene copy) exhibit a mosaic phenotype in which some skin patches lack sweat glands and other patches have a normal distribution of sweat glands. Because this gene is X-linked, only females (XX) carry two copies of this gene and can be heterozygous. The mosaic phenotype in these heterozygous females indicates that some skin patches express the mutant gene and other skin patches express the wild-type gene. This is best explained by random inactivation of one X chromosome in embryonic cells during development, which generally occurs in mammalian females. Depending on which X chromosome is inactivated, some embryonic cells will contain an active copy of the wild-type X chromosome and some will contain an active copy of the X chromosome with the mutation. Accordingly, some of these cells will divide and differentiate into patches of skin cells exhibiting either wild-type expression (normal sweat gland formation) or expression of the mutation (no sweat gland formation). (Choice A) Because the individuals described in the question are heterozygous for the X-linked gene, they must be female (have two X chromosomes) and would not carry a Y chromosome (ie, be male). (Choice B) Silencing of paternally inherited X chromosome genes would cause expression of only maternal X chromosome genes in all cells of the female organism. For females heterozygous for the mutation, this would cause expression of a single (not mosaic) phenotype. (Choice C) If no mutant genes on X chromosomes were expressed, the individuals with this mutation would express only the wild-type gene and exhibit normal sweat gland formation. Concept:
Q29
Just figure 1 analysis. pretty easy. Educational objective: 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.
Q1
Look at 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.
Q14
Mismatch repair system: When DNA polymerase incorporates an incorrect nucleotide into the newly synthesized strand and an exonuclease comes to excise the wrong nucleotide and maybe a couple others and then DNA pol incorporates the appropriate NTs and DNA ligase catalyzes the formation of new phosphodiester bonds to seal the gap bw the strands. MMR machinery corrects only replication errors within DNA strands.
Q30
Q is basically testing if you can recognize post-transcriptional modifications so removal of introns (alternative splicing), adding polyA tail, and methylating 5'phosphate end. All helps transport mRNA from nucleus to cytoplasm to ribosome for translation w/o being degraded. concept: 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.
Q29
Need to know that the coding strand is the DNA strand that is not growing the mRNA strand. The mRNA strand is growing from the DNA template or anti-coding strand. The Q is asking for what the DNA coding strand would look like from the given mRNA strand so it would be the exact same except U would be replaced with T. concept: In transcription, RNA polymerase reads the noncoding DNA strand to produce a complementary mRNA transcript (with uracil replacing thymine). The sequence of the mRNA transcript is identical to the sequence of the coding DNA strand (again, with uracil rather than thymine).
Q38
Normal monohybrid cross but just take not that mom is recessive allele carrier and doesn't express the condition so she's heterozygous Educational objective: For the cross between a female carrier of an X-linked recessive trait and an unaffected man, only male children will express the trait. The probability of a child having the trait is given by P(child inherits affected X-chromosome from mother) × P(child is male) = 0.5 × 0.5 = 0.25. Therefore, 25% of all offspring will express the trait.
Q17
Passage states "The genes Abl, Enabled (Ena), and receptor protein tyrosine phosphatase (PTP) are involved in axonal guidance during neuronal development and are all believed to operate in the same pathway." 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.
Q12
Passage states that ANGPTL4 codes for angiopoietin-like 4, a protein that inhibits the action of LPL. Also states that LPL degrades triglycerides in blood. So increase of ANGPTL4 mRNA -> increase in angipoetin like 4 activity -> decrease in LPL activity -> increase in triglycerides levels no ANGPTL4 mRNA -> no angipoetin like 4 activity -> increase in LPL activity -> decrease in triglycerides levels concept: 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.
Q5
Population bottleneck reduces pop size drastically in response to a sudden uncontrollable disaster (flood, famine, human-induced catastrophe). This substantially alters the genetic diversity and allele frequencies of the population in a way that doesn't necessarily follow the principles of natural selection (where organisms with beneficial alleles are more likely to survive). 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.
Q21
Q was asking abt telomeres which I understood but did not know telomeres were made up of heterochromatin (transcriptionally inactive and tightly condensed DNA wrapped around histones) Concept: 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.
Q23
Restriction enzymes cleave DNA molecules at specific nucleotide sequences, known as restriction sites. Each restriction enzyme has a unique restriction site that typically consists of a DNA palindrome, in which the sequence of one strand is the reverse of its complement strand. If a restriction enzyme cuts anywhere other than directly in the middle of the palindrome, a few nucleotides will be left on the end of the DNA strand without any complementary bases paired with them. These leftover nucleotides are known as overhangs or sticky ends because they can anneal or "stick to" their complementary bases in another DNA strand. 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′]). Concept: 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.
