DAT biology question bank q's

Which of the following originated first? A. Heterotrophic prokaryotes B. Primitive eukaryotes C. Autotrophic prokaryotes D. Complex eukaryotes E. Protobionts

-Life developed on earth following a timeline. First, the primordial atmosphere was a reducing environment with many inorganic gases (except for oxygen). Some inorganic gases formed a primordial sea as the earth began to cool down. Gradually these inorganic compounds int he sea conglomerated into organic monomer, which then gave rise to more complex organic polymers. -then, biological prototypes called protobionts arose, followed by heterotrophic prokaryoties, and then autotrophic prokaryotes. The autotrophic prokaryotes started to release oxygen, which transformed the Earth from a reducing to an oxidizing environment. As a result, primitive eukaryotes and eventually complex eukaryotes formed. -Heterotrophic prokarytoes are the simplest lifeform to have been introduced to the earth,and they consumed the organic matter found in the primordial sea. Autotrophic prokaryotes developed as hetertrophic prokaryotes evolved to make their food. Thus, both came after protobionts, so (A) and (C) are incorrect. -Primitive eukaryotes formed after smaller prokaryotes were taken up by larger cells (endosymbiotic theory) to handle the newfound oxygen in the atmosphere. AS the environment continued to change, the eukaryotes became more complex. Thus, choice (B) and (D) came after prokaryotes (which came after protobionts), so they are incorrect choices

In vertebrates, what stimulates ligand-gated Na+ channels to open at a neuromuscular junction? A. potassium B. magnesium C. glutamate D. epinephrine E. acetylcholine

-To transfer an action potential from the presynaptic motor neuron to the postsynaptic muscle fiber, acetylcholine must be released from the neuron into the neuromuscular junction and then must bind to ligand gated Na+ channels on the muscle fiber. Think of it as a synapse, except the postsynaptic neuron is relplaced by a postsynaptic muscle fiber. Thus, we can see that the correct answer choice is (E). -Potassium is an important ion in action potential propagation whose outflow is responsible for repolarization of the membrane, but does not act as a neurotransmitter. Answer choice (A) can be eliminated. -Magnesium is the central atom in chlorophyll, allowing plants to convert light into energy. It is not present at the neuromuscular junction, so answer choice (B) can be ruled out. -Glutamate is the most excitatory neurotransmitter in the central nervous system. Since neuromuscular junctions are part of the peripheral nervous system, answer choice (C) can be eliminated.

Which group of organisms form a blastopore that turns into the mouth? A. Protostomes B. Eumetazoans C. Echinodermata D. Chordates E. Deuterostomes

A -Protostomes include the animal phyla that have a blastopore during development that turns into the mouth and deuterostomes include the animal phyla that have a blastopore that turns into the anus. It is useful to remember that Echinodermata and Chordata are deuterostomes while the other animal phyla are protostomes. Thus, the correct answer choice is (A). Also, we can rule out answer choices (C), (D), and (E). -Eumetazoans represent a clade containing many animal phyla including protostomes and deuterostomes so we can rule out anwer choice (B).

Bacterial cloning requires each of the following compenents, EXCEPT one. Which is the EXCEPTION? A. Processed mRNA B. Taq polymerase C. Reverse trascriptase D. Competent cells E. Restriction enzymes

B. -Bacterial cloning is the technique that clones eukaryotic gene products in prokaryotic cells. It does this by following these steps: 1. Isolate processed mRNA for the eukaryotic gene of interest 2. Treat processed mRNA with reverse transcriptase to produce cDNA 3. Incorporate the cDNA into a plasmid using a restriction enzyme and DNA ligase. The plasmid will act as the transfer vector. 4. The vector is taken up by competent bacterial cells 5. The bacterial cells will undergo transformation 6. Use antibiotic resistance or color change methods to find the gene of interest - as you can see choices (A), (C), (D), and (E) are all components of bacterial cloning and therefore can be eliminated *Taq polymerase is important in PCR bc it remains functional at high temperatures

Which of the following best describes an ecological community including its abiotic factors? A. habitat B. ecosystem C. biosphere D. environment E. population

B. -An ecological community is a term for all the populations of living organisms that interact in a given area. Abiotic factors are the nonliving chemical and physical parts of the environment such as temperature, wind, and soil composition. The term for all living organisms and nonliving factors present in an environment is known as an ecosystem. -A habitat describes the location that a specific organism lives in and includes the biotic and abiotic factors around the organism. The environment is most synonymous with an organism's habitat, also referring to an organism's surroundings. Since they both refer to only the environment of an organism and not the community of organisms as well, we can rule out answer choices [A] and [D]. -The biosphere is a term for the entire Earth and all the ecosystems that the Earth encompasses. Since the biosphere is much larger than an ecological community, we can eliminate answer choice [C]. -A population is a group of organisms that are of the same species that live together. A population does not include abiotic factors, so we can rule out answer choice [E].

Innate immunity provides protection via which of the following mechanisms? A. Opsonization via immunoglobulin-G B. Memory of previous pathogenic infections C. Clonal proliferation of B cells D. Tagging of pathogens via antibodies E. Inflammatory response

E. -An importatn part of innate immunity is the inflammatory response, which is generated in quick response to an injury or pentetration of your bodys innate barriers. -Opsonization via immunoglobuline-G is a specific way of tagging pathogens via antibodies for subsequent phagocytosis, and antibodies are an important part of adaptive immunity. Thus we can rule out answer choice (A) and (D)> -Adaptive immunity creates memory B and T cells to counter another infection of the same pathogen, so we can rule out answer choice (B). -B cells are part of adaptive immunity, so we can eliminate answer choice (C).

Which of the following statements is correctly associated with epigenetic changes? A. DNA methylation increases gene expression B. Histone methylation increases gene expression D. Heterochromatin condensation suppresses gene expression E. Epigenetics involve heritable changes to the DNA sequence

D. -Heterochromatin is DNA strands that are tightly wrapped, and is not easily accessible by transcription factors. Therefore, gene trasncription and expression are suppressed. Choice (D) is the correct answer. -Choice (A) is incorrect. DNA methylation will induce the formation of heterochromatin, which will in turn suppress gene expression (not increase it). -choice (B) is incorrect. This is not always true. Histone methylation can either decrease or increase gene expression, depending on the location and number of methyl groups added within the histone proteins. -Choice (C) is incorrect. Euchromatin means DNA strands that are loose. Euchromatin is the opposite of heterochromatin. Gene transcription is increased within the more easily accessed euchromatin. -Choice (E) is incorrect. Epigenetic modifications are indeed heritable. However, they are not changes in DNA sequences. Epigenetics refer to changes that are "on top" of DNA sequences, hence the name "epi".

If a brown haired homozygous dominant rat (BB) is crossed with a grey haired true breeding recessive rat (bb), what will the genotypic ratio of BB:Bb:bb be in the resulting generation? A. 1:0:1 B. 1:2:1 C. 1:0:0 D. 0:1:0 E. 0:0:1

D. -by drawing a simple punnett square you can see all offspring are heterozygous Bb

If a nematode was sliced in half and its cross-section studied, which of the following would be microscopically observed? A. A fully developed mantle B. Pseudocoelom partially lined by mesoderm C. Dorsal hollow nerve cord D. One germ layer E. Absence of endoderm

_Nematoda and Rotifera are the two animal phyla that you must remember are pseudocoelomates. A pseudocoelomate has a body cavity that is only partially surrounded by the mesoderm while a true coelomate has a body cavity completely surrounded by the mesoderm. -The mantle is a tissue layer found in the animal phylum Mollusca that is used to secrete calcium carbonate for producing shells. Thus, we can rule out answer choice (A). -The dorsal hollow nerve cord is a structure in the animal phylum Chordata that develops into the central nervous system, so we can rule out answer choice (C) -Nematodes are triploblasts that develope three germ layers (endoderm, mesoderm, ectoderm) so we can eliminate answer choices (D) and (E). It is helpful to remember that all animal phyla except Porifera and Cnidaria are triploblasts.

Which of the following would increase the likelihood of a chiasma forming between linked genes? A. separation by more map units B. separation by fewer centimorgans C. A loss of function mutation D. a deletion between the genes E. genes with multiple alleles

A. -A chiasma is the poing where homologous chromosomes cross over in prophase I of meiosis. Therefore, we are looking for the answer that increases the chance of separating linked genes through crossing over. -The Golden rule of thumb is: the more physically separated the genes are on a chromosome, the more likely they are to be separated through crossing over. "map units" is a scientific use to talk about the distance between genes. Hence. the higher the map unit, the more separated are the genes. -Choice (B) is incorrect. Centimorgans is also a unit that can be used interchangeably with map units. But this choice illustrates the opposite of what we want, "fewer" centimorgans would decrease the chance of cross over. -Choice (C) is incorrect bc loss-of-function mutation is unrelated to crossing over and the chances of separating linked genes -Choice (D) is incorrect. A deletion mutation between the genes would only shrink the distance between them. -Choice (E) is incorrect. Again, genes having multiple alleles is irrelavent to how they would be separated.

which of the following structures develops from the mesoderm? A. notochord B. spinal cord C. nervous system D. liver E. adrenal medulla

A. -A gastrula has three germ layer. The ectoderm is the outermost layer, the endoderm is the innermost layer, and the mesoderm is the middle of those two. The mesoderm gives rise to skeletal bones, muscles, the cardiovascular system,gonads, adrenal cortex, spleen, and the notochord. Therefore, the notochord develops from the mesoderm and (A) is our correct answer choice. -The central and peripheral nervous systems are produced from ectodermal tissue, as is the adrenal medulla. There is one trickly point to note however. The notochord results from mesodermal tissue, and it induces the formation of the spinal cord. The spinal cord is a part of the CNS which consists of ectodermal tissue. Therefore choices (B), (C), and (E) are all incorrect. -The endoderm develops the lining of the digestive, respiratory, and excretory systems. It also develops into serveral vital tissues, which we can remember using the acronym PLTT: Pancreas, Liver, Thyroid, and Thymus. Therefore the liver results from endoderm (not mesoderm), meaning that choice (D) 9s correct.

Which of the following might decrease hemoglobins affinity for oxygen? A. Low pH B. Low CO2 C. Low temp D. Low 2,3-BPG E. Fetal hemoglobin

A. -A low pH corresponds with a high concentration of H+ ions. Oxyhemoglobin (HbO2) can become reduced hemoglobin (H+Hb) in the presence of H+ ions. This process decreases hemoglobins affinity for oxygen and results in oxygen being displaced. Thus, we can see that answer choice (A) is correct. -CO2 reacts with water in the presence of carbonic anhydrase to produce bicarbonate anions and H+ ions, so low CO2 would mean less H+ ions to decrease hemoglobins affinity for oxygen. Answer choice (B) can be ruled out. -Higher temperatures correlate with increased cellular respiration, resulting in decreased affinity for oxygen. Since the opposite is true for lower temperatues, answer choice (C) can be eliminated. -2,3-biphosphoglycerate (2,3-BPG) levels increase when oxygen is low in cells, causing decreased affinity for oxygen in hemoglobin so cells can absorb more of the of the released oxygen. We can rule out answer choice (D). -Fetal hemoglobin has a higher affinity for oxygen so we can eliminate answer choice (E).

Which of the following modifications to histone proteins will increase the rate of DNA transcription? A. Acetylation B. Protonation C. Methylation D. Oxidation E. Phosphorylation

A. -Acetylation of histones will remove their positive charge, hence loosening up the attraction between histones and the negatively charged DNA. Acetylation generated euchromatin ('loose' DNA), which increases rate of transcription due to its easily accessible structure. -choices (B) and (D) are incorrect bc protonation and oxidation are not histone modifications - choice (C) is incorrect because histone methylation can either decrease or increase gene expression depending on the location and number of methyl groups added within the histone protein. -choice (E) is incorrect. Phosphorykation is indeed a type of histone modification. However it does not directly impact transcription level. Histone phosphorylation is more of an intermediate step that occurs in various cell activities (Cell division, transcription, etc)

Which of the following exemplifies apparent competition? A. an increase in mice causes an increase in foxes and a decrease in rabbits B. two lions compete for the same zebra carcass C. short beaked birds are more successful at eating seeds than long beaked birds D. a tick feeds on the blood of a dog E. the lion is the ultimate predator in the deserts of Africa

A. -Apparent competition describes a situation where indirect competition is taking place, specifically through a predator-prey relationship. Apparent competition is a complex relationship between a predator and its different types of prey. -If because of some changing factor one type of prey increases, this can result in an increase in the number of predators. This occurs as the increase in prey number means there is now more biomass (food) to support the predator species. An increase in the number of predators means that other types of prey will face more predation, and consumption. This can lead to a decrease in that other prey type. -Answer choice [A] is correct because it most accurately fits the previous description of apparent competition, whereby an increase in one prey (mice) causes a decrease in another prey (rabbits) through shared predators (foxes). The prey populations are affecting each other not through competition of resources but rather by having shared predators. -Two lions competing for the same zebra carcass is an example of intraspecific competition, or competition for resources among individuals in the same population. Thus, answer choice [B] is incorrect. Short-beaked birds having an advantage over long-beaked birds can be described by natural selection, through which adaptations present an advantage. We can eliminate answer choice [C], as this is direct competition. -A tick feeding on the blood of a dog is an example of parasitism, an example of a long-term biological interaction (symbiosis) that involves one organism benefiting at the expense of another organism. We can rule out answer choice [D]. -The apex predator is the term for the predator at the top of food chain such as the lion in answer choice [E], which we can eliminate since it is not an example of apparent competition.

