DAT Booster Practice Exam #1 BIOLOGY Flashcards
Which of the following will occur when a red blood cell is placed in a dish of pure water that has been distilled of all solutes?
burst
Tonicity is defined as the relative concentration of two solutions that are separated by a semipermeable membrane. The two solutions being compared are often the cytoplasm within a cell and the extracellular environment. There are three terms used to categorize tonicity: hypertonic, isotonic, and hypotonic. Concerning the question, when the red blood cell is placed in distilled water, the distilled water is hypotonic relative to the cell. This means that the distilled water has a lower solute concentration than the cytoplasm within the cell. As a result, water will diffuse into the cell in an attempt to equalize the solute concentration on both sides of the plasma membrane which will cause the cell to swell and eventually lyse. Therefore, Option B. Burst is the correct answer. Here is an illustration to help visualize this concept:
CC BY 4.0
Process of Elimination
Option A. Plasmolysis – This option is incorrect because plasmolysis is the detachment of a cell wall due to the cell being placed in a hypertonic solution. This process can only occur in cells with a cell wall, for example, plant cells. Because red blood cells don’t have a cell wall and the red blood cell in question is being placed in a hypotonic solution, not a hypertonic solution, this option is incorrect.
Option C. Shrivel – This option is incorrect because when a cell is placed in a hypotonic solution, water will diffuse into the cell resulting in the swelling or bursting of the cell (cytolysis). Remember that when a cell is placed in a hypotonic solution, the concentration of solutes inside the cell is greater than the concentration of solutes in the extracellular environment. As a result, water will diffuse into the cell via osmosis in order to equalize the concentration gradient. A cell would shrivel if it was placed in a hypertonic solution.
Option D. Formation of an endospore – This option is incorrect. Recall that an endospore is a dormant and highly resistant structure that bacteria form under times of stress. Endospores protect the genetic material of the cell in the presence of extreme temperatures, lack of nutrients, and harmful chemicals. Endospores are primarily formed from gram-positive bacteria but can also be produced by some gram-negative bacteria. This question does not specifically state that the cell in question is a bacterium. A hypotonic solution would also not lead to the formation of an endospore; therefore, this option is incorrect.
Option E. Remain the same –This option is incorrect. A red blood cell would remain the same size and morphology in an isotonic solution, not a hypotonic solution. In an isotonic solution, the concentration of solutes is equal on both sides of the semipermeable membrane, therefore, there would be no net movement of water into or out of the cell. The question specifically asks what would happen when a red blood cell is placed in a hypotonic solution, not an isotonic solution. In a hypotonic solution, water would enter the cell via osmosis and cause the cell to swell and burst (cytolysis).
A significant spike in which hormone causes ovulation?
LH
The menstrual cycle can be broken down into 2 distinct phases: the follicular phase, and the luteal phase. These phases are separated by the event known as ovulation. While there is no beginning or end of the menstrual cycle (because it is a cycle after all), we will regard the beginning of the phase to be the start of the follicular phase.
The menstrual cycle can be broken down into 2 distinct phases: the follicular phase, and the luteal phase. These phases are separated by the event known as ovulation. While there is no beginning or end of the menstrual cycle (because it is a cycle after all), we will regard the beginning of the phase to be the start of the follicular phase.
The follicular phase begins with the release of gonadotropin-releasing hormone (GnRH) from the hypothalamus. This tropic hormone causes the release of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) from the anterior pituitary. The main function of FSH is to stimulate the development of ovarian follicles, one of which will become the Graafian follicle which is where an egg will be released to be fertilized. The main function of luteinizing hormone is to cause ovulation.
During the follicular phase, the levels of estrogen continually rise. These rising estrogen levels cause the endometrium, a mucous membrane lining the uterus, to thicken in preparation for the implantation of a fertilized egg (more specifically, the blastocyst). The increasing estrogen levels also lead to a spike in luteinizing hormone, known as the LH surge. This surge in luteinizing hormone is what ultimately leads to ovulation. Therefore, answer Option B. LH is the correct answer.
A detailed diagram is shown below to help visualize hormonal levels and endometrial thickness through each phase of the menstrual cycle.