Q17
Smooth ER - lipid synthesis (testes/ovaries), drug/pison detox (liver) and calcium ion storage (muscle). Rough ER - produces proteins Concept: In eukaryotic cells, the nucleolus is found within the nucleus and is the primary site of ribosomal RNA (rRNA) transcription by RNA polymerase I. Ribosomal proteins synthesized in the cytoplasm are transported into the nucleolus, where they combine with rRNA to form 40S and 60S ribosomal subunit precursors. These precursors are exported from the nucleus to fully mature in the cytoplasm.
polymerase chain reaction
Standard polymerase chain reaction (PCR) measures DNA amplification after all the thermal cycles are complete, whereas real-time PCR quantifies product amplification as the reaction progresses in real time
Q16
Telomeres are noncoding DNA sequences that protect the rest of the chromosomes from degradation by degrading themselves. Repeating sequence: TTAGGG RNA (not DNA) can self-complement (complementary base pair w/ ITSELF ex tRNA) Restriction enzymes typically cut DNA at sequences known as palindromes, in which the complementary strands are identical when read from 5′ to 3′ ends
Q26
The cDNA in the passage was obtained from zebrafish; therefore, to detect the cDNA, the microarray must contain single-stranded DNA from zebrafish genes. Because cDNA is complementary to the sense strand of genomic DNA, the single-stranded DNA must correspond to (ie, have the same sequence as) the sense strand of zebra fish genes, allowing it to hybridize with cDNA. (Choice A) Double-stranded DNA is already hybridized and would be unable to anneal to the cDNA probe. The cDNA would be washed away from all wells, and no signal would be detected. (Choice B) Proteins perform many different functions within cells, but only a subset (eg, transcription factors) bind to DNA. Accordingly, the cDNA would be unlikely to hybridize with zebrafish proteins, and proteins would not be useful in analyzing gene expression. (Choice C) Antibodies typically bind proteins, not DNA. Antibodies may be used in protein identification techniques such as ELISA but would not hybridize with cDNA. Concept: 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.
Q36
The data in the passage shows that captive male cheetahs have relatively higher susceptibility to diseases despite having increased genetic diversity. As a result, it can be concluded that other factors besides genetics determine the fitness of captive male cheetahs. Indeed, according to Figure 3, captive cheetahs have increased fecal cortisol levels. Therefore, the decreased fitness can be attributed to increased cortisol, which is a marker of increased environmental stress. In response to stress, the hypothalamus stimulates the pituitary gland to release adrenocorticotropic hormone (ACTH). As a result of ACTH release, the adrenal cortex secretes glucocorticoid hormones such as cortisol. Cortisol mediates stress responses by increasing blood glucose levels and decreasing inflammation and protein synthesis. Chronic exposure to cortisol can negatively impact immunologic responses, increase the susceptibility for diseases, and affect reproductive health (fecundity). Educational objective: 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.
Q3
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. DNA does not have disulfide bonds to be reduced (reducing substances are used to break disulfide bonds in protein, not DNA). Also denaturing substances separate double stranded DNA into ss by breaking H-bonds bw two DNA strands. Denaturing substance being added wasn't mentioned in the passage protocol. 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.
Q34
The passage states that the event that occurred 12,000 years ago resulted in inbreeding depression (decreased heterozygosity and reduced fitness) of Namibian cheetahs. Typically, individuals within a sufficiently large population will outbreed (mate with nonrelatives) because of the additional fitness that outbreeding confers. Outbreeding has an evolutionary advantage because the behavior introduces new genetic material and reduces the probability of passing familial genetic abnormalities to offspring. 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. Educational objective: Inbreeding results in decreased heterozygosity (genetic diversity), reduced fecundity, and reduced fitness. Species that mate with non-relatives (outbreed) increase their fitness because the introduction of new genetic material results in increased heterozygosity.
Transformation
Transformation occurs when prokaryotes pick up foreign genetic material from their surroundings. Eukaryotes do not participate in transformation.
Q34
Understood that FL-SMN and snRNA needed to be at the same levels but did not read the Q properly bc it states in the next sentence "In motor neurons of SMA patients" so this means that SMNdelta7 would be high and FL-SMN and snRNA would be lower but equal. concept: 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).
Q28
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. Concept: 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.
Q36
if the large ribosomal subunit is impaired that means the whole ribosome is impaired which means that it will not bind to the rough ER. Also AD is in humans = eukaryotes so ribsome in euks Concept: 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.