What activates pancreatic proteases? A. Enteropeptidase B. Pepsinogen C. Chymotrypsinogen D. Hydrochloric acid E. Bicarbonate ion

A. -Enteropeptidase activates trypsin, which in turn activates chymotrypsin. Both trypsin and chymotrypin are pancreatic proteases. -Choices (B) and (D) are incorrect bc pepsinogen is a zymogen that is released into the stomach.It is activated by stomach acid (which contains a high concentration of hydrochloric acid) -Choice (C) is incorrect bc chymotrypsin is a pancreatic zymogen, and it does not activate another pancreatic protease -Choice (E) is incorrect bc bicarbonate is a basic ion that neutralizes the gastric juice (which is extremely acidic) as it reaches the small intestine

All of the following substances appear in the urine produced by a normally functioning kidney EXCEPT one. Which one is the EXCEPTION? A. Proteins B. Sugars C. Creatinine D. Drugs E. Vitamins

A. -Filtration occurs from the glomerular capillaries and into the Bowman's capsule. Blood enters the glomerulus, and the small substances (such as water and solutes) from the blood flow into the Bowman's capsule. Bowman's capsule has podocytes (long, foot-like proesses) that wrap around the glomerulus in a way that they form fenestrations (means "window" or "opening"), or slits. These fenestrations are large enough for certain substances to pass through but small enough to prevent others. -Large substances (such as proteins and blood cells) are too big and cannot flow out of the glomerulus. -All of the other choices are examples of small substances and can therefore be eliminated

Where does hematopoiesis predominantly occur? A. Epiphysis B. Diaphysis C. Synarthroses D. Lamellae E. Cortical bone

A. -Hematopoiesis is the process by which new blood cells are produced and occurs in red bone marrow. Epiphyses are the end parts of long bone that are distinct from the shaft bc they contain red bone marrow. -The diaphysis is the term for shaft of the long bone that contains yellow bone marrow (fat stored for energy) in adults, so we can eliminate answer choice (B). -Synarthroses are a type of joint that does not move due to its highly fibrous nature, containing connective tissue and not red bone marrow. Thus we can rule out answer choice (C). -Cortical bone is the dense, protective outer layer of bone. Cortical bones' functional unit is an osteon, which is cylinder-like and made up of layers called lamellae that give the bone strong resistance to strain. Hematopoiesis does not occur within the dense cortical bone, we can eliminate answer choices (D) and (E).

Which of the following leukocytes fucntions as an antigen presenting cell? A. Macrophage B. Basophil C. Neutrophil D. Lymphocyte E. Eosinophil

A. -Leukocytes (WBCs) function as antigen presenting cells when they present antigens to adaptive immune cells for an adaptive immune response. Macrophages and dendritic cells are the main antigen presenting cells in the body. -(B) and (E): Basophils and eosinophils are leukocytes that are known for having granules, which are released to kill pathogens and induce inflammation. -(C): Neutrophils are the most abundant type of leukocye and work to kainly phagocytose pathogens. -(B): Lymphotcyes and B cells, T cells, and natural killer cells. B and T cells are part of the adaptive immune response and natural killer cells target virally infected cells or cancer cells.

What is the major difference between optical and electron microscopy? A. Optical techniques can view living samples, while electron techniques always kill the sample B. Optical microscopy uses a vacuum, while electron microscopy uses atmospheric air C. Optical microscopy allows for a higher resolution than electron microscopy D. Optical techniques produce 3D images, while electron techniques produce 2D images E. Optical microscopy utilizes reflected light, while electron microscopy utilizes light phase changes

A. -Optical micrscopy views a sample by shining a light (not by using a vacuum) on it. Many optical microscopy techniques can be used to directly observe living cells through the optical lens of the microscope. This technique uses reflected light, similar to how mirrors work. This helps us understand that this technique produces 2D images (not 3D images) -Electron microscopy views a sample by bombarding it with electrons (not atmospheric air or light phase changes). A vacuum is used in electron microscopy to prevent the electrons from deviating from their path. It requires the sample cell to be fixed and stained, and killed before it can be indirectly observed (using a computer). *note: the microscopy technique that uses light phase changes is phase-contrast microscopy, a type of optical microscopy* -Therefore we can eliminate choices (B), (D), and (E), and also conclude that choice (A) is the correct answer. -Electron microscopy produces a much higher resolution than optical microscopy because the wavelength of an electron is smaller than that of a light. A higher resolution means that it can be used to observe smaller objects. There fore choice (C) is incorrect.

The let-down reflex refers to: A. increased contractions of the breast to eject milk B. drop in FSH and LH in response to rising progesterone and estrogen levels C. increased contractions of the uterus during labor D. increased contractions stimulated by prostaglandins to propel sperm E. the pathway of sperm from the seminiferous tubules down to the penis

A. -Prolactin functions in lactation (milk production). The more an infant suckles on the nipple, the more prolactin is producted, which leads to more lactation. Oxytocin increases smooth muscle contractions int he breast to release that milk to the baby. The more lactation, the more milk will be released. In this way, prolactin and oxytocin follow a positive feadback loop. Oxytocin is specifically inolved with the release of milk, which is known as the milk let-down reflex, and (A) is our correct answer choice. -Progesterone, estrogen, FSH and LH are hormones that are involved in the menstrual cycle. They are not directly involved with milk let-down, so (B) is incorrect. -In addition to its involvement with milk release via the let-down reflex, oxytocin also functions to increase contractions during labor via a positive feedback mechanism. However, this additional effect of oxytocin is not releated to the let down reflex bc it is not related to milk release. Therefore, chocie (C) is correct. -Prostaglandins are secreted by seminal vesicles to aid in the propulsion of sperm through a mans urethra. Recall that sperm follows the SEVEn UP pathway. However, sperm movement is unrelated to milk release via the let-down reflex, so (D) and (E) are incorrect choices.

Which of the following is a common feature between reptiles, birds, and mammals? A. they each have an amnion B. they undergo holoblastic cleavage C. the yolk sac is the sole source of nutrients to the young fetus D. fertilized eggs will develop a gray crescent E. they undergo spiral cleavage

A. -Reptiles, birds and mammals are amniotes, which means that they each produce an amnion. The amnion is the inner layer that is closest to the growing embryo. It forms a membrane around the embryo and secretes the amniotic fluid which gives a water cushion and protects the embryo from damage. -Holoblastic cleavage is a complete cleavage that evenly divides the entire embryo into distinct blastomeres. Typically this happens in emrbyos without a lot of yolk, such as humans and sea urchins. Reptiles and birds undergo meroblastic cleavage, a partial cleavage that happens in parts of the embryo. Therefore, the entrie embryo does not evenly divide. This usually happens in embryos with a lot of yolk, such as birds, fish and reptiles. Since reptiles and birds undergo meroblastic cleavage, option (B) is incorrect. -In egg laying animals, such as bird and reptiles, the yolk sac is the sole source of nutrients to the young fetus. In mammals, however, the chorion and allantois form the placenta and umbilical cord, respectively which helps with the exchange of gases, nutrients and wastes. This makes option (C) incorrect. -The development of a gray crescent is a feature of the embryonic development of frogs. Reptiles, birds and mammals do not develop a gray crescent, option D is incorrect. -Spiral cleavage is a feature of protostomes, such as flatworms, mollusks and anthropods. Spital cleavage results in misaligned cells that deviate away from the axis, meaning that if you look at it from a top-down view, you can see that the bottom cells are shifted compared to the top cells. Reptiles, birds and mammals are deuterostomes, which undergo radial cleavage. Radial cleavage results in cells that are aligned on the verticle axis, with the top cells directly overlapping the bottom cells. As mammals, reptiles and birds are chordates that undergo radial cleavage,option (E) is incorrect.

Which layer protects the skin from UV radiation? A. Basale B. Granulosum C. Merkel D. Desmosomal E. Papillary

A. -The basale layer of the epidermis protects the skin from UV radiation. This is achieved by the melanocytes (they are also responsible for skin pigmentation_ -The mnemonic "Come Let's Get Some Beers" helps us remember the layers of the epidermis: Corneum, Lucideum, Granulosum, Spinosum, and Basale. Therefore we can eliminate choices (C), (D) and (E) bc they are not even layers of the epidermis. -Choice (C) is incorrect bc the main function of the granulosum layer of the epidermis is to waterprrof the skin (made possible by the lamellar bodies found in this layer). -Choice (C) is talking about Merkel cells, which are found in the basale layer of the epidermis. They are mechanoreceptors that pick on light touch stimuli and send messages to the CNS. -Choice (D) is talking about desmosomes found in the spinosum layer of the epidermis. They are cell adhesion proteins that hold keratinocytes together. This is what makes the skin strong and flexible. -Choice (E) is talking about the papillary dermis, a layer of the dermis

Which limbic structure is responsible for the consolidation of memories? A. Hippocampus B. Hypothalamus C. Cerebellum D. Dura Mater E. Parietal lobe

A. -The limbic system is located right below the cerebrum and is involved in motivation, emotion, learning and memory. The hippocampus is the most directly related part of the limbic system for learning and memory retention, especially for consolidation and retainng memories. -The hypothalamus is another limbic structure, but it controls many bodily needs such as body temperature, hunger, hormones and the autonomic nervous system. We can eliminate answer choice (B). -The cerebellum is located in the back of the brain and is needed for motor control/ fine tuning, and balance so we can rule out answer choice (C). -The dura mater is the outermost layer of the meninges, which covers the brain and the spinal cord to protect the central nervous system. We can eliminate answer choice (D). -We can rule out answer choice (E) bc the parietal lobe is the part of the brain that integrates sensory and spatial infomration, but not memoreis. Also, its not part of the limbic system.

Which hormone significantly increases in concentration during the luteal phase of the menstraul cycle and maintains the thickened endometrium? A. Progesterone B. Pregnenolone C. Oxytocin D. Estrogen E. Prolactin

A. -The mentraul cycle is a series of changes in the female reproductive system preparing for a possible pregnancy. It is regulated by a number of hormones, some of which control the thickness of the endometrium (inner lining of the uterus) in accordance to the timing of the menstrual cycle. -Progesterone levels greatly increase during the luteal phase of the menstraul cycle to thicken and maintain the endometrium. It is secreted by the corpus luteum, which develops from the remnants of the dominant follicle after it has released the egg during ovulation. The corpus luteum also secretes moderate amount of estrogen, but estrogen levels are decreased relative to their peaks in the follicular phase, and it is progesterone that is primarily responsible for maintaining the endometrium. Therefore answer choice (A) is correct and we can eliminate answer choice (D). -pregnenolone is synthesized from cholesterol and mainly acts as a precursor molecule to the other steroids, but does not directly stimulate the endometrium. Thus we can eliminate answer choice (B). -Oxytocin uses a postitive feedback loop to carry out unterine contractions during labor triggers milk letdown during breastfeeding. We can rule out answer choice (C). -Prolactin stimulates milk production during breastfeeding and mammary gland development. Maintenance of the endometrium is not regulated by its actions so we can finally eliminate answer choice (D).

Nitrogen-fixing bacteria have a symbiotic relationship with legumes by converting A. atmospheric nitrogen to ammonium B. ammonium to nitrite C. ammonium to nitrate D. nitrate to atmospheric nitrogen E. nitrate to ammounium

A. -The nitrogen cycle is a commonly tested cycle on the DAT, and you should know the roles of the 3 main types of bacteria in the cycle: 1. Nitrogen fixing bacteria converts gaseous atmospheric nitrogen (N2)-> ammonium (NH4+) 2. Nitrifying bacteria converts ammonium (NH4+)-> nitrite (NO2-)->nitrate (NO3-). Nitrate is the absorbable version of nitrogen taken up by plants. 3. Denitrifying bacteria converts nitrate (NO3-) -> atmospheric nitrogen (N2) to complete the cycle -notice that (E) is not accomplished by bacteria. Nitrate cannot be converted back to ammonium without going through the full cyle.

Each of the following is true of arteries and veins, EXCEPT one. What is the EXCEPTION? A. Arteries always carry oxygenated blood, veins always carry deoxygenated blood B. Arteries travel away from the heart; veins travel toward the heart C. Arteries have a thicker tunica media than veins D. Veins contain one-way valves; arteries do not E. Veins are wider than arteries and hold more blood volume

A. -The question is asking about exceptions, so we are looking for a wrong statement. Recall from the diagram of the heart that the pulmonary artery carries deoxygenated blood to the lungs and the pulmonary vein carries oxygenated blood back to the heart. This represents a case where an artery does not carry oxygenated blood, and a vein does not carry deoxygenated blood. This directly contradicts answer choice (A). -Arteries are defined as blood vessels that carry blood away from the heart while veins are defined as blood vessels that carry blood back toward the heart. Thus, we can rule out answer choice (B). -The tunica media contains smooth muscle that determine the flexibility of blood vessels, giving arteries their elasticity to carry blood under high pressure away form the heart. We can rule out answer choice (C). -Since veins lack the higher blood pressure of arteries, they require one-way valves to block blood from flowing the wrong way. Answer choice (D) can be ruled out. -Veins are known as capacitance vessels, and they are under less pressure than arteries and hold most of the bodys blood almost like a storage. Answer choice (E) can be ruled out.

What is the order og integument layers on a forearm, from superficial to most deep? A. corneum, granulosum, spinosum B. corneum, lucidum, granulosum C. basale, spinosum, granulosum D. dermal, hypodermal, epidermal E. hypodermal, epidermal, dermal

A. -There are two orders of layers that we should remember: 1. the first is the layers of the integument. The order goes: epidermis, dermis,and hypodermis (from superficial to deep). *choices (D) and (E) can be eliminated bc they do not follow the correct order. 2. the second is the layers of the epidermis specifically. The mneumonic "Come Lets Get Some Beers" helps us remember the layers of the epidermis: Corneum, Lucidum, Granulosum, Spinosum and Basale (from superficial to deep). -Choice (C) can be elminated bc it does not follow the correct order. -The lucidum layer has a star next to it bc it has some special exceptions- it is only found on the palms of the hands and the soles of the feet. The lucidum layer would not be found in the forearm so choice (B) is eliminated.