Process of Elimination
Option A. FSH – This option is incorrect. While there is a small spike in follicle-stimulating hormone (FSH) levels during ovulation, it is the surge of luteinizing hormone (LH) that directly causes ovulation. Recall that FSH is responsible for the development of follicles in the ovary.
Option C. Estrogen – This option is incorrect. Recall that estrogen levels increase during the follicular phase of the menstrual cycle. This increase in estrogen levels is responsible for the thickening of the endometrium in preparation for the implantation of the fertilized egg. The rising estrogen levels also lead to positive feedback on the anterior pituitary to release luteinizing hormone (LH) and follicle-stimulating hormone (FSH). FSH is responsible for the development of follicles in the ovary. The LH surge is responsible for inducing ovulation. An estrogen spike does not directly cause ovulation, the LH surge does, therefore, this option is incorrect.
Option D. Progesterone – This option is incorrect. Recall that after an egg is released from the Graafian follicle during ovulation, the leftover follicle develops into the corpus luteum. The corpus luteum is responsible for releasing progesterone and some estrogen. Progesterone and estrogen maintain the endometrium lining during the luteal phase of the menstrual cycle. Progesterone levels do not significantly increase until after ovulation, during the luteal phase. Progesterone does not cause ovulation; therefore, this option is incorrect. because progesterone increases after ovulation during the luteal phase.
Option E. hCG – This option is incorrect. Recall that human chorionic gonadotropin (hCG) is released from the placenta after a blastocyst has implanted into the uterine wall. hCG is responsible for maintaining the corpus luteum which continues to release progesterone and estrogen in order to maintain the endometrium. hCG is not released until after implantation, therefore, this option is incorrect.
Adaptive radiation is best described as __________.
Adaptive radiation is a process in which multiple diverse species evolve rapidly from one ancestral species. It occurs when members of the ancestral species start occupying different niches with different environmental conditions, triggering these members to evolve in order to adapt to their new environment. Adaptive radiation can occur with or without the presence of a geographical barrier. As long as there are different ecological conditions (niches) that require separate traits necessary for survival, speciation via adaptive radiation is possible
The most well-known example of adaptive radiation is finch speciation on the Galapagos islands. As the ancestral species of finch fled the mainland and began occupying separate islands, they began evolving to accommodate the new ecological conditions of these islands. This led to the creation of new species that arose from one single ancestral species.
Therefore, Option A. The emergence of multiple lineages from a single ancestral species best describes the concept of adaptive radiation. The image below shows an example of the Galapagos Finches. It can clearly be seen how different species evolved from one species depending on the niche occupied.
The majority of energy in an ecosystem is lost due to __________.
The majority of energy in an ecosystem is lost due to heat. Approximately 90% of the energy at every trophic level is lost as heat due to metabolic processes or to detrivores when they die. Therefore, only about 10% of the energy is passed onto the next trophic level. For example, let’s say there is 1000 kJ of energy amongst the primary producers. About 90% of that energy will be lost at that trophic level, leaving only 10%, or 100 kJ of energy for the next trophic level (primary consumers). Another 90% of energy is lost at the primary consumer level, leaving only 10 kJ of energy for the secondary consumers. 90% of energy is then lost again, leaving only 1 kJ of energy for tertiary consumers. As a result, there is much more biomass at the lower trophic levels to accommodate for this loss of energy. After all, organisms at the higher trophic levels (including humans) would not be able to survive without the abundance of biomass at the lower trophic levels.
Organisms at the lowest trophic level are known as primary producers. The vast majority of primary producers are photosynthetic autotrophs that use photons from the sun’s rays to produce organic material from carbon dioxide and water. Keep in mind, photosynthetic eukaryotes have mitochondria in addition to chloroplasts in order to perform photosynthesis as well as cellular respiration. 90% of the energy created via cellular respiration is lost to heat while 10% is absorbed by primary consumers (herbivores) when they eat primary producers. Once again, 90% of this energy is lost to metabolic processes necessary for survival (transportation of nutrients, cellular growth/repair, maintenance of core body temperature, etc.). The 10% of energy left is absorbed when secondary consumers (primary carnivores) consume the primary consumers. This process is repeated again when tertiary consumers (secondary carnivores) consume secondary consumers. Organisms at the highest trophic level are known as apex predators (organisms that are not preyed upon by any other organism). Apex predators contain the lowest amount of biomass among the whole ecosystem (which is why the population of apex predators is so low compared to any other trophic level). Finally, decomposers consume and recycle dead matter from each trophic level and recycle the nutrients through the ecological pyramid.