Q15
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
Q7
AUG - start UGA- stop codons UAA UAG Concept: 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.
Conjugation
Conjugation is the exchange of genetic information between prokaryotes, typically in the form of plasmid DNA. It does not occur in eukaryotes.
Q28
Did not read the Q properly didn't realize that there were multiple generations. Look at image Educational objective: 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.
Q15
Did not understand that the Q was asking for allele separation and thought when they said recombination didn't occur that they meant Anaphase I. I was wrong The formation of gametes (ie, eggs, sperm) requires cellular division via meiosis rather than mitosis. Meiosis consists of two major stages, meiosis I and meiosis II. Like mitosis, each stage of meiosis is divided into prophase, metaphase, anaphase, and telophase. The homologous chromosomes first align at the metaphase plate in metaphase I and then move to opposite poles of the cell during anaphase I. Unlike anaphase of mitosis, which separates sister chromatids, anaphase I of meiosis separates homologous chromosomes. 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. The formation of gametes (ie, eggs, sperm) requires cellular division via meiosis rather than mitosis. Meiosis consists of two major stages, meiosis I and meiosis II. Like mitosis, each stage of meiosis is divided into prophase, metaphase, anaphase, and telophase. The homologous chromosomes first align at the metaphase plate in metaphase I and then move to opposite poles of the cell during anaphase I. Unlike anaphase of mitosis, which separates sister chromatids, anaphase I of meiosis separates homologous chromosomes. 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. 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.
Q37
Easy bc of inbreeding Educational objective: 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.
Q40
Easy but have to know that the more similar a species is to its common ancestor the more recently or later on it diverged. So the one that is least similar is the one that diverged first bc it accumulated more mutations over time. Educational objective: 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.
Q31
Loss of exon 7 is the SMA pathology and B is the only one that shows that concept: 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.
Q49
Q is asking for the complementary strand to the one in the passage. My ******* read it 3'-5' and wrote it that way. Dumb mistake Concept: 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.
Q9
Q is asking what is the structure of the complementary base pair for cytosine? its guanine and also need to know that in passage they say the samples are DNA samples not RNA so you would need one less OH on the 2'C on the sugar backbone. Guanine is a purines with two rings and one carbonyl. concept: 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.
Q19
The passage and Table 1 state the chromatin domain that the H1 is acting on so the researchers were interested in assessing how the mobility of H1 is influenced by chromatin compaction and organization. The passage states that H1 and HMG proteins have similar binding sites on chromatin and therefore compete for the same regions (ie, if one binds, the other cannot). II. was not correct bc GFP is a fluorescent tag that allows tracking of H1 movement within the cell, there is no competition bw the two molecules. Concept: 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.
Q41
Did not realize that "distantly related" with similar characteristics meant converging evolution Educational objective: Convergent evolution leads to similar characteristics in distantly related species that are exposed to similar environmental pressures. Parallel evolution also leads to similar characteristics in species found in similar environments but occurs in species with a more recent common ancestor. Divergent evolution leads to unique characteristics in somewhat closely related species that face contrasting environmental pressures.
Q2
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.
Q24
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.
Q43
Q is asking what happens to DNA replication enzymes in BLM-T99A and BLM-K695T cells treated with Aph? So had to look at figures 2 and 3. According to the passage, BLM-T99A and BLM-K696T cells were stably transfected. Figure 3 shows that after Aph treatment, BLM-T99A cells have more new sites of replication than BLM+ cells (phenotypically similar to wild-type). Therefore, compared to wild-type cells, BLM-T99A cells would have increased activity of the following replication enzymes. Need to know fxn of single-strand DNA-binding protein binds to each strand to prevent spontaneous reannealing of unwound single-stranded DNA. A, C, D wrong bc there would be an increase, decrease and decrease in all fxns. concept: 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.
Q1
Small nuclear RNAs (snRNAs), a group of molecules confined to the nucleus, function in the splicing of pre-mRNA. snRNAs associate with a specific set of proteins to form small nuclear ribonucleoproteins (snRNPs). Various snRNPs then assemble with other proteins to form a spliceosome, the large protein-RNA complex involved in splicing. Concept: 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.
Q14
**All cells that comprise a multicellular organism contain the same DNA (ie, genome). Therefore, the SRY gene is present in all cells, and this comparison can be made.** 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.