What is the function of the endosperm of a seed? A. provides embryo with nutrients B. protect the seed from mechanical stress C. Develops into the root D. Carries out photosynthesis E. Absorbs water to start germination

A. -We neeed to remember that the endosperm is a nutritious tissue structure found within seeds. The endosperm nourishes the growing plant embryo. Hence, (A) is the correct answer. -Choice (B) is incorrect bc the seed coat is the hard structure that protects the seed from mechanical stress. Endosperm does not have this function. -Choice (C) is incorrect. The radicle of the embryo is the structure that develops into the root. *Mnemonic: Radicle -> Root* -Choice (D) is incorrect. Remember , photosynthesis occurs within green structures, as the chlorophyll has green pigmentation. Photosynthesis only takes place in the young leaves that emerge from the seed. The endosperm does not carry out photosynthesis to obtain food, it already has stored nutrients for its initial development. -Choice (E) is incorrect bc the seed itself absorbs water passively through a process called imbibition. Endosperm does not help the seed absrob water from the surroundings.

Each of the following accurately compares the sympathetic (SNS) and parasympathetic (PNS) nervous systems, EXCEPT one. What is the EXCEPTION? A. SNS is part of the somatic; PNS is part of the autonomic B. Both release acetylcholine from preganglionic nerve fibers C. SNS has long postganglionic nerves; PNS has short postganglionic nerves D. SNS utilizes epinephrine and norepinephrine; PNS does not E. SNS decreases peristalsis; PNS increases it

A. -You must remember that the peripheral nervous system is composed of the somatic nervous system (voluntary movement) and the autonomic nervous sytstem (involuntary movement) -the autonomic nervous system is further divided into the SNS and PNS. Thus, we can see that answer choice (A) is the exception because BOTH the SNS and PNS are actually part of the autonomic nervous system (the SNS is not part o fthe somatic nervous system). -Preganglionic nerve fibers come before the ganglia (groups of neuronal cell bodies) and both the SNS and PNS rely on acetylcholine as a preganglionic neurotransmitter. However, the SNS is different because it uses epinephrine and norepinephrine as its postganglionic (after the ganglia) neurotransmitter, whereas the PNS uses acetylcholine as its postganglionic neurotransmitter. Thus, we can eliminate answer choices (B) and (D). -The PNS has ganglia that are close to effector organs, so it has short postganglionic nerves. On the other, the SNS has ganglia that are further away from effector organs, so it has long postganglionic nerves. We can rule out answer choice (C). -the PNS represents the "rest and digest" response, so it increases peristalsis (digestive contractions), while the SNS is the "flight or fight" response that must decrease peristalsis so that more energy is put into movement, not digestion. We can finally elminate answer choice (E).

By which method do bacteriophages contribute to bacterial genetic diversity? A. tranduction B. sexual reproduction C. vertical gene transfer D. pilus conjugation E. transformation

A. -here, we need to know immediately that bacteriophages are a type of virus that infects bacteria. When virus contribute to bacterial genetic diversity, it is through transduction. -choice (B) is incorrect bc bacteria cannot go trough sexual reproduction. They only reproduce asexually through binary fission. -choice (C) is incorrect bc bacteria do not obtain genetic diversity through vertical gene transfer (down a generation) since they do not reproduce sexually. Rather, bacteria achieve genetic diversity through horizontal gene transfer (across the same generation). -choices (D) and (E) are incorrect. Pilus conjugation and transformation are indeed methods for bacteria to acquire genetic diversity. However, they do not involve bacteriophages.

Acute stress leads to an increase in activity in which of the following? A. adrenal medulla B. corpus luteum C. pineal gland D. parafollicular cells E. leydig cells

A. acute (or short term) stress puts us into the "fight or flight" mode, which is mediated by catecholamines, specifically epinephrine and norepinephrine. These two hormones are secreted by the adrenal medulla. Therefore making choice (A) the correct answer. -Choice (B) is incorrect. The corpus luteum produces progesterone and estrogen during luteal phase of the menstrual cycle -choice (C) is incorrect. The pineal gland secretes melatonin. Melatonin regulates your circadian rhythm. -Choice (D) is incorrect bc the parafollicular cells secrete calcitonin. Calcitonin "tones down blood calcium" and stimulates bone building -Chocie (E) is incorrect bc the Leydig cells produces testosterone in males (remember, LH acts on Leydig cells to make men Large and Hairy)

Two different species of butterflies look nearly identical, but only one species is poisonous when eaten. Which of the following best describes this phenomenon? A. Mullerian mimicry B. Batesian mimicry C. Camouflage D. Aposematic coloration E. Divergent evolution

B. -Batesian mimicry: occurs whenever a non-poisonous species evolves to mimic the aposematic coloring of another poisonous species to avoid predation. Our question stem tells us that both butterfly species appear nearly identical. However, only one species is poisonous. -Mullerian mimicry: occur whenever two posionous species evolve to look alike when they share a common predator. BC one of our butterfly species is non-poisonous in the question stem, choice (A) is incorrect. -Camouflage: is a tactic that species use to blend into their environment to avoid predation. Camouflage does not tell us anything about mimicry or poison, so choice (C) is incorrect. -Aposematic coloration (warnig coloration) is the vibrant colors poisonous species use to warn predators that they are posisonous choice (D) is incorrect bc it does not tell us about a non poisonous species mimicry. -Divergent evolution- occurs when species evolve to become more dissimilar within similar environments.

How do oral contraceptive birth control pills prevent ovulation? A. early stimulation of the LH surge B. Inhibition of GnRH production C. Suppression of Leydig cells D. Activation of Sertoli cells E. Development of follicles

B. -Birth control pills release synthetic estrogen and progesterone, inhibiting GnRH production during the menstrual cycle through negative feedback and thus preventing the menstrual cycle from causing ovulation. -The hypothalamus is sensitive to estrogen and progesterone in the blood. It will secrete GnRH when estrogen and progesterone levels are low bc estrogen and progesterone have a negative feedback effect on GnRH. Normally, estrogen and progesterone levels are low during the beginning of the menstrual cycle (follicular phase) to stimulate GnRH production form the hypothalamus, causing FSH (follicle stimulating hormome) and LH (luteinizing hormone) to be released from the anterior pituitary. Birth control pills will reverse its effects. -In the absence of birth control pills, FSH stimulates the development of follicles, which will secrete estrogen. Estrogen prepares the uterus for implantation of the embryoby thickening the endometrium. During the middle of the menstrual cycle (ovulation phase), negative feedback from estrogen to GnRH switches to positive feedback due to estrogen reaching a certain threshold amount. This will cuase a surge in LH, intiating ovulation. -The LH surge causes ovulation so early stimulation of the LG surge would not result in the prevention of ovulation. Choice (A) is incorect. -Leydig cells are present in the male testes and produce testosterone in response to LH. They are a major contributor to sperm maturation. Choice (C) is incorrect. -Sertoli cells are cells that envelope and nourish developing sperm cells in the testes. Choice (D) is incorrect. -Follicle development results in the secretion of estrogen, which is needed for the LH surge and thus ovulation. Choice (E) is incorrect.

Which of the following point mutations will be most likely to damage protein function? A. Conservative missense B. Early nonsense C. third base substitution D. duplication E. late insertion

B. -Early nonsense mutation is the most severe one. This is bc a nonsense mutation leads to a stop codon, hence terminating translation. If this occurs early in the process, the remaining chain of mRNA is not translated at all, leading to severe protein dysfunction. This is why (B) is the correct answer. -choice (A) is incorrect bc conservative missense is a mild form of mutation. Conservative missense leads to another type of amino acids with roughly the same properties as the original one. Hence, it doesnt impact overall function very much. -choice (C) is incorrect A substitution at the third base would have the minimal deleterious effects, bc of the degeneracy in the mRNA code. Redundancy in the codons at the third base mean that often a change in the third base of a codon doesnt change the amino acid coded for. -choice (D) is incorrect. Even though duplication leads to a frameshift mutation, it does not stop translation prematurely. The impact could be big or small depending on many factors. So, comparatively, ti is not as bad as an early nonsense mutation. -choice (E) is incorrect bc a late insertion would cause a frameshift mutation only near the end of the mRNA chain, thus not making a huge impact to the protein product compared to early nonsense mutation.

The process of living organisms turning into fossils is called: A. ossification B. petrification C. sedimentation D. catastrophism E. hybridization

B. -Evolution is the gradual change in heritable traits, seen in populations over many generations. The following five pieces of evidence support the theory of evolution: fossils, biogeography, embryology, comparative anatomy, and biochemical. This question asks us about fossils produced from an organisms actual remains. Pettrification occurs when inorganic minerals (found in the sediment) replace a corpse's organic materials. -Ossification refers to bone formation in animals, and is not involved in fossilization. Therefore (A) is incorrect. -Sedimentation is the process where particles in a suspension settle to the bottom. Sedimentation acts to bury dead organisms over time, aiding the process of petrification. Choice (C) is a tricky distractor, but it does not describe the fossilization process (petrification) , so it is incorrect. -Catastrophism says sudden and violent catastrophes (floods, earthquakes) caused mass extensions and geographical changes. Catastrophism does NOT describe the process of fleshy organisms hardening into fossils. Choice (D) is incorrect. -In reproductive biology, hybridization refers to the process where a hybrid offspring results from two different species mating. In molecular biology, hybridization is the process where two complementary single strands of nucleic acid come togehter to form a double-strand. Neither process relates to fossil formation. SO choice (E) is incorrect.

All of the following would disrupt genetic equilibrium EXCEPT one. Which of the following is the EXCEPTION? A. polyploidy B. isolated population C. genetic drift D. inbreeding E. natural selection

B. -Genetic (Hardy-Weinberg) equilibrium is a condition where there are no changes in allele frequencies (evolution). For genetic equilibrium, the population will need to be Large, mate Randomely, and experience no Mutations, natural selection, or Migration. Remember these conditions using Large, Random M&M (read & like "n"). -Isolated populations do not migrate. Therefore, choice (B) does not disrupt genetic equilibrium. Isolated populations do not migrate. Therefore, choice (B) does not disrupt genetic equilibrium. -Polyloid inidividuals have more allele copies than usual. Moreover, they can only mate with other polyploids. If mating does occur, an entirely new species can form. Therefore, choice (A) can disrupt genetic equilibrium. -Genetic drift changes allele frequencies by chance events/changes and it usually affects small populations. As a result, choice (C) disrupts genetic equilibrium. -Inbreeding occurs when individuals mate with their family members. Therefore, choice (D) disrupts genetic equilibrium bc mating is not random. -Natural selection describes the increase or decrease in allele frequency due to envrionmental adaptations where there is survival of the fittest. Therefore choice (E) disrupts genetic equilibrium.

Which of the following best exemplifies LaMarck's theory of Evolution? A. Natural selection causes a better camouflaged insect to escape predation B. A man develops his ability to spring produces offspring that are faster runners C. A pair of organisms produce more offspring than they need to replace themselves D. An animal with genes for a thick fur coat survives a harsh winter and produces offspring with thick fur coats E. Ostriches have wings even though they are not able to fly

B. -LaMarch believes that whatever characteristic an organism acquires throughout its life (through use and disuese) will be passed onto its offspring. His theory of evolution is also known as the theory of use and disuse. This theory is incorrect bc environmentally acquired characteristics are actually not heritable. They are changes to the organism, but dont represent a heritable change bc their use/disuse doesnt change the genetic code, ie. the DNA. A runner passing on his developed ability to spring to his offspring is an example of Lamarck's theory of evolution, making option (B) the correct answer. -Answers (A) and (D) are both incorrect bc they are examples of natural selection, part of the theory of evolution proposed by Charles Darwin. His theory of evolution refers to heritable changes in populations of species over generations. More specifically, evolution refers to the changes in allele frequencies in populations over time. For example, the allele that codes for white fur coat will become more common as a population of foxes begins to live in the arctic. -An animal's fitness is measured by its ability to produce viable and fertile offspring. This does not fit into LaMarck's theory of evolution. Option [C] is incorrect. -Vestigial structures are structures that exist without a function. Vestigial structures are often homologous to structures that do serve a purpose in other species (ostrich wings vs. eagle wings). They are a result of evolution but do not exemplify Lamarck's theory. Option [E] is incorrect.

Which type of antibody provides passive immunity to a developing fetus? A. IgA B. IgG C. IgM D. IgD E. IgE

B. -Passive immunity refers to the transfer of ready-made antibodies from one source to another. You must recall that immunoglobulin G (IgG) is the only antibody that can cross the placenta to give a developing fetus passive immunity. -IgA is mostly found in bodily secretions and can be transferred to newborns via breast feeding (not a developing fetus), so we can eliminate answer choice (A). -IgM is the first antibody to be secreted in humoral immunity and mainly found in blood and lymph. Answer choice (C) can be ruled out. -IgD is present in small amounts and little is known about its function. Its not passed through the placenta. Answer choice (D) can be ruled out. -IgE is an important antibody secreted during allergic reactions and is present on the surfaces of basophils an mast cells, but it is not passed through the placenta. We can elminate answer choice (E).

The cohesion-tension theory explains which of the following phenomenon? A. Capillary action B. Transpiration C. Root pressure D. Opening and closing of stomata E. Thigmotropism

B. -The cohesion-tension theory in plants explains how water is pulled up from the root to the leave, eventually escaping the leaves through transpiration. Transpiration is when water evaporates from leaves' openings (stomata). Water molecules are naturally cohesive to each other due to their polarity; cohesion is the stickiness between two identical particles (i.e. water and water). -Therefore, when one water molecule moves out of the stomata due to evaporation, it pulls on its surrounding water molecules due to cohesion, which creates tension, pulling the entire column of water upwards. Cohesion-tension is the theory behind the mechanism of transpiration in plants. -Choice (A) is incorrect. Capillary action is the result of adhesion forces, NOT cohesion-tension. Adhesion is the stickiness between two different particles (i.e. water and glass). You can think of capillary action as moving up a small glass tube. -Adhesive forces between water and glass will cause water to climb up the glass tube, explaining capillary action. -Choice (C) is incorrect. Root Pressure builds up when water is passively moving into the root through osmotic force. When the pressure is high enough, water moves up the water column (towards stem and leaves), relieving the pressure. -Choice (D) is incorect. Opening and closing of the stomata is NOT caused by cohesion-tension. It is controlled by how the humidity in surrounding enviornment is. -Choice (E) is incorrect. Thigmotropism is a phenomenon where plants grow in response to touch/contact, like when vines climb up a wall. It is controlled by the hormone auxin, not by cohesion-tension.

