Answer:
Nitrogenous;Acid-Base and Fluid
Explanation:
The kidney is referred to as an excretory organ because it excretes nitrogenous wastes. It is also a major homeostatic organ because it maintains the electrolyte Acid-Base, and fluid balance of the blood.
Answer:
Nitrogenous wastes, acid-base, fluid balance.
Explanation:
Kidneys are bean shaped organs present in pairs i vertebrates. These are an excretory organ. Nephron is the basic structural and functional unit of kidneys. It excretes the nitrogenous wastes in the form of urine which is continuously formed. Kidneys are also play major role in body homeostasis by maintaining the electrolyte, acid-base and fluid balance of the blood. Kidneys are act as a hormone because it produces the erythropoietin hormone which help in the formation of red blood cells in the human body.You have made random-sequence synthetic RNA molecules that contains only A and G, and which contain 5 times as much A as G. You use these RNAs to perform in vitro translation. How many different amino acids would you expect to find in the resulting polypeptides? And at what frequency would you expect to find Glutamic acid (Glu) in the resulting polypeptides?
The resulting polypeptide will contains Glycine, arginine ,lysine , and glutamic acid. The probabilities to form a glutamic acid in the coding sequence is 2/9 for each codon.
Genetic code:
It is a three letter code of base pairs in the nucleic acid that code for an specific amino acid.
Here, mRNA is made up of only A and G. Nine code are possible in a mRNA sequence that contains only A and G.
GGC, GGA and GGG code for Glycine. AGA and AGG code for Arginine.AAA and AAG code for Lysine.GAA and GAG code for Glutamic acid.
Therefore, the resulting polypeptide will contains Glycine, arginine ,lysine , and glutamic acid. The probabilities to form a glutamic acid in the coding sequence is 2/9 for each codon.
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Final answer:
In the resulting polypeptides from the random-sequence synthetic RNA molecules, we can expect to find a variety of amino acids. The number of different amino acids will depend on the specific codons present in the RNA. The frequency of Glutamic acid (Glu) will be higher due to the higher frequency of the codon GAA.
Explanation:
In this case, we have made random-sequence synthetic RNA molecules that contain only A and G, with 5 times as much A as G. In terms of translating these RNAs into polypeptides, we can expect to find a variety of amino acids in the resulting polypeptides. However, the frequency of each amino acid will depend on the specific codons that are present in the RNA.
Since there are 61 codons that specify the addition of an amino acid to the polypeptide chain and 3 codons that specify termination, we can expect to find a maximum of 61 different amino acids in the resulting polypeptides. However, it is important to note that some of these codons may code for the same amino acid, so the actual number of different amino acids may be lower.
As for the frequency of Glutamic acid (Glu) in the resulting polypeptides, it will depend on the specific codons present in the RNA. Glutamic acid is encoded by the codons GAA and GAG. Since we have made random-sequence RNA molecules with 5 times as much A as G, we can expect a higher frequency of the codon GAA, and thus a higher frequency of Glutamic acid in the resulting polypeptides.
Los Angeles has a serious problem with smog, particularly photochemical smog. All of these EXCEPT which contributes to the excessive smog in the Los Angeles area?
Answer:
A) Longitude of the city
Explanation:
With the use of site-directed mutagenesis, hemoglobin has been prepared in which the proximal histidine residues in both alpha and the beta subunits have been replaced by glycine. The imidazole ring from the histidine residue can be replaced by adding free imidazole in solution. a) Would you expect this modified Hb with free imidazole to exhibit oxygen binding? Why or why not? b)Would you expect this modified hemoglobin to show cooperativity in oxygen binding? Why or why not?
Answer:
a) No, because the heme group that has the ability to bind oxygen is bound via the imidazole ring of a histidine residue
b) The imidazole in solution bind to the heme group and then facilitates its binding to oxygen; however, protein does not have the proximal histidine residues which are capable of triggering conformational changes in the folded protein structure
After ovulation, ___1___ acts on the remaining follicle cells to promote the development of the ___ 2___ , which produces estrogen and large amounts of ____3____ , which then causes the endometirum to become ____4____ in preparation for ____5____ should fertilization of the ovum occur. If the ovum is not fertilized the corpus luteum ____ 6____ resulting in a drop of estrogen and progesterone levels. The endometrium then sloughs off in the ____7____ . The drop in estrogen and progesterone removes the ____ 8_____ and the GnRH is produced by the _____9____ and the cycle beings again.