Q32
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. Directional selection is a mode of natural selection which means it chooses phenotypes that are beneficial to the species in an environment. (Choice C) is wrong bc in directional selection, one extreme phenotype is favored over all other phenotypes because of the additional fitness it confers. Consequently, the frequency of the related allele increases dramatically within a population. ADPKD does not improve fitness; therefore, it gains no directional advantage over the wild-type phenotype (ie, healthy kidneys). Educational objective: 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.
Q20
**the only thing that reducing agent does is break disulfide bonds** The proteins shown in Figure 1 were analyzed via sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE), which is used to separate proteins by size (molecular weight). During this procedure, samples are first heated with SDS, a denaturing detergent that coats proteins with a negative charge proportional to their length (size). SDS binds hydrophobic regions in the protein core, unfolding and extending the polypeptide into a linear arrangement of amino acids (ie, primary structure). This detergent also solubilizes the proteins and releases them from all associations with other molecules. Proteins are generally smaller than DNA molecules and are therefore loaded on a highly crosslinked polyacrylamide gel instead of an agarose gel (has larger pores). An electric current is applied and negatively coated proteins travel through the gel matrix and toward the anode, denoted as the positive electrode in electrophoresis. (Note that this convention is opposite that used in electrochemical cells, in which the anode is negative.) Smaller proteins travel faster than larger ones, creating lanes of size-separated protein bands. These bands are then stained for visualization using Coomassie blue or the more sensitive silver stain, both of which bind tightly to proteins. Band intensity is an indicator of protein abundance. Concept: 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.
Q33
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 negative impacts of random changes in allele frequencies (genetic drift) that may result in extinction. The other answer choices all beneficial in some way to the species. Gene flow, or the changes in allele frequency due to migration, increases genetic diversity by introducing new alleles to the gene pool. Therefore, small populations with low genetic diversity are more likely to benefit from gene flow and therefore are less likely to go extinct. Educational objective: 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.
Q35
Adaptive radiation is the process of diversifying characteristics in a subgroup of individuals from a single species. Adaptive radiation reduces intraspecific competition, the competition for resources by members of a single species. As a result, fitness is improved for the entire species because the subgroup has a new role within an ecological community (niche) that is different from the rest of the species. Adaptive radiation can eventually lead to speciation, or the formation of a new species, if the subgroup continues to diverge. According to the passage, Namibian cheetahs typically hunt species that are slightly smaller than themselves. In this scenario, the subgroup of cheetahs developed larger claws and teeth that enable them to hunt large prey exclusively. Consequently, the subgroup of cheetahs fills a new ecological niche and reduces competition by hunting prey that differs from the average cheetah hunting behavior. 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. 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. Educational objective: 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.
Q2
Based on the passage, regulation of p21 expression by c-Myc also contributes to B-cell lymphoma development in Eμ-Myc mice. In addition, the passage states that p21 inhibits cell cycle progression, a key feature of a tumor suppressor gene. Tumor suppressor genes regulate DNA repair by repressing or pausing the cell cycle to ensure that only normal cells proceed to the division stage and induce programmed cell death if repair fails. 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. Concept: 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.
Q31
Did not know all the functions of a kidney. (Choice C) is wrong bc production of leukocytes (white blood cells) occurs in the bone marrow, not the kidneys. Leukocytes are immune system cells that protect the body against infectious agents and foreign antigens. Consequently, leukocyte numbers are regulated by the presence or absence of cytokines released during active infection. Educational objective: 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).
Q47
Did not know the diff bw RNA and DNA pol but DNA pol is involved in replication, not transcription. According to the passage, WT1 is present in cells that do not express TERT (healthy somatic cells) and absent in cells that do express it (cancer cells, somatic and germ cells). The opposite occurs with c-Myc, which is absent in cells that do not express TERT and present in cells that do. The presence of WT1 in healthy somatic cells suggests that it inhibits TERT expression in these cells at the transcriptional level. In contrast, c-Myc most likely promotes TERT expression. Based on this information, WT1 must be acting as a repressor and c-Myc must be acting as an activator. Therefore, WT1 and c-Myc must inhibit and facilitate RNA polymerase binding, respectively. **c-Myc is an oncogene (potential to cause cancer)** oncogenes only need one allele to be knocked out **p53, pRb is a tumor suppressor gene, needs both alleles to be knocked out Concept: 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.