What needs to occur in RBCs within alveolar capillaries, in order for carbon dioxide to leave the blood? A. Chloride anions diffuse in B. Bicarbonate anions diffuse in C. Carbonic anhydrase diffuse out D. Diatomic oxygen diffuses out E. Cations (protons) diffuse out

B. -The following equation describes the body's bicarbonate buffering system: *CO2+H2O->H2CO3->HCO3-+H+ -Bicarbonate anion (HCO3-) increase would result in the production of carbon dioxde (CO2) due to the reaction shifting to the left. Thus, bicarbonate anions need to diffuse into red blood cells for carbon dioxide to be produced and leave the blood. This means that answer choice (B) is correct. -Chloride shift is the process by which chloride ions enter red blood cells to displace bicarbonate anions. We actually need bicarbonate to cause CO2 to leave the blood, so answer choice (A) can be ruled out. -Carbonic anhydrase is an enxyme that facilitates the bicarbonate bufferying system, so red blood cells need it for CO2 to leave the cell. Answer choice (C) can be ruled out. -Diatomic oxygen (O2) diffusing out of the cell would increase the hemoglobins affinity for carbon dioxide through the Haldane effect, the opposite of allowing carbon dioxide to leave. Thus, we can eliminate answer choice (D). -Protons are required for the bicarbonate buffering system's reaction to shift to the left because H+ ions are present on the right side of the equation. They must diffuse into the cell, not out, so we can rule out answer choice (E).

Which of the following is true of r-selected species? A. large sized offspring B. little parental care C. many survive to reproductive age D. demonstrate type II survivorship curve E. off spring mature slowly

B. -The r/K selection theory is a theory that describes how certain species undergo population growth. -K-selected species take very long to produce a small number of physically large offspring, but due to high parental investment (during pregnancy and after birth), a high percentage often survive. Many of the offspring do survive to reproductive age and mature slower. Since r-selected species are the opposite of K-selected species, answer choices [A], [C], and [E] can be eliminated. -r-selected species are the opposite, taking very little time to produce a large amount of physically small offspring that usually have close to no parental contact. More offspring are produced to offset a much lower survival rate for r-selected species. Thus, we can see that answer choice [B] is correct because r-selected species have little parental care. -The type II survivorship curve describes the population growth pattern of species that exhibit a relatively constant rate of survival regardless of the offspring's age. These species follow population growth that is between K-selection and r-selection, and we can rule out answer choice [D].

A male stickleback fish attacks other male stickleback fish to protect his property. However, he also attacks models of fish that have a similar red-colored belly. This is an example of: A. instinct B. fixed action pattern C. imprinting D. classical conditioning E. stimulus discrimiation

B. -The situation described in the question involves a behavior carried out by the male stickleback fish that is triggered by a red color. This series of actions is known as a fixed action pattern, which describes a behavioral sequence that continues to completion in response to a certain stimulus. The stimulus is called the releaser or sign stimuli, which is the red-colored belly in this case. Furthermore, this fixed action pattern is inherited through generations and does not have to be learned, so it can be classified as an innate behavior. -Another type of an innate behavior is instinct, which occurs without thought and is also inherited. Fixed action patterns are considered the simplest form of an instinct. However, instinct is typically used to describe general behaviors such as migratory habits or caring for offspring, so answer choice [A] can be eliminated. -Imprinting is also a type of innate behavior that is unique because it is a method through which animals learn other behaviors. During a critical period, the animal can undergo imprinting to gain new behavioral characteristics from a parent figure. Thus, we can rule out answer choice [C] -Classical conditioning is a type of associative learning that pairs a neutral stimulus to a stimulus that already elicits an innate response called an unconditioned stimulus. The neutral stimulus is then converted into a conditioned stimulus. Since classical conditioning is a type of learning rather than an innate behavior, we can eliminate answer choice [D]. -Stimulus discrimination describes how an animal is able to distinguish between a conditioned stimulus and other neutral stimuli. Since the red-colored belly is not a conditioned stimulus in this case, we can rule out answer choice [E].

Which layer of the epidermis allows the skin to stretch and bend? A. Hypodermal B. Spinosum C. Corneum D. Lucidum E. Lamellar

B. -The stratum spinosum provides the skin with strength and flexibility which allows it to stretch and bend -The mnemonic "Come Let's Get Some Beers" helps us remember the layers of the epidermis: Corneum, Lucideum, Granulosum, Spinosum, and Basale. Therefore we can eliminate chocie (A) and (E) as they are not even layers of the epidermis. -choice (A) is talking about the hypodermis, the deepest layer of the integumentary system (under the epidermis and the dermis) -Choice (E) is talking about the lamellar bodies found in the granulosum layer of the epidermis. They are what make the skin waterproof. -Choice (C) can be eliminated bc the function of the stratum corneum is to provide protection against infection, dehydration and physical harm.. -Choice (E) can be eliminated bc the function of the stratum lucidum is to provide protection (similar to the stratum corneum). Additionally, the lucidum layer is found in the palms of the hands and soles of the feet.

Which inner ear structure transduces auditory signals into neural signals? A. semicircular canals B. cochlea C. choroid D. tympanic membrane E. stapes

B. -Transduction is known as the process by which mechanical signals such as sound waves are converted into neural signals (action potentials). The part of the iner ear that carries out transductions is the cochlea, which has a fluid filled cavity that vibrates in specific regions in response to sound waves. These vibrations bend small hairs that produce nerve signals to send to the auditory regions of the brain. Thus, answer choice (B) is correct. -The semicircular canals are located in the inner ear and send signals to the brain about spatial orientation of the body and help with body balance. They do this using a fluid filled cavity and small hairs similar to the cochlea. However, they do not take in auditory signals, so we can eliminate answer choice (A). -the choroid is connective tissue that is located in the eye so we can rule out answer choice (C) - we can rule out answer choice (D) bc the tympanic membrane only transfers sound wave vibrations from the outer ear to the middle ear, it does not perform transduction of those signals. -The stapes is part of the middle ear and is connected to the oval window to assist in transmitting sound vibrations (not tranducing the signals). Thus, we can finally rule out answer choice (E).

All of the following are events that happen after a sperm penetrates a mammalian egg EXCEPT one. Which one of the following is the EXCEPTION? A. the egg exhibits exocytosis of cortical granules B. the corona radiata releases digestive enzymes C. the egg undergoes depolarization D. the zona pellucida separates from the plasma membrane E. the acrosome digests the zona pellucida

B. -When a sperm reaches the egg, it becomes in contact with the eggs outer coat, called the corona radiata in mammals, which is the first layer a sperm will penetrate. The corona radiata is a jelly-like substance that serves the primary function of nourishing the egg as it develops in the follicle. The corona radiata itself does not secrete digestive enzymes, so (B) is our EXCEPTION. It is the sperm that releases digestive enzymes, through the acrosomal reaction. -After a sperm uses its flagella to propel through the outer corona radiata jelly coat, its acrosome will digest the zona pellucida (the zona pellucida is the name given to the vitelline layer in mammels) that lies below the corona radiata. BC the acrosomal reaction occurs after corona radiata penetration, choice (E) is a valid answer choice and can be elminated. -An egg cell depolarizes during the fast block to polyspermy, which occurs after sperm penetration of the corona radiata, acrosomal reaction, and fusion between the sperm and egg plasma membranes. Therefore, (C) is no an EXCEPTION to the question stem, and can be elminated. -An egg will exocytose cortical granules to trigger the zona pellucida to harden and separate from the egg cell plasma membrane; this is the slow block to polyspermy, which occurs after the fast block to polyspermy. Therefor, neither (A) nor (D) can be an EXCEPTION, and both can be eliminated.

What important role do fungi serve within the biosphere? A. Photosynthesizers B. Saprophytism C. Nitrification D. Parthenogenesis E. Primary producers

B. -Fungi absorb dissolved molecules from dead organic matter in their environment using digestive enzymes instead of producing their own food. This method of obtaining food is known as saprotrophic nutrition, giving fungi the classification of being heterotrophic saprophytes. Thus, answer choice (B) is correct. -Photosynthesis is the process by which organisms obtain their own food using light as an energy source, not ingesting organic matter. Primary producers use photosynthesis to fix carbon dioxide into glucose molecules. Answer choices (A) and (E) can be ruled out. -Nitrification is carried out by nitrifying bacteria to convert ammonia in the soil into nitrates for other plants and bacteria to use. This rules out answer choice (C). -Parthenogenesis refers to the process by which an unfertilized egg cell develops into an embryo, something that fungi do not perform. Answer choice (D) can be ruled out.

Which of the following is a process that is used in the regulation of the cell cycle? A. The accuracy of DNA replication is assessed at the end of the G1 phse B. Cell division is inhibited if there is a high cell density in the surroundings C. mitosis stops during prophase if spindle fibers have not attached to chromosomes D. Cells that are ready for division will progress from G1 to G0 of the cell cycle E. cyclin dependent kinases (CDKs) hydrolyze proteins to control cell division

B. -In interphase, there is a checkpoint that occurs during G1 phase, but this checkpoint occurs before the S phase, when genome replication occurs. There is a checkpoint during S phase that ensures that DNA replication has occured without any problems. Therefore choice (A) can be eliminated. -the checkpoint at the end of G1 is called the restriction point. The cell is checked for favorable conditions to grow and divide. If the cell is not ready to divide, it fails this checkpoint and is sent to the G0 stage. Therefore choice (D) can be eliminated. -the checkpoint at the end of G2 ensures that the cell is ready to divide. DNA replication/synthesis happens in the S phase (remember: S for synthesis) so it only makes sense that the check point for the accuracy of DNA synthesis will happen after this stage. Therefore choice (A) can be eliminated. This further confirms that choice (A) is incorrect. -the only checkpoint during the actual cell division (mitosis) portion of the cycle is during metaphase, called the M checkpoint. There is no checkpoint at prophase. It is at the M checkpoint that mitosis stops if the chromosomes are not properly attached to the spindle fibers. Therefore choice (C) can be eliminated. -cyclin dependent kinases control cell division by activating other proteins. CDKs do this by phosphorylating (not hyrolyzing) the proteins. Even if you do not remember right away what CDKs do, you can eliminate this answer choice by remembering that kinases by nature add phosphate groups to other molecules (phosphorylate). Hydrolysis is to break down water molecules. Therefore choice (E) can be eliminated. -this leaves us with choice (B) is our correct answer. This answer choice desvribes the phenomenon of density dependent kinases.

After a person falls and scrapes his knee, what process helps stop the bleeding? A. Erythrocyte fragmentation B. Exposure of collagen proteins C. Breakdown of thromboplastin D. Antibody aggression E. Conversion of fibrin to fibrinogen

B. The clotting cascade describes a series of events that occur after tissue damage that begins with exposed collagen proteins from a damaged blood vessel, which stimulates platelet activation. These platelets form a platelet plug and cause further activation of clotting proteins. -Erythrocytes are red blood cells, which carry oxygen. They do not create the plug to stop bleeding. Thus, answer chocie (A) can be ruled out. -thromboplastin is released by activated platelets to convert prothrombin into thrombin for further plug formation. However, this answer choice says "breakdown of thromboplastin" which would inhibit clotting. Answer choice (C) can be ruled out. -Antibodies are part of the adaptive immune response and are not involved in clotting. Answer choice (D) is incorrect. -Fibrinogen (an inactive precursor) is converted to fibrin to help in blood clot formation, not the other way around. Thus answer choice (E) is incorrect.

Which of the following events in a neuron directly intiates release of neurotransmitters? A. GABA release B. Vesicle endocytosis C. Ca2+ influx D. Na+ outflux E. A graded potential

C, -Action potentials travel down the axon and reach the axon terminal, which contains voltage gated calcium channels. These voltage gated channels respond to the action potentials and open, allowing Ca2+ to flow into the cell, which triggers fusion of the synaptic vesicles with the membrane. These vesicles contain neurotransmitters, which are released into the synapse (gap between neurons). Thus, answer choice (C) is the trigger for neurotransmitter release. -GABA is an inhibitory neurotransmitter used mainly by neurons in the brain so we can elminate answer choice (A). -Vesicle exocytosis describes the process of synaptic vesicle fusion with the release of neurotransmitters into the synapse. This answer choice says endo-cytosis which is the "taking in" of substances. This is the reverse process to the release of neurotransmitters from the neuronal cell. Thus, answer choice (B) can be ruled out. -Na+ outflux occurs when sodium potassium pumps send sodium out of the neuron to maintain the resting membrane potential. During the traveling action potential, Na+ influx occurs as the action potential makes its way down the neuron. NA+ outflux does not occur to initiate release of neurotransmitters, so answer choice (D) can be ruled out. -Whether or not the postsynaptic neuron fires action potentials is determined by graded potential summation. Graded potentials are generated by the binding of neurotransmitters to receptors on dendrites. This occurs on the postsynaptic neuron, not the presynaptic neuron releasing the neurotransmitters. So we can eliminate answer choice [E].