Answer:
The options include
hypothalamus
negative feedback inhibition
menstruations flow
degenerates
Progastrin
LH
nutritive and receptive
corpus luteum
implantation
Explanation:
1)LH (Luteinizing Hormone )
2)CORPUS LUTEUM
3)PROGASTRIN
4)NUTRITIVE AND RECEPTIVE 5)IMPLANTATION
6)DEGENERATES
7) MENSTRUATION FLOW
8)NEGATIVE FEEDBACK INHIBITION
9) HYPOTHALAMUS
From a large-scale screen of many plants of Collinsia grandiflora, a plant with three cotyledons was discovered (normally, there are two cotyledons). This plant was crossed with a normal pure-breeding wild-type plant, and 600 seeds from this cross were planted. There were 298 plants with two cotyledons and 302 with three cotyledons. What can be deduced about the inheritance of three cotyledons
Answer:
The cross is as follows:
P: D/d × d/d
Explanation:
Half of the F1 progeny are mutant. The mutation results in 3 cotyledons is dominant and the original mutant is heterozygous.
If D denotes the mutant allele and d denotes the wild type allele.
The cross is as follows:
P: D/d × d/d
F1 denotes D/d three cotyledons
d/d denote two cotyledons
2 Points
Humans have become dominant in the last ____
_ years.
O
A. 65 million
O
B. 1.8 billion
O
C. 1 million
O
D. 2,000
€ PREVIOUS
Did you mean 200,000?
Explanation:
Humans have domimated the planet for the last 200,000 years.
Answer: 1 million
Explanation: 200,000 is wrong!!
The lab you work in has discovered a previously unidentified extracellular signal molecule called QGF, a 75,000-dalton protein. You add purified QGF to different types of cells to determine its effect on these cells. When you add QGF to heart muscle cells, you observe an increase in cell contraction. When you add it to fibroblasts, they undergo cell division. When you add it to nerve cells, they die. When you add it to glial cells, you do not see any effect on cell division or survival.
A. How can the same signaling molecule cause such diverse affects? Why do the glial cells not respond at all?
B. Devise a signaling pathway that QGF could use to increase contractions in in the heart. (Hint: Ca2+ is very important for muscle contraction)
C. Devise a signaling pathway that QGF could use to lead to apoptosis of nerve cells.
D. Devise a signaling pathway that QGF could use to induce cell division in the fibroblasts?
Answer:
A- It is because said molecule presents binding to different receptors and in the case of glia cells it does not present an answer because they do not have a specific receptor for said molecule.
b-in the case of muscle, this molecule could bind to calcium channels or to the REL itself, generating the opening of said channels or the release of calcium found in REL. In this way, by increasing the intracellular concentration of calcium, it activates the actin head, causing the topoisomerase to move and promoting binding with myosin. Once the activated actin will perform a "rowing" movement to generate the drag of the myosin and the union of the Z lines.
c-In the process of neuronal apoptosis, the cell itself generates biochemical signals (either by positive induction in which the receptors and receptors of the membrane bind to certain substances, or by negative or mitochondrial induction in which the ability to suppress certain substances is lost. substances that would generate the activity of apoptotic enzymes) that cause them to condense and alter the cytoplasm, the cell membrane, the cell nucleus to collapse and DNA to fragment. Finally, the microglial cells end up phagocytizing and eliminating the remains of the dead neurons, so that they do not generate interference for the normative functioning of the brain.
d- This molecule could fulfill a path similar or similar to that of growth factors, that is how they are associated with heparan sulfate of the extracellular matrix, which serves as a warehouse for inactive factors. FCFs contribute to different types of responses, such as wound healing, hematopoiesis, angiogenesis, or embryonic development
Explanation:
Regarding the explanation, it is important to keep in mind that this molecule can play a role both in the nervous system and in other organs, that is why in case it is sought that it does not have this apoptotic effect on neurons, some coarse, macro compound that allows the blood-brain barrier to not allow passage.