Q38 - come backkkkkk
During reverse transcription polymerase chain reaction (RT-PCR), the enzyme reverse transcriptase converts mRNA into double-stranded cDNA, which is then amplified in cycles to yield thousands of copies. The cDNA is initially denatured into single strands using heat, and forward primers and reverse primers anneal to the denatured cDNA strands so that Taq polymerase can elongate the DNA sequence. To identify which tau isoform has the longer half-life (ie, lasts longer in the cytoplasm) in the given scenario, the researchers must first convert tau v1 mRNA and tau v2 mRNA into cDNA. The number of amplified copies (concentration) of tau v1 cDNA can be compared with that of tau v2 cDNA by performing RT-PCR at varying time points. The tau isoform with greater cDNA concentration at the final time point has the longer half-life. By definition, each tau isoform mRNA differs in coding sequence; as a result, their cDNA sequences also differ. This outcome means that tau v1 and tau v2 cDNA molecules need different forward and reverse primers for the primer-annealing step of RT-PCR. (Choice A) Performing RT-PCR at only one time point would yield only one measurement, and would not allow continuous tracking of tau isoform mRNA levels in the cytoplasm. (Choices B and D) Western blot is used to detect proteins, not mRNA. Because the starting material is tau v1 and v2 mRNA, RT-PCR must be performed. However, a different technique that also could be employed in place of RT-PCR is northern blot, which is used to analyze mRNA expression levels. Concept: 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.
Q51
Genes with similar sequences, or high sequence similarity, are considered evolutionarily related. In other words, they are believed to have a common origin in a single gene from an ancestor organism. During the process of molecular evolution, portions of a chromosome may be duplicated by mechanisms such as unequal crossing over. Over time, duplicate genes may undergo mutations, allowing them to carry out distinct roles at various times in an organism's life cycle. The question states that some organisms have multiple TERT genes with high sequence identity. These similar genes most likely arose by gene duplication. (Choice A) Alternate splicing produces multiple protein products from the same gene, not multiple similar genes. (Choice B) Conjugation is the exchange of genetic information between prokaryotes, typically in the form of plasmid DNA. It does not occur in eukaryotes. (Choice D) Transformation occurs when prokaryotes pick up foreign genetic material from their surroundings. Eukaryotes do not participate in transformation. Concept: 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.
Q10
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. Because the Y chromosome is present only in males, its population size is automatically smaller than that of the X chromosome. In addition, the passage states that the Y chromosome has fewer alleles, meaning that Y chromosome alleles will exist at lower frequencies than alleles on the X chromosome. Educational objective: 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.
Q27
Had to notice the word "truncated" in the Q stem bc truncated proteins result from nonsense mutations and a nonsense mutation results from a single nucleotide change into a premature stop codon causing early translation termination and production of a truncated (shortened) protein. Also it says "coding" strand so inserting A after the second nucleotide in the PDK1 coding sequence produces TGA, which corresponds to UGA in mRNA. This is a nonsense mutation that results in truncated PC1. Educational objective: 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.
Q26
Had to understand that Tlr4 gene is closely linked with coat color on chromosome 4 and they mention that in the passage. In the Q stem, they say white is t+ and brown is td so A is incorrect bc F1 mice have one chromosome with a w allele and a tlr4(+) allele, and another with a B allele and a tlr4(d) allele. These linked alleles are more likely to remain on the same chromosome during meiosis. Therefore, F1 mice will produce gametes with unequal haplotype frequencies with a greater number of nonrecombinant w, tlr4(+) gametes and B, tlr4(d) gametes compared to recombinant B, tlr4(+) and w, tlr4(d). C and D are wrong bc all projeny in F1 will be heterozygous dominant and brown so they're wrong Educational objective: 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 non-recombinant genotypes.
Q39
In eukaryotes, reproductive cells (gametes) are produced via meiosis, during which a parent cell (2n) divides to produce four genetically distinct daughter cells containing half the original number of chromosomes (n). Meiosis has four phases (like mitosis), but these phases occur in two successive cell division stages (meiosis I and II). In prophase I of meiosis, homologous chromosomes recognize each other and line up side by side during synapsis. Because each homologous chromosome consists of two identical (sister) chromatids, this adjacent chromosomal alignment forms a tetrad (four chromatids total). Tetrads arise when a synaptonemal complex (protein structure) forms between homologous chromosomes and holds them together tightly. The tetrad structure allows physical contact between the paternal and maternal chromosomes at the chiasma, a point where a chromosome segment can break off and rejoin with the opposite homologous chromosome. This exchange of DNA is the hallmark feature of the process called crossing over. Crossover events result in daughter cells with chromosomes containing combinations of alleles that differ from those in the parent cell, leading to eukaryotic genetic recombination and increased genetic diversity. Because crossover events produce genetically unique gametes, the offspring that develops after fertilization is genetically different from either parent and more genetically diverse. Educational objective: 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.