In a population of rabbits,brown fur color is dominant and white fur color is recessive. 91% of the population is brown and 9% is white. What is the heterozygous frequency? A. 21% B. 30% C. 42% D. 50% E. 70%

C. - This is a hardy-weinberg equilibrium problem, so we need to think about our formulae: *p+q=1 *p^2+2pq+q^2=1 -p: frequency of the dominant allele, which we will call B for brown -q: frequency of the recessive allele, which we will call b for white -p^2: frequency of homozygous dominant (BB) -2pq: frequency of heterozygous (Bb) -q^2: frequency of homozygous recessive (bb) -91% of the population is brown, which means they can be BB or Bb. on the flip side, 9% of the population is white, which can only be bb. -q^2=frequency of bb=9%=0.09 -q=0.3 is the frequency of the recessive allele (b) in the entire population -p+0.3=1 -p=0.7 is the frequency of the dominant allele (B) in the entire population -2pq= frequency of heterozygous (Bb) = 2 x 0.7 x 0.3= 0.42 -0.42 x 100%= 42% (C)

Which of the following is true in the alternation of generations? A. Gametes develop into gametophytes by mitosis B. Gametes develop into gametophytes by meiosis C. Spores develop into gametophytes by mitosis D. Spores develop into gametophytes by meiosis E. Spores

C. -Alternation of generations is when the organism switches between diploid and haploid forms. -When two haploid gametes fuse, they will become a diploid zygote. The zygote will go through mitosis and become a diploid sporophyte. The sporophyte will go through meiosis and give rise to multiple haploid spores. The spores will go through mitosis and become haploid gametophytes. Gametophytes will give rise to haploid gametes and the cycle repeats. *A trick to remember is that the names are exchanged: Spores will become gametophytes while gametes will become sporophyte*

DNA microarrays are used for what purposes? A.Magnifying nucleic acids without fixation B. Sequencing a species entire genome C. Determining gene expression within a cell D. counting the number of base pairs within a genome E. Examining chromosomes during metaphase

C. -DNA microarrays contain thousands of DNA probes that are complementary to a certain gene sequence. Scientists can take a given cell, remove its mRNA, synthesize cDNA from the mRNA, and allow the cDNA to hybridize with the DNA probes on the microarrays. If a cDNA successfully matches and hybridizes with a DNA probe on, fluorescence will be emitted. Successful hybridization indicates that part of the cell's genome matches with at leas one DNA probe, making it easy to determine the gene expression of a cell. -therefore, choices (A), (B), (D), and (E) can be eliminated and choice (C) is our correct answer -DNA microarray is a device that helps identify specific parts of the gene, not the entire genome. This further confirms that choices (B) and (D) are incorrect.

Which of the following maintains normal skeletal muscle tonus? A. Temporal summation of twitches B. Depolarization during muscle fiber relaxation C. variable recruitment of motor units D. tendon attachment of muscle to bone E. myocyte automaticity

C. -Muscle tonus (muscle tone) is the continuous partial contraction of muscle that helps maintain posture and provides resistance to movement. The constant, variable output of action potentials from the CNS provides the signals responsible for muscle tone while the stimulation of different motor units at variable times counters fatigue. Thus, the variable recruitment of motor units helps maintain normal muscle tonus. -Temporal summation of twitches is also known as wave summation, which describes the process by which contraction ends, resulting in a greater overall contraction. This occurs bc another depolarization is triggered while muscle fiber relaxation is occuring but not completed. Since this can result in fatigue, temporal summation is not responsible for maintaining normal muscle tonus. Thus, answer choices (A) and (B) can be eliminated. -Tendons do attach muscle to bone, but without the continuous partial contractions of muscle, normal skeletal muscle tonus would not be achieved. Answer choice (D) can be ruled out. -All cardiac muscle cells (cardiomyocytes) have automaticity, which allows them to initiate action potentials without external help. However, cardiac muscle has no muscle tone, answer choice (E) can be ruled out.

What mechanism allows respiratory inspiration to occur? A. Internal intercostals contract B. External intercostals relax C. Intrapleural pressure decreases D. Intrapleural pressure increases E. Thoracic volume decreases

C. -Respiratory inspriation is triggered by the contraction of external intercostal muscles and the diaphragm, which causes the rib cage to expand and the volume of the thoracic cavity to increase. This increase in volume results in the decrease of intrapleural pressure, which is the pressure in the space right outside of the lungs called the pleural space. Air then flows into the lungs bc the decrease of intrapleural pressure causes the lungs to expand. -Internal intercostal muscles cotract to compree the rib cage, resulting in respiratory exhalation. We can rule out answer choice (A). -We can rule out answer choice (B) bc we are looking for the contraction of external intercostals, not relaxation. -Intrapleural pressure increases during respiratory exhalation bc air needs to be forced out of the lungs. Thus, we can rule out answer choice (D). -Thoracic volume decreases during respiratory exhalation due to compression of the rib cage from internal intercostal muscles. Thus we can rule out answer choice (E).

Quillworts and ferns are examples of: A. angiosperms B. gymnosperms C. seedless tracheophytes D. detritus E. bryophytes

C. -Seedless tracheophytes are amongst the earliest plants that formed forests on Earth. Tracheophytes are plants with a vascular system (xylem and phloem) to transport food and water all around. Quillworts, ferns, and club mosses are signiture examples of seedless tracheophytes. -Choice (A) is incorrectbc angiosperms are flowering plants that can produce fruits. Quillworts and ferns are note part of this category. -Choice (B) is incorrect bc gymnosperms are seeded tracheophytes with their seeds exposed and unprotected. Famous examples of gymnosperms are your winter conifers (pine trees, fir trees, spruce trees). Conifers are plants that produce cones. -Choice (D) is incorrect. Detritus is decayed dead plants and animals, it is irrelevant to the categories of plants. -Choice (E) is incorrect. Bryophytes are plants without a vascular system. They are typically very short and grow in moist habitats to absorb water. Famous examples of bryophytes are mosses, hornworts, and liverworts.

If a plant contains 100 untis of energy, approx. how much of that energy would be expected to be converted into organic tissue of a secondary consumer? A. 0.001 units B. 0.01 units C. 1 unit D. 10 units E. 100 units

C. -The amount of biomass that is stored in each trophic level can be visualized using a food pyramid. In the food pyramid, the producers (make their own energy) are at the very bottom and the order goes up following primary consumers, secondary consumers, and finally tertiary consumers at the top. -These consumers must feed on organisms below their trophic level for energy. Only ~10% of the energy from one trophic level can be converted into the organic tissue or substance of the next trophic level. -Therefore, a plant (primary producer) that contains 100 units of energy will transfer approximately 10 units of energy to primary consumers (10% of 100). Subsequently, the primary consumers will transfer 1 unit of energy to the secondary consumers (10% of 10). Since 1 unit of energy is converted to the organic tissue of secondary consumers, answer choice [C] is correct.

All of the following accurately describe the dermis, EXCEPT one. Which one is the EXCEPTION? A. deep to the epidermis B. glandular C. avascular D. contains hair E. dense, irregular tissue

C. -The dermis is not avascular. In fact, it contains a lot of blood vessels, which we can see in both the papillary and reticular dermis layers. -Choice (A) can be eliminated bc the dermis is deeper than the epidermis. The order goes: epidermis, dermis and hypodermis (from superficial to deep).

Which of the following conditions is most likely to be seen in a male whose somatic cells contain 45 chromosomes? A. Downs syndrome B. Klinefelter's syndrome C. Monosomic aneuploidy D. Trisomic aneuploidy E. Barr body formation

C. -The first thing to remember after reading the question is: normal diploid humans have 46 chromosomes. A man having 45 chromosomes means that he is missing one chromosomes somewhere. Monosomic (mono-meaning one, -somic meaning soma ie. cell) aneuploidy is an abnormality where a diploid organism is missing a copy of its chromosome. -choice (A) is incorrect bc downs syndrome is a trisomy (three copies) at chromosomes 21. People with Downs syndrome would have 47 chromosomes. -choice (B) is incorrect. Klinefelters syndrome only occurs in males, it is a sex chromosome trisomy syndrome. Males suffering from this disease would have XXY instead of XY, giving 47 total chromosomes. -choice (D) is incorrect. Down's syndrome and Klinefelters syndrome are both examples of trisomic aneuploidy. In the question, we are asked about when a person is missing one chromosome, not having an extra chromosome. -choice (E) is incorrect. Barr body formation leads to X chromosome inactivation. Barr body formation, the condensation of an X chromosome only happens in females, bc their second X chromosome is able to compensate for the inactivation. While a male with Klinefelters syndrome could have Barr body formation, a male with Klinefelters syndrom wouldnt have 45 chromosomes.

Where is vitamin K produced? A. stomach B. small intestine C. large intestine D. liver E. pancreas

C. -The large intestine has 3 major functions: water absorption, mineral absorption, and vitamin production and absorption. -Our body does not produce vitamin K on its own. Vitamin K is produced by symbiotic bacteria that live in our large intestine. The large intestine houses a rich bacteria commnity which benefits rom us by having an environment to lvie in and we benefit from them through vitamin production. -The other answer choices are not where vitamin K production occurs

Which region of the heart is typically most muscular? A. Left atrium B. Right atrium C. Left ventricle D. Right ventricle E. Aorta

C. -The most muscular part of the heart will likely be the section that needs to create the most pressure. We must recall that the left ventricle pumps blood out of the entire body, so it will require the most muscle. Thus we can see that answer choice (C) is correct. -The left atrium takes oxygenated blood in from the pulmonary veins and delivers the blood to the left ventricle. We can rule out answer choice (A). -The right atrium takes low pressure blood from the body and delivers it to the right ventricle. Answer choice (B) can be ruled out. -The right venticle accepts the low pressure blood from the right atriuma nd sends it to the lungs. Thus, answer choice (D) can be ruled out. -Although the aorta has a large amount of smooth muscle to accomodate high blood pressure blood entering it, it is the left ventricle that is responsible for pumping the blood and creating high pressure. The left ventricle is more muscular and we can rule out answer choice (E).

What is the function of the rough ER? A. synthesis of lipids and steroid hormones B. packaging of proteins for exocytosis C. synthesis of proteins D. synthesis of ATP E. breakdown of nutrients, bacteria, and cellular debris

C. -The rough ER is responsible for the synthesis of proteins in eukaryotic cells -the smooth ER is responsible for the synthesis of lipids and steroid hormones. It also helps with cell detoxification. In contrast to the rough ER, smooth ER is not associated with ribosomes and protein synthesis. Answer choice (A) is incorrect. -the golgi apparatus is like the mailroom in a cellular factory; it is responsible for the packaging of proteins for export and for transport throughout the cell. The Golgi consists of flattened sac called cisternae (these look like pancakes). In general, products secreted by the golgi tend to travel to the cytosol, cell membrane, extracellular environment (via exocytosis), into a lysosome (another organelle), or into a vacuole (another organelle). Answer choice (B) is incorrect. -the rough ER is continous with the outer membrane of the nuclear envelope, which means the ER lumen is continuous with the perinuclear space. The rough ER membrane appears "rough" bc its surface contains ribosomes. Ribosomes attached to the cytoplasmic side of the rough ER translate proteins into the lumen, which mechanistically looks similar to threading a string through a button. -The synthesis of ATP in eukaryotic cells takes place in the mitochondria. Mitochondria are organelles in eukaryotic cells (plants included) that produce a lot of ATP through the use of a proton gradient that powers ATP synthase. Answer choice (D) is incorrect. -Lysosomes can breakdown a cell's unneeded/defective components (debris) in a process called autophagy. Lysosomes are membrnae-bound organelles that hydrolyze (breaks down) substances Lysosomes contain acidic hydrolytic digestive enzymes that are designed to function at a low pH. They can hydrolyze substances taken up by the cell via endocytosis), as well as break down nutrients, bacteria, and other cellular debris. Answer choice (E) is incorrect.

What do chloroplasts and mitochondria have in common? A. they generate ATP for the cell by cellular respiration B. An electron transport chain occurs along their inner membrane C. they posses their own DNA D. they have a high pH in the intermembrane space E. they synthesize carbohydrate for energy metabolism

C. -chloroplasts and mitochondria are similar in that they possess their own DNA. Another similarity they share is that they both have a double membrane. Both of these qualities are thought to be a result of the endosymbiosis. The endosymbiotic theory suggests that some membrane-bound orgranelles, such as mitochondria and chloroplasts, were actually once free living prokaryotes. Likely through means of phagocytosis, these free living prokaryotes become engulfed by other cells, and began to function as organelles within the cell. -choice (A) is incorrect because chloroplasts generate ATP via photosynthesis, not cellular respiraiton. Mitochondira generate ATP via cellular respiration. -choice (B) is incorrect because the electron transport chain of chloroplasts is found along the thylakoid membranes, not along the inner membrane of the chloroplasts themselves. The electron transport chain of mitochondria is found along the inner membrane of the mitochondria. -choice (D) is incorrect because in chloroplasts, the H+ ions accumulate in the thylakoid lumen, not the intermembrane space. The intermembrane space of chloroplasts does not have any significant function. In mitochondira, H+ ions accumulate in the intermembrane space. This however would result in a low pH, not a high pH. -choice (D) is incorrect because only chloroplasts synthesize carbs (via photosynthesis). Mitochondria produce ATP (via cellular respiration)

In prokaryotic transcription: A. transcripts are synthesized in the nucleus B. promoter regions contain TATA boxes C. RNA polymerases directly bind to DNA D. transcription factors bind to enhancers and silencers E. termination sequences contain a poly-A signal

C. -in prokaryotic (bacteria), RNA polymerases can directly bind to DNA. Recall that they require a sigma factor for specific targeting of the promoter. Eukaryotic RNA polymerases require specific transcription factors to bind DNA and recruit them to the promoter. Therefore, (C) is the correct answer. -choice (A) is incorrect bc prokaryotes do not have a nucleus - they do not have membrane- bound organelles in general. As a result, transcription is carried out in the cytosol in prokaryores. -choice (B) is incorrect bc oly promoter regions of eukaryotes contain TATA boxes, which are sequences rich in thymine and adenine base pairs. -choice (D) is incorrect. Again, transcription factors are only found in eukaryotes, not prokaryotes. -Choice (E) is incorrect bc poly-A tail is only found in eukaryotes.