A graphic displaying the energy transfer between trophic levels is shown below. This graphic is known as an ecological pyramid or food pyramid.
Adipose tissue would be classified as which of the following tissue types?
There are four categories of tissues:
- Epithelial (outer layer of the epidermis, covering of organs)
- Connective tissue (blood, bones, and components of the dermal layer of skin)
- Muscular (cardiac, skeletal, and smooth muscles)
- Nervous tissues (Brain, neurons, and the spinal cord)
Out of these categories, adipose tissue belongs to the “connective tissue” category. Remember, the main role of adipose tissue is to store energy in the form of fat, although it also cushions and insulates the body. Be sure to know the functions of the remaining types of tissues for the DAT! Here is an illustration of the different types of tissues in the human body to help visualize this concept:
Enzymes are macromolecular biological catalysts that accelerate chemical reactions. Enzymes can either be proteins or RNA macromolecules (ribozymes). They allow reactions to occur faster by taking an alternative reaction pathway that has lower activation energy. As a result, enzymes allow equilibrium to be achieved at a faster rate. It is vitally important to remember that enzymes only change the energy of the transition state of a reaction. Enzymes are not capable of making a non-spontaneous reaction into a spontaneous one. Enzymes are also not used up during a reaction, therefore, they are not considered reactants. Instead, enzymes have an active site where reactants are able to bind. Enzymes are substrate-specific and can only be bound to only by certain substrates.
Always remember, an enzyme does NOT change the free energy (ΔG) of a reaction! Although it does NOT change the equilibrium state or the enthalpy of formation (ΔH), it does INCREASE the rate constant of the forward and reverse reactions! Here is an illustration to help visualize this concept:
CC BY 4.0
Notice in the graphic that the Gibbs free energy of the products and reactants is the same whether or not an enzyme is present. This explains why the change in enthalpy (ΔH) and change in Gibbs free energy (ΔG) remains constant with or without an enzyme. The only aspect of the reaction that is changed is the amount of activation energy required for the reaction to proceed.
For the DAT, it’s also important to know that enzyme activity can be affected by other molecules like inhibitors (molecules that decrease enzyme activity) and activators (molecules that increase activity). An enzyme’s activity decreases markedly outside its optimal temperature and pH, and many enzymes are denatured when exposed to excessive heat, losing their structure and catalytic properties.
Enzyme kinetics is a high-yield topic on the DAT so be sure to understand the details of how enzymes function for the test.
Process of Elimination
Option A. Does not change the rate at which equilibrium is achieved – This option is incorrect as an enzyme lowers the activation energy of a reaction which means that less energy is required for the reaction to reach equilibrium. This allows equilibrium to be reached at a faster rate.
Option B. Increases the activation energy – This option is incorrect because an enzyme lowers the activation energy, thus making the reaction faster. If the activation energy was increased, the reaction would proceed at an even slower rate.
Option C. Affects the overall energy change of the reaction (ΔG) – This option is incorrect because an enzyme does not affect the free energy change of a reaction (ΔG). Instead, enzymes increase the rate constant of the forward and reverse reactions.
Option D. Ensures that the final state of the reaction is the same as the initial state – This option is incorrect as an enzyme does not affect the products and reactant concentrations or free energy in a reaction.
For the DAT, it is important to know the order that blood flows from the heart and how it returns:
Aorta → Arteries → Arterioles → Tissue Capillaries → Venules → veins → Superior and Inferior Vena Cava
Arteries contain the highest blood pressure that fluctuates due to systole and diastole, which is generated by the heart when it pumps out blood and refills. Veins have the lowest blood pressure and thus, require valves to prevent the backward flow of blood.