Q25
In genetic studies, hybridization refers to the annealing of two complementary nucleic acid strands. Specifically, mRNA is a single-stranded nucleic acid that is present only in the cytosol of cells that express the corresponding gene. To determine whether a cell expresses a specific gene, tissues may be incubated with a labeled RNA strand that is complementary to the mRNA of interest. This labeled RNA hybridizes with the mRNA in cells that express it, but is washed away from other cells, allowing researchers to visualize which cells express a particular gene. 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 alsoexpress plp1A, then cldnk is expressed in oligodendrocytes. Concept: 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.
Q22
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)]. (Choice B) If the mutant allele were recessive, F1 mice would exhibit a full response to LPS, and measurements observed in the F1 generation would be indistinguishable from measurements observed in the He-N mice [tlr4(+)/ tlr4(+)]. (Choice C) Penetrance refers to the portion of offspring that express a genetic trait. If the tlr4(d) mutation had reduced penetrance, only a fraction of the F1 offspring would show decreased response to LPS compared to He-N mice. As seen in Table 1, immune response was relatively uniformly decreased among F1 mice. (Choice D) Expressivity refers to the range of symptoms observed in individuals with a given genetic condition. If the tlr4(d) allele had variable expressivity, the mutation would result in a range of different phenotype characteristics. Educational objective: Co-dominance refers to the co-expression of two alleles at the same locus, where both allelic products are observable in the phenotype of heterozygous individuals.
Q6
Just dumb. Deoxyribonucleotide triphosphates, not ribonucleoside triphosphates, are used during PCR to synthesize new DNA strands. dNTPs are free nucleic acids that contain deoxyribose (sugar) whereas ribonucleoside triphosphates are used during the transcription of RNA as they contain ribose (sugar). Polymerase chain reaction (PCR) uses thermal cycling to amplify small DNA fragments, which can then be screened for mutations or used in further genomic analysis. PCR reagents include the following: A source DNA template (containing deoxyribonucleotides) that includes the target region to be amplified and its adjacent flanking sequences (Choice C) Primer pairs designed from the oligonucleotide sequence of the regions flanking the target sequence A thermostable DNA polymerase (ie, not denatured at high temperatures) to replicate the DNA template using a pool of supplied deoxyribonucleotide triphosphates (dNTPs) A buffer solution with positively charged ions (cations) to provide an optimal environment for DNA polymerase to function (ie, cations bind the negatively charged phosphates on the DNA backbone and those on dNTPs, neutralizing the negative charge of DNA and stabilizing primer-template binding) In PCR, thermal separation (denaturation) of the DNA template is accomplished by exposing the sample to high temperatures. Subsequent cooling allows primers to anneal to the single-stranded flanking ends of the target region. DNA polymerase uses the primers to elongate the new daughter strands in the 5′ to 3′ direction. These steps are repeated to obtain millions of copies of the target DNA segment in a short time. Deoxyribonucleotide triphosphates, not ribonucleoside triphosphates, are used during PCR to synthesize new DNA strands. dNTPs are free nucleic acids that contain deoxyribose (sugar) whereas ribonucleoside triphosphates are used during the transcription of RNA as they contain ribose (sugar). 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.
Q18
Knew this and accidentally chose the wrong answer. Homologous chromosomes can exchange genetic information in a process called crossing over, or recombination, during prophase I of meiosis. Crossing over can produce new combinations of alleles within a chromosome by moving some alleles of the maternal copy to the paternal copy and vice versa. New combinations of alleles are called recombinant, whereas combinations that already existed in the parent are called parental. 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.
Q27
Knockout organisms have one or more genes that have been made inactive. Genes may be knocked out by introducing mutations that result in a truncated gene product, a frame-shifted gene product, or no gene product, each of which results in genes that produce nonfunctional proteins. The biological function of a gene can be inferred by knocking it out and observing the differences between wild-type organisms and knockout organisms. If cldnk is required for wild-type myelination of oligodendrocytes, then knocking the gene out should result in impaired or abnormal myelination. Therefore, comparing myelination of oligodendrocytes in wild-type zebrafish to myelination in cldnk knockout zebrafish would show whether cldnk is required for proper myelin formation. Concept: A gene's biological function can be inferred by comparing the differences in organisms with the gene knocked out (inactivated) to wild-type organisms.
Q4
Positive controls are often employed to ascertain test validity and allow researchers to correctly interpret scientific data. As part of proper experimental design, positive controls allow investigators ensure that the dependent variable to be studied is present or that the treatment meant to manipulate the dependent variable functions properly. 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. Concept: 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.