At what point of the cell cycle does DNA replication occur? A. cytokinesis B. G1 phase C. S phase D. Mitosis E. G2 phase

C. -mitosis is where cell division takes place. You can also see this as the stage where the replicated DNA is separated into two separate cells, meaning the DNA has to be replicated before mitosis takes place. As you may recall, cytokinesis is the last part of mitosis where the cytoplasm divides and you officially form two separate cells. Therefore choices (A) and (D) can be eliminated. -in between each mitosis cycle is an interphase stage. This is where the cell prepares for cell division. Interphase can be divided into three separate stages: G1, S, and G2 phases -the G1 phase is the stage where the cell increases in size. It also serves as the checkpoint for the entire cycle-it makes sure everything is ready for DNA synthesis before it lets the cell go on to the next phase. Therefore choice (B) can be eliminated. -the S phase is the stage where DNA synthesis (or replication) occurs. Hint: Synthesis happens in the S phase. -the G2 phase is the stage where the cell grows rapidly, prepares for cellular division (which happens in mitosis), and replicated its organelles. Therefore choice (E) can be eliminated.

Which of the following would be most directly affected by a mutation that causes major histocompatibility complex (MHC) class II receptor to be non-functional? A. Natural killer cells B. Cytotoxic T cells (CD8+) C. Helper T cells (CD4+) D. Memory B cells E.Plasma antibodies

C. MHC class II is a type of surface receptor that presents antigens on its extracellular side. Helper T cells become activated when their T cell receptors (TCRs) bind to MHC class II on antigen presenting cells. This process is required for helper T cells (CD4+) to perform actions such as cytokine release and assisting in clonal expansion of B and T cells. -Natural killer cells do not use antigen presentation to kill cancerous and virally infected cells, so we can eliminate answer choice (A). -Cytotoxic T cells are activated by MHC class I antigen presentation, not MHC class II so we can eliminate answer choice (B). -Memory B cells can undergo activation and differentiation into plasma B cells that secrete antibodies, relying mainly on immunoglobulinand not MHC class II. These plasma antibodies can carry out their immume functions without MHC class II involvement. Thus we can rule out answer choice (D) and (E).

All of the following are structures found on angiosperms EXCEPT one. Which one is the EXCEPTION? A. Fruit B. Stamen C. Pollen tube D. Cones E. Pistil

D. -Angiosperms are flowering plants that can produce fruits. For angiosperms, their seeds are protected in their fruits. Cones are structures that are typically found in gymnosperms-seeded, vascular plants that have their seeds unprotected. Hence, cnes are the exception in the list of structures, making (D) our answer. -Choice (A) is not an exception bc fruits are signatures to angiosperms. -Choices (B), (C), and (E) are also NOT exceptions bc they are all structures found in flowers: Stamen is the male sex organ in plants; pollen tube is the structure that guides spores to the plants ovary; pistil is the female sex organ in plants.

Which of the following most accurately describes a function performed by complement system proteins? A. Inhibition of blood clotting B. Phagocytosis of pathogens C. Signals non-infected cells about viral attacks D. Membrane attack complex formation E. Vasoconstriction of local blood vessels

D. -Complement system proteins mainly tag pathogens for phagocytosis, amplify the inflammatory response by binding to mast cells for histamine release, and create the membrane attack complex (MAC) to create holes in pathogen membranes. Thus, answer choice (D) is correct. -Answer choice (B) can be ruled out bc this is not an immediate function of complement proteins, but rather a downstream effect. Complement proteins can tag pathogens, but do not phagocytose pathogens themselves. -Basophils release heparin, an anticoagulant that inhibits blood clotting, but are not part of the complement system. We can eliminate answer choice (A). -Interferons are secreted by cells that are virally infected and signals to non-infected cells to prepare them for a virus attach. Interferons are not part of the complement system so we can rule out answer choice (C). -Complement proteins do not cause vasoconstriction and (E) can be eliminated.

Which of the following correctly matches an extraembryonic structure with its primary function? A. Amnion-waste storage B. Allantois - nutrient supply C. Yolk sac-protection from physical trauma D. Chorion-develops into the placenta E. Inner cell mass- develops into the three germ layers

D. -Extraembryonic structures provide protection and nourishment to the developing fetus, and they originate from the trophoblast of the blastocyst. The 4 critical extraembryonic structures to remember for the DAT are chorion, amnion, allantois, and yolk sac. -The chorion is the outer extraembryonic that surrounds the embryo. The chorion forms the fetal half of the placenta (in placental animals), which is responsible for gas, nutrient, and waste exchange between the mother and fetus. As a result, we can see that choice (D) is our correct answer. -The amnion is the inner extraembryonic layer closest to the embryo and it secretes fluid that cushions and protects the fetus. The amnion is not involved with waste storage, so choice (A) is incorrect. -In placental mammals, the allantois stores and transports the wastes from the fetus to the placenta. The allantois does not supply nutrients. Hence, choices (B) is incorrect. -The yolk sac functions to provide nutrients to a developing placental fetus, only until the placenta is ready to take over that role. The yolk sac does not function as a physical trauma protector of the fetus, so choice (C) is incorrect. -The inner cell mass does develop into the three germ layers: endoderm, mesoderm, and ectoderm. However, the inner cell mass forms the embryo, it is not an extraembryinic structure. Choice (E) is incorrect.

A scientist examines a fertilized frog egg and observes a gray crescent and dorsal lip. The scientist can accurately conclude that the frog egg is: A. undergoing spiral cleavage B. in the morula stage C. in the blastula stage D. in the gastrula stage E. without embryonic yolk

D. -Frog embryos have an uneven distribution of yolk, which means that it has polarity, containing an animal pole and vegetal pole. -The animal pole is dark, with little yolk, and it displays rapid cleavage. The vegetal pole is pale, with abundant yolk, and divides slowly. During the fetilization of a frog embryo, the animal and vegetal poles will mix at the exact opposite side of where the sperm penetrated. -The mixing creates a gray crescent (dark color + pale color= gray) -Another structure called the dorsal lip of the blastopore forms at the site of the grey crescent. The dorsal lip acts as the structure that initiates gastrulation in frogs. Bc the scientist sees the dorsal lip, the scientist is observing an early gastrula, and choice (D) is our correct answer. Therefore, we can also see that choices (B), (C), and (E) are incorrect. -Radial cleavage results in cells aligned on the vertical axis with the top cells directly overlap the bottom cells. Spiral cleavage results in misaligned cells that deviate away from the axis, meaning that if you look at it from a top-down view, you can see that the bottom cells are shifted compared to the top cells. -Deuterostomes (which includes the amphibians like frogs) undergo radial cleavage, wheras protostomes undergo spiral cleavage. Eliminate (A), as frogs would not undergo sprial cleavage (and in fact, by gastrulation the cleavage stage has ended)

Thousands of years ago, an island housed a single species of gray crickets. Now, the same island only houses white and black crickets. Which of the following best explains this trend? A. stabilizing selection B. directional selection C. artificial selection D. disruptive selection E. macroevolution

D. -Natural selection is a theory put forth by charles darwin, which says that the best-adapted organisms will acquire their necessary resources, allowing them to produce viable and fertile offspring in other words, the survival of the fittest. The three differetnt types of natural selection are disruptive selection, stabilizing selection, and directional selection. -Distruptive selection is the form of natural selection that selects for extremes. These extremes have selective advantages in different enviornments. In our scenario, white crickets thrive in certain environments (for ex, sandy beaches), while black crickets thrive in others (for ex, the forest). Grey crickets do not blend in either environment. -In stabilizing selection, there is selection for the intermediary trait. bc we see an evolution away from intermediate grey crickets, choice (A) is incorrect. -Directional selection occurs when there are evolutionary changes that favor one extreme. For example, white crickets replacing grey crickets. Bc this scenario tells us there was an evolution to two extremes n either end of the spectrum, choice (B) is incorrect. -Artificial selection is not natural selection, instead, it is a type of selective breeding usually carried out by humans to create new species (like domesticated dog species). Therefore, choice (C) is incorrect. -Macroevolution usually looks at the entire tree of life instead of just one species. Hence, choice (E) is incorrect bc we are only looking at evolutionary changes in one cricket species.

Where does filtration take place in the excretory system? A. proximal tubule B. the ureters. C. descending limb D. renal corpuscle E. efferent arteriole

D. -Nephrons are the functional units of the kidney and are involved in four major processes: filtration, reabsorption, secretion and excretion -Filtration happens first at the glomerulus and bowmans capsul which are collectively known as the renal corpuscle -Blood enters the nephron via the afferent arteriole and enters into the glomerulus which is located right next to the bowmans capsule. This fluid in the Bowmans capsule is now called the filtrate. -The capillaries of the gloermulus exit the Bowman's capsule via the efferent arteriole. Therefore choice (E) is incorrect. -The filtrate travels from the Bowman's capsule to the proximal convoluted tubule, the loop of Henle (down the descending limb and up the ascending limb), and then the distal convoluted tubule. -The loop of Henle is a major area of reabsorption of water and solutes from the filtrate into the blood. Therefore choice (C) is incorrect. -The two convoluted tubulesare major areas where secretion of solutes from the blood into the filtrate happens. Therefore choice (A) is incorrect. -By the end of filtration, reabsorption, and secretion, the filtrate is ready for excretion. All the nephrons empty then filtrate into collecting ducts -> renal pelvies -> ureter -> bladder -> urethra -> out of body. Choice (B) is incorrect.

The polymerase chain reaction (PCR) enables scientists to: A. separate nucleic acids by size and charge B. use prokaryotic cells to synthesize proteins C. test for antigens associated with diseases D. rapidly clone millions of DNA fragments E. analyze the movements of radioactive proteins

D. -Polymerase chain reaction (PCR) by definition is a biotechnology process that quickly creates millions of copies of DNA. It does this by: *Denaturing (opening up) the DNA by heating it *allowing DNA primers to hybridize onto the open DNA *Adding nucleotides and Taq polymerases. Taq polymerase is the enzyme that incorporates nucleotides one by one to synthesize DNA. *Repeat -Choice (A) is incorrect bc this describes gel electrophoresis, which separates DNA fragments by size and charge. Smaller fragments travel further toward the positive end of the gel. -Choice (B) is incorrect for two reasons. The first is that PCR is automated-it does not require any cells. The second is that PCR synthesizes DNA fragments, not proteins. -Choice (C) is incorrect bc this describes enzyme-linked immunosorbent assay (ELISA), a technology to determine if a specific antigen exists in a person. If any of the antibodies bind to the antigen that is being tested for, the microtiter plate will change color, indicating that that person does have the disease/antigen (bc they have an antibody for it) -Choice (E) is incorrect bc this describes pulse chase experiments. This technique occurs in two phases: the pulse phase and the chase phase. During the pulse phase, a ton of radioactively labeled amino acids are added to a cell so that synthesized proteins are radioactively labeled. This is followed by the chase phase, where a ton of non-radioactively labeled aminoacids are added to a cell so that all proteins that are synthesized from this point on are not radioactively labeled. This allows the observer to see the movement of the labelled proteins.

What do Porifera, Cnidaria, and Platyhelminthes share in common? A. Closed circulatory system B. Tripoblastic tissue orgnaization C. Bilateral body symmetry D. Lack of coelom cavity E. Gastrovascular cavities

D. -Porifera (sponges), Cnidaria (hydra, jellyfish, sea anemone, coral), and Platyhelminthes (flatworms, flukes, tapeworms) represent the three phyla in the kingdom animalia that lack coelom. Now, we can calssify Platyhelminthes as being acoelomates, but we should be careful not to classify Porifera or Cnidaria as such, as they are not trilaminar organisms, and you must have three germ layers in order to be classified in terms of coelom. -the other animal phyla Annelida and most Chordates have a closed circulatory system (used a fluid medium such as blood travelling within vessels to exchange gases), so we can rule out answer choice (A). -Triploblastic tissue organization is composed of three germ layers, which Porifera and Cnidaria do not have. Thus,we can rule out answer choice (B). -Porifera have an asymmetrical body plan, Cnidaria have radial body symmetry, and only Platyhelminthes have bilateral body symmetry. This rules out answer choice (C). -A gastrovascular cavity is a digestive system with only one opening through which food and waste move through. Out of all the animal phyla, this is only present in Cnidaria and Platyhelminthes, so we can rule out answer choice (E).

Which of the following accurately describes the Bohr effect? A. Increased capacity to form myoglobin when pH is high B. Increased capacity to form carbaminohemoglobin when PO2 is low C. Increased capacity to form reduced hemoglobin when PO2 is high D. Decreased capacity to form oxyhemoglobin when PCO2 is high E. Decreased capacity to form carboxyhemoglobin when PCO is high

D. -The Bohr effect describes the conditions of high CO2 and high H+ ion concentrations due to CO2 being converted into bicarbonate anions and H+ ions via carbonic anhydrase. The H+ ions then convert oxyhemohlobin into reduced hemoglobin. Thus the Bohr effect is associated with high PCO2 and a lowered capacity to form oxyhemoglobin. -Myoglobin is used by cardiac and skeletal muscle cells to move oxygen into these cells bc it has a much higher affinity for oxygen than hemoglobin. We can rule out answer choice (A) since it is not related to the Bohr effect. -The Haldane effect,not the Bohr effect, describes the formation of carbaminohemoglobin, so we can eliminate answer choice (B). -Although reduced hemoglobin is involved in the Bohr effect, PO2 should be low, not high. We can rule out answer choice (C). -Carbon monoxide (CO) is not involved in the Bohr effect, so we can rule out answer choice (E).

What slows the automaticity of the sinoatrial (SA) node? A. Atrioventricular (AV) node B. Gap junctions C. Epinephrine D. The vagus nerve E. The sympathetic trunk

D. -The SA node is the pacemaker of the heart and can be influenced by the nervous system. The vagus nerve is an important part of the parasympathetic nervous system, which slows down the automaticity of the SA node. Thus, answer choice (D) is correct. -The AV node is located at the bottom of the right atrium and adds a delay between atrial contraction and ventricular contraction, but does not affect the rate of SA node automaticity. Answer choice (A) can be ruled out. -Gap junctions connect the cytoplasm of cardiomyocytes to allow action potential propogation throughout the heart and coordinated contraction. Answer choice (B) can be ruled out. -Epinephrine is a hormone that is part of the sympathetic nervous system and when secreted , increases heart rate. The sympathetic trump is a group of nerve fibers that are fundamental part of the sympathetic nervous system and also can increase heart rate. Thus, answer choice (C) and (E) can be ruled out.