Furthermore, the propagation of blood within veins is permitted by the skeletal muscle pump. The skeletal muscle pump pushes blood through veins via the constriction of skeletal muscles. As muscles constrict, they apply pressure to the walls of veins that push blood forward toward the heart
Natural selection
refers to the increase and decrease of allele frequencies to improve the fitness of a species. Organisms that are best suited for their environment will be able to survive and pass down their traits to offspring. There are three types of natural selection that you need to know for the DAT: stabilizing selection, directional selection, and disruptive selection.
Stabilizing selection
is observed when the intermediate traits are favored. In this question, stabilizing selection would be observed if an average moth was neither very dark nor very light but rather, an intermediate shade.
Directional selection is
observed when evolution begins to favor one extreme. In this question, directional selection would’ve been observed if the moths started evolving in a way that all of them became very dark.
Disruptive selection
Disruptive selection occurs when the environment favors traits on either extreme, selecting against intermediate traits. In the question, the moths were originally an intermediate shade of grey, but after a logging event in their environment, the moths were either light or dark grey, suggesting disruptive selection.
Bottleneck Effect
when a population decreases in size dramatically
makes population susceptible to genetic drift
Founder Effect
describes a change in allele frequency due to migration of individuals to a new location
can leave a population vulnerable to genetic drift
the cortical reaction directly functions by
formation of an impenetrable fertilization membrane
During sexual reproduction, there are mechanisms in place to prevent polyspermy. Polyspermy occurs when
multiple sperm fertilize one egg
results in an inviable zygote
fast block - to prevent polyspermy steps:
immediate, temporarily blocks additional sperm from fertilizing the egg
when sperm has fused with eggs plasma membrane, Na+ channels open and allow for the diffusion of Na+ into the egg cell
depolarization of the egg cell prevents any other sperm from binding with the egg cell shell.
only for a few seconds
the cortical reaction is part of the ________ component of polyspermy
slow block
describe slow block of polyspermy prevention
during slow block, calcium is released into the eggs plasma membrane
the calcium ions cause a release of cortical granules which inactivate the zona pellucida and render it impenetrable
cortical granules also separate the zona pellucida from the plasma membrane
this prevents any other sperm from reaching the egg’s plasma membrane
much longer lasting mechanism to prevent polyspermy
It’s important to keep in mind that the fast and slow block to polyspermy occurs after the
acrosomal rxn
which is the process by which a sperm cell will penetrate the zona pellucida in order to fuse with the egg’s plasma membrane.
Hydrolytic enzymes are released from the sperm cell during the acrosomal reaction. These hydrolytic enzymes are responsible for ‘chewing’ through the zona pellucida and allowing the sperm to fuse with the egg’s plasma membrane. The cortical reaction only occurs after this process leads to fertilization.
Glycolysis –
glycolysis is the metabolic pathway that converts glucose to pyruvate. It is the first pathway that is observed during cellular respiration. The process of glycolysis converts one glucose molecule into two pyruvate molecules, yielding two ATP and 2 NADH in the process. Glycolysis does not involve the transfer of protons across a membrane
Krebs Cycle
The Krebs cycle is the process that occurs immediately after the pyruvate manipulations. Recall that the Krebs cycle is the process by which two acetyl-CoA molecules generate 4 CO2, 6 NADH, 2 FADH2, and 2 GTP. The Krebs cycle does not involve the transfer of electrons across a membrane; therefore, this option is incorrect.
Fermentation –
Fermentation is an anaerobic pathway used to oxidize NADH into NAD+ so the process of glycolysis can continue. Fermentation does not involve the transfer of protons across a membrane
Pyruvate dehydrogenase complex –
the pyruvate dehydrogenase complex is a complex of three enzymes that converts pyruvate into Acetyl-CoA, which can later be utilized in the Krebs cycle.
Blood is composed of
plasma, buffy coat, and red blood cells, which make up 55%, less than 1%, and 45% of total blood volume respectively.
Contained within the buffy coat, are
white blood cells and platelets, also known as thrombocytes.
White blood cells fall into two groups:
- Granulocytes have granules in their cytoplasm. They include neutrophils, eosinophils, basophils.
- Agranulocytes do not have granules in their cytoplasm. They include monocytes and lymphocytes.