A. Globin is a red blood cell protein that is responsible for oxygen transport. The amino acid
sequence for a portion of the globin protein is Proline, Glutamic Acid, Glutamic Acid, Lysine. Write
the mRNA sequence of the amino acids for these amino acids in the space below (note, you don't
have to write all possible mRNA combinations for each amino acid, simply choose one correct codon
each amino acid specified above).
B. How many nucleotides would it take to code for the four amino acids in the above question.
Explain your answer.
Answer: A
Explanation:
Proline:CCG
Glutamic Acid: GAG,GAA
Lysine: AAG
Answer: B
Explanation: Each group of three nucleotides code for one amino acid. This means it would take 12 (3 times 4 equals 12) nucleotides to code for, four amino acids.
The mRNA sequences for Proline, Glutamic Acid, Glutamic Acid, and Lysine could respectively be CCU, GAA, GAA, AAA. Twelve nucleotides are needed to code these four amino acids.
Explanation:A. The mRNA sequences for the amino acids Proline, Glutamic Acid, Glutamic Acid, Lysine could be CCU, GAA, GAA, AAA respectively.
B. Each amino acid is coded for by a set of three nucleotides known as a codon, therefore to code for four amino acids, one would need 3 x 4 = 12 nucleotides. This is because in the genetic code, every sequence of three nucleotide bases corresponds to one specific amino acid.
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What can a planetary nebula form around a dwarf star?
A planetary nebula sheds the outer layers during the evolution into a white dwarf around a dwarf star. This process marks the final stages of stellar evolution for stars with masses less than about 8 times that of the Sun.
A planetary nebula can form around a dwarf star as part of the star's late-stage evolution. When a low to intermediate mass star (like a dwarf star) reaches the end of its life, it undergoes several stages of nuclear burning. Eventually, as the star runs out of fuel, it enters a phase where it sheds its outer layers into space.
For a dwarf star specifically:
1. Red Giant Phase: The star expands into a red giant as it exhausts its core hydrogen fuel. During this phase, the outer layers of the star become loosely held due to the diminished gravitational pull.
2. Planetary Nebula Formation: As the star sheds its outer layers, a shell of ionized gas and dust, known as a planetary nebula, forms around the remaining core of the star. This nebula is illuminated by the hot core of the star, creating a beautiful and often intricate structure visible to telescopes.
3. White Dwarf: The core of the star that remains after the nebula disperses becomes a white dwarf. This is a dense remnant of the star's core, composed mainly of carbon and oxygen, which gradually cools over billions of years.
Which of the following describes respiration?
The process of using carbohydrates and oxygen for energy
The process of using sunlight to produce carbohydrates
The process of moving water through the plant
The process of absorbing water through the leaves
Answer:
process of using carbohydrates and oxygen for energy
Answer:
The process of using carbohydrates and oxygen for energy
Explanation:
Respiration is the process in which cells in organisms make use of oxygen and glucose to produce energy. Respiration is a biochemical process whose reactant is oxygen and glucose and the output is ATP(ENERGY), carbondioxide and water.
6O2+C6H1206----6CO2+6H20+ATP
We have types of cellular respiration Aerobic respiration and anearobic respiration.
Aerobic respiration takes place in the presence of oxygen while anaerobic takes place in the absent of oxygen. Product of anerobic is ethanol,Atp and CO2.
Give the order for the life cycle of the hydrozoan, Obelia, starting with the reproductive polyp form. Answers Selected Answer Growth into mature polyp 1. Reproductive polyps form. Cells grow into Planula. 2. Growth into mature polyp Reproductive polyps form. 3. Medusa buds. Medusa buds. 4. Meiosis occurs. Meiosis occurs. 5. Fertilization occurs. Fertilization occurs. 6. Cells grow into Planula.
Answer:
3. Medusa buttons. Medusa buttons.
4. Meiosis occurs. Meiosis occurs.
5. Fertilization takes place. Fertilization takes place.
6. The cells grow in Planula.
1. Form of reproductive polyps.
2. Growth in mature polyp Form of reproductive polyps.
Explanation:
The Obelia life cycle begins when two jellyfish buttons of opposite sexes undergo meiosis forming the female and male gametes. The male gamete will fertilize the female gamete, forming a zygote. The cells of the zygote will grow into a structure called Planula that will develop and form a mature popolis (a structure that looks like a stem with roots that hold it to the rocky part of the ocean), the mature propolis means that it is able to reproduce itself. if so, it will form reproductive propolis.