Which of the following represents the correct order of skeletal muscle structures, from the most superficial to most deep? A. myofilament, sarcomere, endomysium B. Fascicle, epimysium, muscle fiber C. sarcoplasm, myofibril, endomysium D. perimysium, myofibril, myosin E. muscle fiber, sarcolemma, fasicle

D. -The largest is the muscle itself, which is covered in a sheath called the epimysium -next are muscle fascicles that are covered in a sheath called the perimysium. Inside of muscle fascicles are muscle fibers, which have a sheath around them called the endomysium. -Lastly, muscle fibers have strands of contractile proteins called myofibrils, which contain thin actin filaments and thick myosin filaments -Thus, we can see that answer choice (D) is correct bc it lists a skeletal muscle structure arrangement from most superficial (largest) to most deep (smallest) -Actin filaments and myosin filaments are types of myofilaments that are contained within units called sarcomeres, which make up myofibrils. Answer choice (A) can be ruled out bc its entire order should be reversed. -The epimysium should be the largest structure so answer choice (B) can be eliminated -Myofibirls are smaller than the endomysium, which covers entire muscle fibers, so answer choice (C) can be ruled out. The sarcoplasm is the cytoplasm of a muscle fiber. -The sarcolemma is the cell membrane of a muscle fiber which should be deeper than a muscle fasicle. Thus, answer choice (E) can be eliminated.

What is the primary role of the posterior pituitary gland? A. produce melatonin B. produce oxytocin C. store tropic hormones D. release direct hormones E. hypophyseal portal communication

D. -The posterior pituitary stores and releases two hormones in particular: ADH and oxytocin -ADH is an antidiuretic. It does this by inserting aquaporins into the collecting ducts of our nephrons, which allows us to reabsorb water from the urine before it is excreted. -Oxytocin is a hormone that increases uterine contraction during childbirth and also increases milk letdown during breastfeeding. -ADH targets the nephrones, and oxytocin targets the uterus and mammary glands. Bc the two hormones directly stimulate other organs, they are considered direct hormones. Therefore choice (D) is our correct answer. -The posterior pituitary glands stores and releases hormones.... it does not produce any hormones. We can eliminate choices (A) and (B) -tropic hormones are hormones that target other endocrine glands. We can eliminate choice (C). -The hypophyseal portal system describes the connection between the anterior pituitary and hypothalamus. We can eliminate choice (E).

Which part of the postsynaptic neuron directly receives signals from a presynaptic neuron? A. axon hillock B. node of ranvier C. soma D. dendrite E. myelin sheath

D. -The postsynaptic neuron is the neuron that recieves the signal at the synapse, or junction between two nerve cells, while the presynaptic neuron is the neuron that is presenting the signal. In addition, dendrites are extensions of the neuronal cell body that are the part of the neuron that recieves these signals. -The axon hillock is responsible for integrating the signals that are recieved by the neuron to let the neuron know whether to fire an action potential or not, and is located between the cell body and the axon. Thus, answer choice (A) can be eliminated. -The axon is a large extension of the neuronal cell body that sends action potentials toward another neuron, and specialized cells cover certain regions of the axon to form the myelin sheath (Schwann cells in the PNS and olgiodendrocytes in the CNS). The myeling sheath helps the signal travel faster through insulation while nodes of ranvier are uncovered regions that permits ion exchange across the membrane for action potentials to continue.Thus we can eliminate answer choice (B) and (E). -The soma ie another term for the neuronal cell body, which dendrites emerge from. It is the dendrites that recieve the signals and pass them onto the soma, so we can rule out answer chocie (C).

Which type of biome most significantly contributes to the earths atmospheric oxygen? A. freshwater biome B. taiga C. tropical rainforests D. saltwater biome E. temperate deciduous forests

D. -You must recall important attributes that distinguish Earth's terrestrial and aquatic biomes from one another. Earth's aquatic biomes are known to contribute to most of the Earth's atmospheric oxygen because they cover 3/4 of Earth's surface area and contain a large amount of photosynthetic algae. -Furthermore, we know that saltwater biomes most significantly contribute to Earth's atmospheric oxygen because only a small percentage of aquatic biomes are freshwater biomes - the majority are saltwater. -The taiga is the largest terrestrial biome and is known for being characterized by coniferous trees. -Tropical rainforests have the most diversity of organisms out of all the biomes and are littered with tall trees covering the landscape. Although rainforests do produce a large amount of oxygen, it is a fraction of what is produced by the aquatic biomes (just based on the huge size of saltwater biomes that cover the Earth). -Temperate deciduous forests are known for their seasonal changes (warm summers, cold winters) and broad plant life.

Genetic variation of gamaetes is increased during which of the following stages of cell division? A. anaphase II of meiosis B.metaphase of mitosis C. prophase of mitosis D. prophase I of meiosis E. prophase II of meisosis

D. -one of the main differences between meiosis and mitosis is that meiosis produces four unique gamete cells while mitosis produces two identical somatic cells. Mitosis does not exhibit genetic variation and do not produce gametes. This gives us two reasons to eliminate choices (B) and (C). -meiosis is broken up into two meiosis I and meiosis II. Meisosis I (specifically in prophase I) is where tetrads/bivalent form and synapse. It is during synapsis that the homologous chromosomes cross over and genetic variation occurs. Therefore choices (A) and (E) can be eliminated, and choice (D) is our correct answer. -Anaphase (regardless of whether it is for mitosis, meiosis I or meiosis II) is when the genetic material are pulled away from each other to opposite sides of the cell. Anaphase is where separation occurs. *Tip: Anaphase and away both start with the letter "a". Crossing over cannot happen when the chromosomes are being separated.

How do peptide hormones relay messages to target cells? A. exocrine signaling B. DNA binding C. voltage-gated ion channels D. secondary messengers E. direct stimulation

D. -peptide hormones are water soluble, which has two implications: 1. they can travel freely through the bloodstream (do not require a protein carrier) 2. they cannot pass through the phospholipid bilayer of thier target cells -the first implication helps us eliminate choice (A). Endocrine signaling is the secretion of hormones into the bloodstream, which is how peptide hromones travel. However, this answer choice states exocrine, which is the secretion of substances into ducts to be released outside the body. -the second implication helps us eliminate choices (B) and (E). Peptide hormones cannot bind to DNA bc they do not go into the cell. Bc they dont get into the cell, peptide hormones indirectly stimulate their target cells in one of two ways. -One way peptide hormones relay messages to their target cells is by binding to a cell surface receptor. The binding to cell surface receptors activates secondary messengers (such as cAMP, IP3, DAG and calcium ions). This is the method described in (D) which is our correct answer. -The other way is by binding to ligand gated ion channels (not voltage gated). This helps us eliminate choice (C).

What might a starfish have in common with a bird? A. spiral cleavage B. meroblastic cleavage C. holoblastic cleavage D.Indeterminate cleavage E. Determinant cleavage

D. -starfish are part of the animal phylum echinodermata and birds are part of the animal phylum Chordata. Both phyla are deuterostomes, which have indeterminant cleavage. On the other hand, the rest of the phyla in the animal kingdom are protostomes, which have determinant cleavage. -Indeterminant cleavage refers to the rapid embryonic cellular division early in development. During indeterminant cleavage, a removed blastomere is able to form a complete separate organism if separated from the rest of the cells. Compare this with determinant cleavage, which refers to a removed blastomere being unable to produce a separate organism. This leads us to the correct answer choice (D). Also, answer choice (E) can be ruled out. -Deuterostomes have radical cleavage while protostomes have spiral cleavage, so answer choice (A) ca be ruled out. -Meroblastic and holoblastic cleavage refers to how the embryo divides. Meroblastic cleavage happens in embryos with abundant yolk such as bird, and describes uneven division. In meroblastic cleavage, an animal pole (rapidly dividing) and vegetal pole (negligilble division) form within the embryos. Holoblastic cleavage occurs in embryos with little yolk such as starfish, and describes even cleavage throughout the embryo. Thus, we can rule out answer choices (B) and (C).

Which of the following is true about the lac operon? A. it generates proteins which can sythesize lactose B. it is activated by high levels of glucose and fructose C. its gene products are constitutively expressed D. It is an inducible operon that is usually inactive E. Allolactose binds to and activates the promoter region

D. -the lac operon is responsible for producing proteins that metabolize lactose in E.Coli bacteria -but remember, lactose is not the preferred energy source like glucose. Therefore, the cell wants to preserve energy and only turn on lac operon transcription when it needs to. As a result, the lac operon is a type of inducible operon which is normally inactive, and only active when turned on. Choice (D) is our answer. -choice (A) is incorrect bc it generates proteins to break down lactose, not synthesize lactose -choice (B) is incorrect. Again, cells much prefer glucose to lactose for its energy source. Therefore, to save unnecessary energy, the lac operon is rendered inactive under high levels of glucose. -choice (C) is incorrect. When a gene product is constituitvely expressed, it means that the gene is always actively going through transcription and translation, which is not the case for the lac operon. On the other hand, however, the lac operon repressor gene (lacl) is constitutively expressed. -choice (E) is incorrect bc allolactose is an isomer of lactose that binds directly to the lac repressor protein. Allolactose binding will indrectly activate lac operon transcription, it does not directly interfere with the promoter region.

Which of the following is most likely to occur if a neuron was mutated such that the voltage-gated Na+ channgels always remained open? A. Hyperpolarization B. Constant repolarization C. Faster action potential propagation D. increased intracellular concentration of K+ E. Increased consumption of ATP

E. -At resting membrane potential, sodium potassium pumps (Na+/K+ ATPase) use ATP to pump three Na+ ions out of the cell and two K+ ions into the cell. Thus, the inside of the cell is more negative than the outside, giving it a negative resting membrane potential. -To answer this question, we need to realize that voltage-gated Na+ channels are unique channels, and are separate from the Na+/K+ ATPase. -If voltage fated Na+ channels always remained open to let sodium into the cell( more positive coming into the cell), the sodium potassium pump needs to work harder to maintain its negative resting membrane potential. As the Na+/K+ ATPase is powered by ATP, there would be an increased consumption of ATP. Thus, answer choice (E) is correct. -The continous flow of sodium back into the cell would ause increased depolarization (a more positie membrane potential). Since hyperpolarization and repolarization describe membrane potential becoming more negative, answer choice (A) and (B) are eliminated. -Increased depolarization in this cell would make it easier for action potentials to be triggered but not at a quicker propagation rate. It is mainly myelination and axon diameter affect the speed of action potentials propagarion. Additionally if voltage gated sodium channels remained open, this would greatly affect the ability to send a coordinated action potential. We can eliminate answer choice (C). -More sodium inside the cell would increase the membrane potential, causing K+ ions to be more likely to leave the cell following its electrochemical gradient, so we can rule out anser choice (D)

Phototropism, the phenomenon of a plant growing toward a light source, is dependent on which hormone? A. Ethylene B. Cytokinins C. Gibberellins D. Abscisic Acid E. Auxins

E. -Auxin is the hormone that i sresponsible for various tropisms (movement to a specific stimulus) in plants. Therefore, choice (E) is the correct answer. Some common tropisms are phototropism (as described in the question), gravitropism (controlling growth directions opposed to gravity), and thigmotropism (growth on contact). -Choice (A) is incorrect. Ethylene is a caseous hormone that stimulates fruit ripening. -Choice (B) is incorrect. Cytokinins are hormones that affect cell growth and delay agining of plants. Thats why sometimes they are sprayed on cut flowers to prolong their lives! -Choice (C) is incorrect. Gibberellins generally promote plants growth. It can also cause bolting (the rapid elongation of stem) after spring rain. -Choice (D) is incorrect. Abscisic acid has the opposite effect of gibberellins. It is a growth inhibitor and maintains the seed at dormancy.

In binary fission, two identical copies of a cell are separated by a: A. cell wall B. middle lamella C. vesicle D. cleavage furrow E. septum

E. -Binary fission is a form of asexual reproduction that occurs in prokaryotes and some organelles, such as mitochondria and chloroplasts. Binary fission begins as one parent chromosome replicates into two daughter chromosomes. Then, the chromosomes move to opposite ends of the cell before a septum separates them, partitions the parent cell into two daughter cells, which then split into separate cells. Therefore, chocie (E) is our correct answer. -Cell walls cover plant, fungal, bacterial, archaeal, and some protist cells. The purpose of a cell wall is to provide a cell with structural support, protection and filter what molecules enter and exit a cell. Choice (A) is incorrect. -The middle lamella is a stickly substance that holds plant cell walls together. Choice (B) is incorrect. -Vesicles move substances throughout cells. Specifically, in plant cell division, the Golgi apparatus secretes vesicles that separate the two daughter nuclei during telophase. Ultimately these vesicles fuse during cytokinesis to form a new cell wall that separates the two daughter cells. However, this does not occur in binary fission, so choice (C) is incorrect. -A cleavage furrow forms in eukaryotic animal cells to separate the cytoplasm of the two daughter cells through cytokinesis. Bc we are talking about the sexual reproduction of prokaryotic cells (and specific organelles like mitochondria and chloroplasts) in binary fission, choice (D) is incorrect

A bell is rung every time a dog is given food. The dog eventually becomes classically conditioned to salivate whenever it hears a bell ring. Which of the following can be concluded about this scenario? A. Food: conditioned stimulus B. Food: neutral stimulus C. Bell ring: unconditioned stimulus D. Bell ring: unconditioned response E. Salivation: conditioned response

E. -Classical conditioning involves pairing a neutral stimulus with an unconditioned stimulus so that the neutral stimulus ends up eliciting a physiological response. -Initially, the bell ring is referred to as the neutral stimulus, which causes no physiological response in the dog. The unconditioned stimulus is represented by the food, which does cause a physiological response (salivation) without prior classical conditioning. -After classical conditioning, the neutral stimulus becomes a conditioned stimulus, which causes salivation. In addition, the salivation is known as the conditioned response because it is elicited by the conditioned stimulus. Hence, we can see that answer choice [E] is correct. -As mentioned previously, the food acts as the unconditioned stimulus because no classical conditioning is required to link this stimulus with a physiological response. Thus, we can eliminate answer choices [A] and [B]. -The bell ring is a neutral stimulus that becomes a conditioned stimulus through classical conditioning as discussed earlier. We can eliminate answer choice [C] and [D].