Recall that white blood cells (also known as leukocytes) are responsible for protecting the body against
foreign pathogens
Platelets (thrombocytes) are responsible for forming the
platelet plug during the first step of the blood clotting cascade
Podocytes, which are cells found in the
Bowman’s capsule
The Bowman’s capsule, along with the glomerulus, are crucial structures of the nephron that are responsible for filtration. Podocytes wrap around the glomerulus, forming slits that filter substances based on size. Smaller molecules (mostly water and small solutes) are allowed passage while larger molecules (blood cells, proteins, etc.) are prevented from entering and remain in the bloodstream.
The ventral body cavity contains the
thoracic cavity and abdominopelvic cavity
The dorsal cavity contains the
cranial cavity and vertebral cavity.
The thoracic cavity contains the
heart, lungs, esophagus, trachea, and thymus.
The diaphragm is the lower border of the thoracic cavity and is the dividing structure between the
thoracic cavity and the abdominopelvic cavity.
During inhalation, the diaphragm contracts and increases the volume of the thoracic cavity while simultaneously decreasing
the pressure of the cavity.
Photosynthesis is a process that is required for the sustainability of all life on earth. What are the products of the light-dependent reactions of photosynthesis?
Photosynthesis is an autotrophic process that utilizes
photons from the sun to create glucose , the fuel source for the organism
during photosynthesis, CO2 and water are used to create
sugar and atmospheric oxygen with the help of sunlight
equation for photosynthesis
(T/F) photosynthetic organisms contain mitochondria
true
it converts the glucose made from photosynthesis into ATP for energy
photosynthesis releases atmospheric oxygen via the
photolysis of water
The process of photosynthesis begins when photons from the sun are captured from
photosystem II in the thylakoid membrane
when this occurs, a water molecule will be split into 2 protons, 2 electrons, and 1/2 O2
the 2 electrons will be passed tp P680, the rxn center chlorophyll molecule of photosystem II
After the 2 electrons will be passed tp P680, the rxn center chlorophyll molecule of photosystem II, the electrons will get excited to be passed through an ETC to _________
photosystem I
while the electrons are traveling down the ETC
protons are being pumped into the thylakoid lumen from the stroma creating an electrochemical gradient
the gradient is relieved when these protons travel through ATP synthanse, phosphorylating ADP to create ATP via chemiosmosis
The electrons mentioned previously are now passed to P700, the reaction center chlorophyll molecule of photosystem I. The 2 electrons become excited once again and travel down another ETC until they combine with NADP+ and H+ to form NADPH, completing the process of non-cyclic photophosphorylation.
The second stage of photosynthesis is the production of glucose from CO2 via the light-independent reactions (Calvin Cycle).
A marine biologist discovers an organism in the ocean with the following characteristics: bilateral symmetry, body segmentation, and multiple pairs of jointed appendages. Which of the following animal phyla would this organism best be categorized as?
As shown in the image below, the general order of stages in pre-embryonic development is the following:
- A zygote results after the fertilization of an egg and sperm cell that occurs inside the oviduct (also known as the fallopian tube).
- Upon fertilization, the zygote will begin rapidly dividing via mitosis in a process called cleavage, which results in the formation of the morula (a solid ball of 16-32 cells called blastomeres).
- As the morula travels towards the uterus, the cells continue to divide until it forms a hollow ball of cells called the blastula with a liquid-filled cavity called the blastocoel.
- In humans, the blastula then divides further to form two layers: the trophoblast and inner cell mass. Once the inner cell mass has formed, the blastula is now considered the blastocyst. (Note, in humans, the blastocyst is the sole term for this structure. ‘Blastula’ is not a term used during human embryonic development).
- The blastocyst then implants itself within the endometrium (the inner uterine lining) for further embryological development to occur.
For the DAT, it is important to know these stages of pre-embryonic development. Be sure to review the following figure
autosomal dominance
all affected individuals must have at least one affected parent
if the 2 parents are unaffected, all offspring must be affected (homozygous recessive)
if 2 parents are affected, they must have offspring who are unaffected (if both parents are heterozygous)
autosomal recessive
If two parents show a trait, all children must also show the trait (homozygous recessive)
An affected individual may have two normal parents (if parents are both heterozygous carriers)
X-Linked Dominance
– If a male shows a trait, so must all daughters as well as his mother
– The disorder is more common in females
X-Linked Recessive
– If a female shows the trait, so must all sons as well as her father
– The disorder is more common in males
From analyzing the pedigree, boxes represent
males
From analyzing the pedigree, circles represent
females
In order to answer this question, we simply need to count the DNA strand with the greatest amount of cytosine and guanine nucleotides. Seeing as answer choice D has the greatest number of cytosine/guanine nucleotides, this is the correct answer choice.