Q22
Q asks which graph corresponds to A1 being injected into cells w/ B1 already, and shows them competing with one another with a t80 which is defined by the passage as the average amount of time required to recover 80% of the pre-bleach fluorescence (ie, the time it takes for HMGB1-GFP to move back and restore 80% of the original fluorescence to the photobleached region). 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. A is wrong bc this graph shows almost no difference in the t80 values of both states. The results show that HMGB1 and HMGA1 do not compete with each other, indicating that each protein may bind different sites on chromatin. Concept: 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.
Q42
Q is asking what conclusion can be drawn from the following graph? So you can already see that rates of DNA synthesis differ bw the two. Now paragraph 1 states that there is an excess of somatic recombination events in the setting of BS, which leads to genomic instability. Therefore, compared to PSNF5 (BLM+) cells, a lack of BLM in PSNG13 cells led to increased somatic recombination events (genomic instability) characteristic of BS and also resulted in a lower rate of DNA synthesis (ie, lower replication fork recovery). Accordingly, during the replication recovery fork period, PSNF5 and PSNG13 cells will have differing rates of somatic recombination and DNA synthesis. C is incorrect bc PSNG13 cells would be more sensitive to the effects of Aph (causes DNA arrest) bc it lacks BLM so it would take longer for it to recover. Concept: 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.
Q45
Q is asking what happens to DNA after 85min and the addition of new dNTPs in the new 20min replication period. The passage states that CldU is incorporated only for 85 mins which means that in the new 20min replication period it won't be present. Also DNA grows 5'->3' direction so 3' OH from last NT on growing strand would attack 5' PO4 group of incoming dNTP Based on the AT/CG rule of complementary base pairing, a free dNTP enters the catalytic site of DNA polymerase and forms hydrogen bonds with a complementary nucleotide on the parent (template) DNA strand. The 3′ OH from the last nucleotide of the new growing strand "attacks" the 5′ PO4 group of the incoming dNTP (Choice D). As a result, two phosphate groups are cleaved off the dNTP, resulting in a nucleoside monophosphate molecule and in the release of water, pyrophosphate (PPi), and energy (exergonic process). PPi is further cleaved into two phosphate groups to release more energy. The energy generated from these reactions is used to form a covalent phosphodiester bond between the last nucleotide of the growing strand and the incoming dNTP (endergonic process). Once the phosphodiester bond is formed, the enzyme proceeds to couple the next free dNTP to the new growing strand. (Choices A and B) dNTP cleavage is exergonic, but phosphodiester bond formation is endergonic. CldU is no longer incorporated into the growing DNA strand after 85 minutes; therefore, it is not involved in the reactions necessary for nucleotide coupling during DNA synthesis in the new 20-minute replication period. concept: 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.
Q11
Q stem states "significantly" so look for significance in graph concept: 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.
Q39
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. The western blot in Figure 3 displays the concentrations of initiation factor eIF4E, elongation factor eEF2K, and tau proteins extracted from cells treated with eIF4E-targeting siRNA or a negative control (lacking eIF4E siRNA). As expected, eIF4E siRNA promoted eIF4E mRNA degradation, leading to markedly decreased eIF4E protein concentration. 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. (Choice A) The western blot shows unaffected tau protein expression in cells exposed to both eIF4E siRNA and the negative control. This finding indicates that eIF4E siRNA does not affect tau mRNA initiation and that eIF4E is not required because tau protein is translated in a cap-independent process via IRES-ribosome binding. (Choice B) siRNA signals for the degradation of mRNA molecules instead of proteins. Elongation factors are proteins that facilitate the lengthening of the amino acid chain during translation. The western blot illustrates that the decreased concentration of eEF2K protein does not affect the translation of tau v2 protein. (Choice C) Paragraph 4 states that eIF4E mRNA is translated in a cap-dependent manner. Concept: 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.