Which of the following offspring indicates that they came from the fittest organism? A. 2 out of 3 offspring survive B. 5 out of 8 offspring survive C. 5 out of 10 offspring D. 7 out of 10 offspring survive E. 17 out of 20 offspring survive

E. -Fitness is an organism's ability to survive to produce offspring that are viable (meaning they live) and fertile (meaning they can reproduce). In this way, the fittest organisms are those that produce the most offspring that survive to sexual maturity, such that those offspring can have children of their own. -We need to look for the answer choice that has 1) the most offspring that survive and 2) the best ratio of living to nonliving offspring. In this way, we see that choice (E) came from the fittest organism bc this choice yielded the most living offspring (17) in the best ratio of living to non-living offspring (17/20= 8.5/10=0.85). Choice (E) is correct.

Which of the following best describes the relationship between fungi and slugs? A. both break down living mater B. slugs consume fungi C. both are considered detritivores D. fungi are considered detritivores; slugs are considered saprophytes E. fungi break down dead matter; slugs consume it

E. -Fungi are an example of saprophytes, a type of decomposer that breaks down dead organic material and converts it into detritus (organic wastes). On the other hand, slugs are an example of detritivores, a type of decomposer that consumes detritus to release even more dead organic material for saprophytes to decompose. Thus, answer choice [E] is correct because it best fits the above descriptions.

Which cell type is responsible for secreting gastric juice? A. Kupffer cells B. Goblet cells C. Chief cells D. Mucous cells E. Parietal cells

E. -Gastric juice (hydrochloric acid, HCl) is released into the stomach when food reaches the stomach. It is extremely acidic and helps to chemically break down food and activate enzymes. -Choice (A) is the irrelevant answer choice. Kupffer cells are phagocytes that destroy old/ useless red blood cells and bacteria. They are involved in maintaining the blood stream as a function of the liver. Kupffer cells are not directly involved in the digestive system. -Choice (B) is incorrect bc goblet cells secrete mucus in the small intestine. Chief cells, mucous cells, and parietal cells all contribute to the digestion that happens in the stomach. Howeber, choice (C) is incorrect bc chief cells secrete gastric lipases and pepsinogen, not gastric juice. -Choice (D) is incorrect bc mucous cells secrete mucus

A Gram-positive bacterial cell will have which of the following features? A. Pink after counterstain B. Thin peptidoglycan C. Second membrane D. Lipopolysaccharide E. Teichoic acids

E. -Gram Staining is used to distinguish Gram-positive bacteria from Gram-negative bacteria by detecting the presence of peptidoglcan in the bacterial cell wall. Gram- positive bacteria have a much larger amount of peptidoglycan in their cell walls than gram negative bacteria. Teichoic acids work together with the peptidoglycan in Gram-Positive bacteria to give support and connect the cell wall to the plasma membrane. Thus, answer choice (E) is correct. Also, answer choice (B) can be ruled out. -Gram positive bacteria will be purple after staining with dye while gram negative bacteria will be pink from a second staining called the counterstain. This rules out answer choice (A). -Next, you must remember that Gram-negative bacteria have a second membrane on the outside containing lipopolysaccharide (LPS). Gram-positive bacteria do not have this outer membrane so we can rule out answer choices (C) and (D).

Which of the following structures are connected by the plantar fasica, the longest ligament in the foot? A. Endosteum and periosteum B. Endosteum and heel bone C. Lumbrical muscles and talus bone D. Soleus muscle and metatarsal bones E. Heel bone and metatarsal bone

E. -Ligaments connect bones to other bones -The question itself tells you that the plantar fascia is a ligament, and bc we know that ligaments connect bones to other bones we can just junt for an answer choice that satisfies this. -The endosteum and periosteum are membranes made up of fibrous connective tissue. The endosteum is located between cortical and cancellous (spongy) bone. On the other hand, the periosteum covers all cortical bone except at the joints of long bones. They are not structures connected by the plantar fascia, so answer choices (A) and (B) are incorrect. -Recall that tendons connect muscles to bone, not ligaments. Thus, answer choices (C) and (D) can finally be eliminated.

Which of the following describes normal breathing? A. Vital capacity B. Functional residual capacity C. Expiratory reserve D. Residual volume E. Tidal volume

E. -Normal breathing refers to resting inhalation and exhalation without extra effort put into breathing. By definition, tidal volume is the volume of air displaced during normal breathing. -Vital capacity describes the max volume of air that can be exhaled after doing a max inhalation or, in other words, how much air your lungs have to work with. -After a normal exhalation, the volume of air remaining in the lungs is the functional residual capacity. If a maximal exhalation is performed following the normal exhalation, the volume of air exhaled is known as the expiratory reserve volume. In addition, after a maximal exhalation , the volume of air left in the lungs to keep it from collapsing is known as the residual volume. None of these terms characterize normal breathing, so we can rule out answer choices (B), (C) and (D).

Which of the following would be used to visualize the amount of protein in a sample? A. a hemocytometer B. A DNA probe C. Northern blotting D. Southern blotting E. Western blotting

E. -The quickest way to answer this question is remembering the mnemonic SNOW DROP. This will help us remember that Southern blotting is associated with DNA, and Northern blotting is associated with RNA. Therefore we can eliminate choices (C) and (D). -SNOW DROP will also help us remember that proteins are associated with Western blotting. Therefore choice (E) is our correct source. -A DNA probe is a labeled tool that hybridizes with complementary DNA sequences and helps observers identify those sequences. It binds to DNA sequences, not proteins. Therefore we can eliminate choice (B). -A hemocytometer is a device that is used to manually count red blood cells. Therefore we can eliminate choice (A).

What trait would a lancelet, shark, kangaroo, human, and alligator all have in common? A. Four chambered heart B. Veretebrae C. Bony Skeleton D. Homeothermy E. Notochord

E. -These animal are all examples of Chordates, an animal phylum that is distinguishable from the other animal phyla based on 4 features: Notochord, Dorsal hollow nerve cord, Pharyngeal gill slit, and Muscular post-anal tail. -Lancelets are chordates that do not have a heart, vertebrae, or a true skeleton. Thus we can rule out answer choices (A), (B), and (C). -Homethermy refers to animals that retain a stable internal body temperature no matter what the edxternam temperature is. Reptiles such as alligators are known to be cold blooded or poikilothermic, having internal temperatures that change along with the environment. We can eliminate answer chocie (D).

What role do zymogens play in the human digestive system? A. Support mechanical emulsification B. Production of epithelial cells C. Digest starches into simple sugars D. Postpone gastric emptying E. Prevent cellular autodigestion

E. -Zymogens are the inactive form of an enzyme. They are converted to their active form when they are in the ideal conditions to be active. -For example, Pepsinogen is the zymogen of the enzyme pepsin (active form). Pepsin is a protease that is released to the stomach to digest proteins-however pepsinogen is inactive so it cannot digest anything. This inactive enzyme precursor is stored within cheif cells, and secreted as pepsinogen. It is only converted to the active form pepsin when it comes into contact with stomach acid. This way , the enzyme will only start digesting when it encounters food in the stomach acid. If it were stored within cheif cells in its active form (as pepsin) it could start digesting proteins within the very cells that are secreting it (which would be bad). -Zymogens are inactive and therefore do not perform any mechanical or chemical digestion. This helps us eliminate choices (A) and (C). -Choice (D) describes bile and pancreatic lipase. These two emulsify fats, which is an example of mechanical digestion. Bile is a fluid, and pancreatic lipase does not have a zymogen form. -Choice (C) describes salivary amylase and pancreatic amylase. They chemically break down start to maltose. Neither of those enzymes have a zymogen form. -Choice (B) is incorrect bc it describes a function of crypt cells. -Choice (D) is incrrect bc it describes one role played by the bormone cholecytokinin.

Which of the following is a functional limitation that will encourage cell division? A. A very large genome/volume ratio B. Cell entry into the G0 phase C. Density dependent inhibition D. A very large surface area/ volume ratio E. a very large cell volume

E. -choice (A) is incorrect bc as a cell grows, the genome/volume ratio decreases. This is because a cell's genome will never increase-only the volume increases. As a cell grows larger (and as its genome/volume ratio decreases), the cell starts to become too large for its genome to regulate its cellular activities. At this point, the cell will divide so that the new cells will have a more favorable genome/ volume ratio that will allow them to regulate their activities. -therefore a small genome/volume ratio means a bigger cell, which means the cell is encouraged to divide. On the flip side, a large genome/volume ratio means a smaller cell, which means that the cell does not have to divide to survive. Thereore choice (A) can be elminated. -choice (B) is incorrect because G0 is the phase that cells go into when conditions are not favorable for cell growth and cell division. Therefore if a cell enters the G0 phase, it means that the cell is prevented from cell division. -choice (C) is incorrect because density dependent inhibition describes the phenomenon where cells stop dividing when the surrounding cell density reaches a max. -choice (D) is incorrect because a large surface is needed for the cell to be able to easily interact with the environment and survice. A large surface area/ volume ratio is favorable for the cell and the cell does not need to divide to survive. Therefore choice (D) can be eliminated. -along that same logic, a very large cell volume would make for a small surface area/ volume ratio. This is not favorable for the cell, and would encourage the cell to divide. therefore choice (E) is our correct answer.

In an experimental animal, a null mutation is introduced into one allele of a p53 gene. What would be expected to be found in the animal after this experimental modification? A. a proto-oncogene B. A malignant tumor C. An increase in DNA replication errors D. A gain in function mutation E. Tumor suppression

E. -p53 is an important tumor suppressor gene, that plays a variety of roles in reducing the changes of developing cancer. This ene can be subject to a loss-of-function mutation, a mutation that decreases or elimates a gene product or gene function. Keep in mind that tumor suppressor genes are haplosufficient, meaning that both alleles need to be nullified for loss-of-function to occur, also known as the "two hit hypothesis". In this questions scenario, only one copy is rendered useless, so that p53 gene would still function normally, hence resulting in continued tumor suppression. -Proto-oncogenes regulate the cell cycle and are haplosufficient. Choice (A) is incorrect as p53 is a tumor suppressor gene, not a proto oncogene. They are two separate things! -choice (B) is incorrect. We said that p53 would still be functional even with one copy knocked out. Therefore, it will be functioning to prevent malignant tumor formation. -choice (C) is incorrect. There wouldnt be an increase in DNA replication errors if p53 is doing its job normally. -choice (D) is incorrect. Tumor suppressor genes are always related to loss-of-function mutations, it is proto-oncogenes that are prone to gain-of-function mutations (in increase in gene activity leading to cancer promotion).

Two organisms from different species mate, and the offspring of their offspring have reduced fitness. This is an example of: A. hybrid mortality B. behavioral isolation C. bottleneck effect D. disruptive selection E. hybrid F2 breakdown

E. -Prezygotic and post zygotic mechanisms are in place to ensure reproductive isolation between different species. In prezygotic isolation, fertilization cannot occur between the two different species do mate. The mating of two separate species is known as hybridization. If the children of a hybridization event lives (is viable), then they are known as F1 individuals. -If two F1 hybrids reproduce (fertile), there is a good chance the F2 generation will have reduced fitness bc of new and untested allele combinations. The above mechanism describes hybrid F2 breakdown, which is a postzygotic isolation mechanism that assures reproductive isolation between the original parent (P) generation containing the two different species. Therefore, choice (E) is our correct answer. -Hybrid mortality is a form of postzygotic isolation. Here the original F1 hybrid is not viable, so it will not survive to reproductive age. Hence, Choice (A) cannot lead to the production of an F2 generation, so it is incorrect. -Behavioral isolation is a form of prezygotic isolation where the two parent species have different mating rituals, preventing the formation of an F1 hybrid. Therefore, chocie (B) is incorrect bc if an F1 hybrid cannot form, an F2 hybrid cannot form either. -The bottleneck effect refers to the genetic drift that occurs as disasters kill of the majority of a population, therefore reducing (or completely eradicating) specific alleles in an abrupt fashion. Choice (C) is incorrect bc it is not a reproductive isolation mechanism. -Disruptive selection is the form of natural selection that selects for extremes. These extremes have selective advantages in different environments. The hybrid F2 generation from out question stem has reduced fitness, so choice (D) is incorrect.

Why do photosynthetic plants split water molecules? A. to produce NADPH in the calvin cycle B. to increase Gibb's free energy (deltaG) C. to allow protons to accumulate in the stroma D. to generate the final electron acceptor of photosynthesis E. to release electrons that become energized

E. -the splitting of water molecules (also known as photolysis) happens during photophosphorylation (noncyclic). This eliminates choice (A) because photolysis does not happen during the calvin cycle. -photolysis produces H+ion, electrons and oxygen -the electrons that are rleased are energized by photosystem II and help transort protons into the thylakoid lumen as they move down the electron transport chain. This helps us confirm that choice (E) is our correct answer -the splitting of water molecules also generate protons, adding to the electrochmical protons, adding to the electrochemical proton gradient in the lumen that powers ATP synthase. ATP synthase uses the energy from the proton movement back into the stoma (negative gibbs free energy) to convert ADP into ATP (positive gibbs free energy). Overall, choice (B) is incorrect bc the splitting of water molecules provides the energy necessary to power the increase in Gibbs free energy caused by the production of ATP. -choice (D) is incorrect bc the final electron acceptor of photosynthesis is NADP+ which is not a product of photolysis