DNA fragments produced by restriction enzymes can be separated from one another using __________.
gel electrophoresis
Gel Electrophoresis
biological lab technique used to separate charged molecules, such as DNA fragments, according to size and charge. Gel electrophoresis is most commonly performed after PCR has been used to create millions of copies of the DNA sequence of interest. Since the question states that restriction enzymes are used, we know that these enzymes have cut the DNA strand at target regions to form DNA fragments of various sizes. Using gel electrophoresis, we can then separate these DNA fragments according to size
DNA samples are loaded into wells at one end of the gel, and then an electric current is applied to move the fragments through the gel. A negative charge is set at the top of the gel (anode) and a positive charge is set at the bottom of the gel (cathode). DNA fragments are negatively charged because of the phosphate groups in their sugar-phosphate backbone, so they move toward the positive anode. Seeing as though there are pores within the gel, the smaller fragments of DNA will travel further down the gel than larger fragments. Using a size ladder for reference, researchers are able to visualize if the fragment of interest is present within their sample.
BACK
If an organism’s DNA contains 30% Thymine, what percentage of guanine would be present in its DNA?
In DNA, there are four possible nitrogenous bases: Adenine (A), Guanine (G), Thymine (T), and Cytosine (C). The nitrogenous bases of each nucleotide will ALWAYS bone with their complementary nitrogenous base on the opposite side of the helix. For the DAT, it is crucial to remember that Adenine will always bond with Thymine (Uracil in RNA) and Cytosine will always bond with Guanine. Knowing this, we can deduce that there is an equal number of Adenine and Thymine in the DNA strand. There is also an equal number of Cytosine and Guanine ([A] = [T] and [C] = [G]). We can use the formula below, known as Chargaff’s rule, to answer this question.
(%A + %T) + (%G + %C) = 100%
From the question, we know that %A = 30. Therefore, %T = 30. (%A + %T) = 60. By plugging that into our equation, (%G + %C) = 40. Since amount of G = the amount of C, we can conclude that %G = %C = 20.
In a population of mice, the coat color varies between dark fur (dominant) and light fur (recessive). The frequency of the homozygous recessive genotype is 16%. Assuming the population is in Hardy-Weinberg equilibrium, which of the following would be the allele frequency of the dominant (dark fur) allele in the mice population?
To answer this question, we must use the following two Hardy-Weinberg equilibrium equations:
- p + q = 1, where p = the frequency of the dominant allele and q = the frequency of the recessive allele
- p2 + 2qp + q2 = 1, where p2 = frequency of the homozygote dominant individuals, qp = frequency of the heterozygote individuals and q2 = frequency of the homozygote recessive individuals
The key step to solving this question is to understand that we are already given the frequency of the homozygote recessive genotypes (q2 = 0.16) and so the question wants us to find the frequency of the dominant allele (p). The question can be answered with the following two steps: - In order to find the frequency of homozygote dominant alleles in the population (p), we must first solve for the frequency of the recessive allele in the population(q) like so: √q2 = √0.16, where we get q = 0.4. We now have the frequency of the recessive allele (q) in the population.
- Lastly, we can find the frequency of the dominant allele (p) using the formula:
p + q = 1
p + 0.4 = 1
p = 0.6.
Therefore, the frequency of the dominant allele in the population is 0.6.
For the DAT, it will be important to understand these two equations and how to use them in order to solve for allele and phenotype frequencies. It is highly recommended that you practice similar questions so you have a firm understanding of how to solve them.
The key step to solving this question is to understand that we are already given the frequency of the homozygote recessive genotypes (q2 = 0.16) and so the question wants us to find the frequency of the dominant allele (p). The question can be answered with the following two steps:
- In order to find the frequency of homozygote dominant alleles in the population (p), we must first solve for the frequency of the recessive allele in the population(q) like so: √q2 = √0.16, where we get q = 0.4. We now have the frequency of the recessive allele (q) in the population.