Question:
Give the order for the life cycle of the hydrozoan, Obelia, starting with the reproductive polyp form.
1. Reproductive polyps form.
2. Growth into mature polyp Reproductive polyps form.
3. Medusa buds.
4. Meiosis occurs.
5. Fertilization occurs.
6. Cells grow into Planula.
Answer:
The order for the lifecycle of the hydrozoan - Obelia starting from the reproductive polyp form is given below:
3. From Reproductive form of the Polyph, it sheds the Medusa buds. The medusa stage of Obelia species is common in coastal and offshore plankton around the world. The blastostyles of the Polyph are the reproductive zooids as they reproduce asexually to give rise to numerous lateral buds called medusa buds or gonophores. These buds develop into third type of zooids of the colony called medusae.
5 After medusa happens, Fertilisation occurs: Sometimes the flagellated sperms swim about in water and fertilize the ova present in female medusae. The fertilisation takes place in water. As medusa is the motile form, it performs two important functions for the colony namely reproduction and dispersal of the gametes.
4. After fertilisation then Meiosis starts to happen. Meiosis is a process where a single cell divides twice to produce four cells containing half the original amount of genetic information. These cells are our sex cells – sperm in males, eggs in females. During meiosis one cell divides twice to form four daughter cells.
6. After meiosis, cells grow into Planula.
The embryo is set free from the egg membrane as a free-swimming larva called the planula. The larva swims about for some time and brings about wide distribution of the species.
After the free-swimming life the planula larva loses its cilia and settles down on the bottom of the sea, gets attached to the substratum by its broader end and undergoes metamorphosis.
1. Reproductive Polyps begin to form.
2. Growth into mature polyp Reproductive polyps form: the Obelia achieves full maturity and ready for the cycle all over again.
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You created a strain with several mutations surrounding (but not within) the sigma32 RBS by changing A-T base pairs to G-C base pairs. Would you predict that these changes have an effect on sigma32 expression levels during high heat stress?
Answer:
no idea ;-;
Explanation:
Enolase is an enzyme that catalyzes one reaction in glycolysis in all organisms that carry out this process. The amino acid sequence of enolase is similar but not identical in the organisms. Researchers purified enolase from Saccharomyces cerevisiae, a single-celled eukaryotic yeast that grows best at 37°C, and from Chloroflexus aurantiacus, a bacterium that grows best at the much higher temperature of 55 C. The researchers compared the activity of purified enolase from the two organisms by measuring the rate of the reaction in the presence of varying concentrations of substrate and a constant amount of each enzyme at both 37°C and 55°C.
Depending on the organism, the optimal pH for enolase to catalyze its reaction is between 6.5 and 8.0. Describe how a pH below or above this range is likely to affect enolase and its catalytic ability
Answer:
pH below or above the optimal range will result in loss of the enzyme activity.Changing the pH above or below optimum range affect the charges that reside on the amino acid molecules. Amino acids present in the active site of enzymes that attracted each other may no longer continue to be attracted.Again, the shape of the active site of the enzyme will change as a result of which substrate binding will be affected.Overall we can say that the enzymes are denatured.
nick normal alveolar ventilation rate (AVR) during mild exercise is 6.0 L/min. She goes snorkeling to a beach, on a warm summer day,water is not too cold. The snorkel volume is 50 ml. Assuming this is a mild exercise for Angina, what should be her respiratory rate to maintain 6.0 L while snorkeling. ( Assume a constant tidal volume of 500 ml and anatomic dead space of 150 ml)
Answer:
RR = 17.14min∧-1
Explanation:
AVR = 6L/min
snorkel volume = 50ml
to calculate the respiratory rate to maintain 6l/min(AVR)
Tidal Volume = 500 ml
anatomic dead space = 150ml
Alvedar ventilation rate = ( TV - Anatomic dead space) x RR
6l/min = ( 500 -150)ml x rr
600ml = 350 x RR
RR = 600ml/mins/350ml
RR = 17.14min∧-1
The European model was the first weather model to correctly predict that Hurricane Sandy would _____, while the American model's early prediction was that Hurricane Sandy would _____.