Q23
The experiment describes a monohybrid cross, or a mating between individuals with different alleles at one genetic locus. The first cross is performed between homozygous He-N [tlr4(+)/tlr4(+)] and He-J mice [tlr4(d)/tlr4(d)], members of the parental or P generation, and produces a heterozygous F1 generation [tlr4(+)/tlr4(d)]. The second cross would be performed between members of the F1 generation and would produce an F2 generation with a mix of genotypes. Depending on the genotype, some F2 offspring would be phenotypically similar to members of the P generation whereas some F2 offspring would resemble members of the F1 generation. For instance, expected F2 genotypes can be used to deduce anti-SRC titers when spleen cell cultures are exposed to SRC plus 1.0 μg Ph-LPS: tlr4(+)/tlr4(+): An estimated 25% of F2 mice would be homozygous for the normal functioning Tlr4 allele. These mice would have an immune response similar to that seen in He-N mice. According to Table 1, this corresponds to an antibody titer of 1605 ± 55 (Number I). tlr4(+)/tlr4(d): An estimated 50% of mice would be heterozygous for the Tlr4 gene mutation. These mice would have an immune response similar to F1 generation mice, or an antibody titer of 1240 ± 58 (Number II). tlr4(d)/tlr4(d): An estimated 25% of mice would be homozygous for the Tlr4 gene mutation. These mice would have an immune response similar to He-J mice, or an antibody titer of 810 ± 17 (Number III). 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.
Q21
The innate and adaptive immune systems are two subdivisions of immune defense: The innate immune system consists of cells poised to attack antigens in a nonspecific manner. These cells include macrophages, dendritic cells, natural killer cells, and granulocytes. This system responds to foreign antigens within minutes to hours and can recognize unique and common motifs present on pathogens. The adaptive immune system contributes specialized or acquired immunity based on learned recognition of specific antigens. Responses of adaptive immune system can be further subdivided into cell-mediated and humoral immunity: In cell-mediated immunity, mainly driven by T cells, receptors on immune cells recognize and bind directly to receptors on target cells. In humoral immunity, B cells produce antibodies specific to a new antigen (primary immune response). These antibodies enable the immune system to respond more quickly if the antigen is encountered later (secondary immune response). 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. (Choice A) Experiment 1 represents an in vitro study, as spleen cells were exposed to LPS and antigen after removal from mice. In contrast, Experiments 2 and 3 were in vivo studies performed on live mice. (Choice C) Experiment 1 allowed researchers to examine the effect of LPS exposure on antibody responses to foreign antigens and to compare these effects among mice with different genotypes. (Choice D) Researchers examined both quantitative phenotypic effects (antibody and interferon levels) and qualitative phenotypic effects (protein expression observed by protein electrophoresis). 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.
Q12
The ovaries are female reproductive structures that produce gametes and secrete sex hormones. Oogenesis or female gamete (ovum) production occurs in the ovaries as follows: In utero, oogonia (ovarian stem cells) of the female embryo rapidly multiply via mitosis to generate primary oocytes, which are surrounded by specialized cells that form a saclike structure known as a follicle. Female gametes must undergo meiosis to mature. Primary oocytes begin the first meiotic division but become arrested at prophase I until puberty. At puberty, hormonal changes during each menstrual cycle result in a single follicle being selected to continue meiosis I. Completion of meiosis I produces one haploid secondary oocyte and one small polar body that ultimately degenerates. The secondary oocyte begins the second meiotic division but is arrested at metaphase II. In the ovulation phase of the menstrual cycle, the follicle ruptures and the secondary oocyte is released into the abdominal cavity. The secondary oocyte enters the fallopian tubes, where it can be fertilized by a sperm cell. If fertilization occurs, the secondary oocyte will complete meiosis II to form one large ovum (fully mature) and a second polar body that degenerates. Educational objective: The ovaries and ovarian follicles play a role in reproduction by facilitating the maturation and release of female gametes.
Q20
When two gene products function in the same biological pathway, altering an upstream product will have an effect on the activity of downstream targets or products. Altering a downstream product in a way that mimics the function of the upstream product can compensate for the loss of an upstream product. 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 PTP does and therefore should lead to the same outcome. (Choice A) Based on the researchers' hypothesis, abnormal axonal guidance occurs when PTP fails to suppress Abl. Therefore, Abl suppression normally helps maintain correct axonal guidance and would be unlikely to increase the severity of the phenotype. 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.
Q25
the passage states "The lethal dose (LD50) was defined as the minimum dose causing death in 50% of injected mice. Compared to He-N mice, results reflected a 10-fold increase in LD50 for F1 mice and a 100-fold increase in LD50 for He-J mice." This means that 100 fold increase is talking about how much more LPS (toxin or whatever) is needed to induce fatality so it needs to be higher. So need to look at where 50 is on graph on y axis and see if there is a diff of 10fold between the two 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). (Choices A and B) These curves correspond to LD50 values near 0.5 mg and 2.5 mg, or only a 5-fold difference. (Choice D) In this curve, the LD50 for the F1 mice is 10 times greater than the LD50 for the He-J mice. 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.