- Lastly, we can find the frequency of the dominant allele (p) using the formula:
p + q = 1
p + 0.4 = 1
p = 0.6.
Therefore, the frequency of the dominant allele in the population is 0.6.
For the DAT, it will be important to understand these two equations and how to use them in order to solve for allele and phenotype frequencies. It is highly recommended that you practice similar questions so you have a firm understanding of how to solve them.
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During the process of gastrulation, which structure is formed even in the most primitive organisms?
gut tube
- gastrulation ( process where the 3 germ layers arise) resulting in a triploblastic embryo - protosomes or deuterostomes
A scientist discovers a new species capable of surviving in dry environments with low water supply. This species produces highly concentrated urine. Which of the following structures would the organism possess that may explain this?
glomerulus
filters small solutes from the blood
proximal convoluted tubule
reabsorbs ions, water, and nutrients; removes toxins and adjusts filtrate pH
descending loop of henle
aquaporins allow water to pass from the filtrate into the interstitial fluid
ascending loop of henle
reabsorbs Na+ and Cl- from the filtrate into the interstitial fluid
distal tubule
selectively secretes and absorbs different ions to maintain blood pH and electrolyte balance
collecting duct
reabsorbs solutes and water from the filtrate
the oocyte will only complete meiosis II if
fertilization occurs, resulting in a newly formed diploid zygote
Which of the following processes are topoisomerases involved in?
transcription
Recall that during these processes, helicase is responsible for separating the two complementary DNA strands. As helicase progresses along the DNA sequence, it creates torsional strain ahead of the helicase enzyme. Topoisomerase is responsible for alleviating that torsional strain breaking/preventing kinks during the unwinding process. Without topoisomerases, helicase would be unable to continue ‘unzipping’ the DNA strand, preventing the ability to replicate DNA altogether. To help visualize this concept, here is an illustration showing topoisomerases working in conjunction with helicase to unwind the DNA strand.
Which of the following direct (non-tropic) hormones has a function that increases the rate of metabolism?
Direct hormones are hormones that act on their target organ directly. In this case, thyroxine (T4) is a direct hormone. Thyroxine (T4) is considered inactive and becomes activated to become triiodothyronine (T3). Thyroxine is a direct hormone and functions to increase cellular metabolism, therefore, Option E. thyroxine is the correct answer.
In order for thyroxine to be released from the thyroid, a cascade of events needs to occur. First, the hypothalamus release Thyrotropin-Releasing Hormone (TRH) which signals the anterior pituitary to release Thyroid Stimulating Hormone (TSH). TSH then stimulates the release of T3 and T4 from the thyroid gland. As you can see, TRH and TSH do not act directly. Recall, they are tropic hormones, so they signal other endocrine glands to release hormones.
Below is an image to help visualize the hypothalamic-pituitary-thyroid axis.
denitrifying bacteria convert
nitrates back into atmospheric nitrogen.
Which of the following is common between the formation of both glycosidic bonds and phosphodiester linkages?
A glycosidic bond is a bond between two alcohol functional groups within organic molecules (or an alcohol and an amine), most commonly seen as the bond joining two monosaccharides together. A glycosidic bond occurs via a condensation reaction, also known as a dehydration synthesis reaction, that results from the alcohol group of one monosaccharide attacking the anomeric carbon of another monosaccharide, as shown in the image below. The result of a condensation reaction is the formation and release of water.
CC BY 4.0
In comparison, a phosphodiester bond is a bond between the 5’ phosphate of one nucleotide and the 3’ hydroxyl of a separate nucleotide.
Phosphodiester bond forms through the condensation reaction between two nucleotides present in the backbone of nucleic acid, as shown in the image below.
In DNA and RNA, the phosphodiester bond is the linkage between the 3′ hydroxyl group of one nucleotide and the 5′ phosphate group of an adjacent nucleotide. Both glycosidic bonds and phosphodiester bonds form via condensation/dehydration reactions which results in the formation of a water molecule, therefore, Option A. Formation of a water molecule is the correct answer.