Choices:
Weakening Before Heading South
Turn Left Towards the Coast
Strengthen Before Moving Away from shore
Turn Right and Head Out to Sea
Answer:
The correct answer is "Turn Left Towards the Coast; Turn Right and Head Out to Sea".
Explanation:
During the Hurricane Sandy disaster of 2012, the European model was famous for predicting that Hurricane Sandy would turn left towards the coast, which was what happened on October 29. This was a disaster for the United States because the hurricane hit the East Coast. The American model had an opposite prediction, it predicted that the Hurricane Sandy would turn right and head out to sea.
The correct answer is "Turn Left Towards the Coast; Turn Right and Head Out to Sea".
The following information should be considered:
At the time of the Hurricane Sandy disaster of 2012, the European model was famous for estimating that Hurricane Sandy would turn left towards the coast, that was happened on October 29. This developed the disaster for the United States since hurricane hit the East Coast. The American model had an inverse estimation it predicted that the Hurricane Sandy would turn right and head out to sea.Learn more: https://brainly.com/question/10309631?referrer=searchResults
Of the 30000000 species on earth,how many are lost each hour?
Harvard biologist E.O. Wilson's research indicates that the planet is losing species at an unprecedented rate of 3 per hour, which is 100 to 1,000 times the natural background rate. This biodiversity crisis could lead to the loss of half of today's species within the century, highlighting the urgency of addressing this global issue.
Explanation:E.O. Wilson, a renowned Harvard biologist, has provided us with alarming insights into the current state of species extinction. His research indicates that we are losing species at an incredibly rapid rate, 3 species per hour and potentially 140,000 species per year. This rate is 100 to 1,000 times the historical average. These extinctions are permanent, leading to a significant decline in biodiversity.
In 2002, Wilson predicted that if the current rates of extinction continue, we might lose up to 50% of today's plant and animal species within this century. This prediction highlights the urgency and potential consequences of this biodiversity crisis. Considering that the natural background extinction rate is only about one per million species per year, the accelerated rate at which we are losing species is of grave concern for conservationists and scientists alike.
It is important to note that the extinction of species has far-reaching effects. It impacts ecosystem function, the potential for medical and technological advancements, and the overall health of our planet. Addressing this issue is crucial for maintaining the integrity of Earth's ecosystems and the resources they provide to humanity.
When electrons move through a series of electron acceptor molecules in cellular respiration ________. When electrons move through a series of electron acceptor molecules in cellular respiration ________. the electrons gain energy as they move from one electron acceptor to another glucose is eventually reduced by the electrons in the formation of water the electrons release large amounts of energy each time they are transferred from one electron acceptor to another oxygen is eventually reduced by the electrons in the formation of water
Answer:
When electrons move through a series of electron acceptor molecules in cellular respiration, oxygen is eventually reduced by the electrons in the formation of water
Explanation:
The electron transport chain is located in the internal mitochondrial membrane. There are three proteinic complexes in the membrane, I, II, and III, that contain the electrons transporters and the enzymes necessary to catalyze the electrons transference from one complex to the other. Complex I contains the flavine mononucleotide -FMN- that receives electrons from the NADH. The coenzyme Q, located in the lipidic interior of the membrane, conducts electrons from complex I to complex II. The complex II contains cytochrome b, from where electrons go to cytochrome c, which is a peripheric membrane protein. Electrons travel from cytochrome c to cytochromes a and a3, located in the complex III. Finally, electrons go back to the matrix, where they combine to H₊ ions and oxygen, to form the water molecule. As electrons are transported through the chain, protons are bombed through the three proteinic complexes from the matrix to the intermembrane space.
In cellular respiration, electrons move through a series of electron acceptor molecules, releasing energy with each transfer. This energy is used to form ATP. The final electron acceptor is oxygen, which is reduced by the electrons to form water.
Explanation:When electrons move through a series of electron acceptor molecules in cellular respiration, they do not gain energy as they move. Instead, each time the electrons are transferred from one electron acceptor to another, they release energy. This released energy is then used in the formation of adenosine triphosphate (ATP), the energy currency of the cell. During the electron transport chain, the last step of cellular respiration, oxygen is the final electron acceptor. This oxygen is reduced by the electrons to form water. Therefore, it's not glucose but oxygen that is reduced during the process.
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You discover a pattern of a 20 amino acid sequence at the C-terminus of over 50 proteins in an organism. You hypothesize that this amino acid sequence is a signal sequence for localization into a specific organelle or cell compartment, but you don't know which one. How might you use microscopy to investigate the location of proteins with this unique C-terminus sequence?
Answer:
It depends on the type of microscope that you have
Explanation:
For example, in the case that you have a fluorescence microscope, it is possible to create a fusion construct of the domain fused to the Green Fluorescence Protein (GFP) and then introduce it into the cell. The observation of fluorescence from this construct enables to confirm the existence of the signal sequence (i.e., the C-terminal domain) in the organelle
Darwin recorded distinct species of finches on different islands in the Galapagos. The images below show two species of finches he observed.
On island X, he saw both species of finches. On island Y, he only saw the finch on the right. Based on these observations, what is the most likely conclusion Darwin would have made?
Answer:
On island y the other finch cant eat the food do to the shape and hardness of his beak therefor he would not survive
Explanation:
Answer:
On island y the other finch cant eat the food do to the shape and hardness of his beak therefor he would not survive.
Explanation:
i did the topic. please like and rate my answer
5. A s-shaped curve is shown when a period of growth is followed by a leveling off as the
population approaches carrying capacity. True or false
Answer:
population expansion decreases as resources become scarce, leveling off when the carrying capacity of the environment is reached, resulting in an S-shaped curve.
Explanation:
Describe what will occur the population if it increases in number when food is scarce?
Answer:
more people will die
Explanation:
If you have a big population and little food mor peple will die.
In goats, a recessive gene causes the goats to "faint" when they are startled. A farmer has a goat that is a carrier for the fainting gene (Gg). Gertrude is mated with one that is homozygous dominant. How many of their offspring will be fainters
Answer:
None
Explanation:
A cross between a heterozygous goat and a homozygous dominant goat would result in no children having the trait.
G G
G GG | GG
-- | --
g Gg | Gg
Answer:
None
Explanation:
This question states that this particular allele (dubbed "g" ) acts in a recessive manner. This means that the corresponding "fainter" phenotype will be found only when the allele is present in a homozygote state (gg).
If a Gg (heterozygous) goat is mated with a GG (homozygous dominant) goat, it would be expected that approximately 50% of their progeny will be GG and the other 50% will be Gg . Barring the occurrence of de novo mutations, one would not expect any of the offspring to possess an homozygous recessive (gg) genotype and therefore develop as "fainters".
SOMEONE PLEASE HELP THIS IS FOR SCIENCE
Planet B has a tilt of 45 degrees. What seasonal changes would be expected on this planet Group of answer choices
A. Extreme temperature changes between seasons
B. Little to no change in temperatures between seasons
C. Seasonal changes along the equator only
D. Seasonal changes at the poles only
Answer:
A. extreme temperature changes between seasons
A transmembrane protein has the following properties: it has two binding sites, one for solute A and one for solute B. the protein can undergo a conformational change to switch between two states: either both binding sites are exposed exclusively on one side of the membrane, or both are exposed exclusively on the other side of the membrane. The protein can switch between the two conformational states only if both binding sites are occupied or if both binding sites are empty, but cannot switch if only one binding site is occupied.
A. What kind of a transporter do these properties define?
B. Do you need to specify any additional properties to turn this protein into a transporter that couples the movement of solute A up its concentration gradient to the movement of solute B down its electrochemical gradient?
Answer:
Explanation:
a) A symporter is simply known as an integral protein membrane that helps in the movement of two or more different molecules or ions across a phospholipid membrane.
b) The glucose–Na+ symport protein uses the electrochemical Na+ gradient to drive the import of glucose.
You are conducting a dihybrid cross. You mate a homozygous-dominant-smooth homozygous-recessive-green pea plant (SSyy) with a heterozygous-smooth homozygous-dominant-yellow pea plant (SsYY). What is the likelihood of obtaining an offspring with the following genotype: SSYy? Note: SS and Ss will result in the smooth phenotype; ss will result in the wrinkled phenotype; YY and Yy will result in the yellow phenotype; yy will result in the green phenotype. Hint: you are crossing the following: SSyy X SsYY. What is the likelihood of getting the following: SSYy?
Answer:
8/16 or 50% is the likelihood of getting SSYy.
Explanation:
To predict the outcome of the occurrence of SSYy, let's make a punnet square:
SY SY sY sY
Sy SSYy SSYy SsYy SsYy
Sy SSYy SSYy SsYy SsYy
Sy SSYy SSYy SsYy SsYy
Sy SSYy SSYy SsYy SsYy
8/16 of the plants will have a chance of having the genotype: SSYy This means that the there will be a 50% chance that the genotype of the offspring will be SSYy.
You are studying mitosis in a species of snake. You notice that during mitosis of diploid cells there are 7 groups of chromosomes on the metaphase plate. Is it surprising that this is an odd number? Why or why not?
Answer:
No, this is not surprising
Explanation:
No, it is not surprising that it is given 7 groups of chromosomes, that is, each chromosome is paired. This makes 7 × 2 = 14 chromosomes in the entire meta phase plate. The human meta phase represents the 23 groups of chromosomes on the plate, that is, 23 × 2 = 46 chromosomes. Every organism on this planet has several chromosomes in diploid or 2N, otherwise an equal distribution of chromosomes on the any phase is not possible.SCENARIO A
Anthropologists have studied the role of quinoa (ancient grain) in the culture of Bolivia. The researchers interviewed people about how frequently they ate quinoa, where they ate it (at home or in restaurants), and during what time of day they ate it. The researchers also asked people about how they prepare quinoa, where they obtain quinoa products, and any beliefs they might have about the quinoa (such as their nutritional value). The data were used to assess the role of quinoa in the culture. The researchers also considered how this relates to global quinoa consumption patterns.
What is the primary field of anthropology addressed in this research?
Linguistic anthropology
Archaeology
Biological anthropology
Cultural anthropology
Question 21 pts
Which field of anthropology (other than the one you indicated above) is most related to this research?
Biological Anthropology
Archaeology
Linguistic Anthropology
Cultural Anthropology
All fields of Anthropology are related to this topic
Question 31 pts
How might this research contribute to a comparative anthropological approach?
This research could be used to compare quinoa consumption in other parts of the world.
This research could be used to compare biological adaptations across Bolivia.
This research could be used to compare quinoa consumption in other primate species.
This research could be used to compare quinoa DNA to other plants.
Answer:
Biological anthropology
cultural anthropology
This research could be used to compare quinoa consumption in other parts of the world.
Explanation:
Anthropology is the science of studying human societies, including language, culture, and the use of other people with previous civilizations. In the case of cultural anthropology, its main purpose is to study the cultural elements unique to each human group; This includes areas such as religion, habits, customs and food. Cultural anthropology is a study of cultures and human society, in the above statements they survey the eating of quinoa, and they clearly state that there is an ancient grain similar to that in Bolivian culture, so they are in cultural anthropology Yes, it may be subject to a comparative approach to all areas of the world because they are eating kvinova, which will be studied by the comparative.which of the following represents a genotype that is homozygous recessive
There are many, and since you didn't specify the choices, some examples include "ff" or "dd" or just two lowercase letters.
Explanation:
Final answer:
A genotype that represents a homozygous recessive condition would consist of two identical recessive alleles, typically represented by two lowercase letters, such as 'aa'.
Explanation:
The genotype that represents a homozygous recessive condition would consist of two identical recessive alleles. In genetic notation, this is often represented by two lowercase letters (e.g., aa). This is because a recessive trait will only be observed when an individual has two copies of the recessive allele, as dominant alleles, if present, will mask the expression of recessive alleles.
For instance, if we consider a gene with two alleles - one dominant (A) and one recessive (a) - the homozygous recessive genotype would be aa for that trait. A homozygous dominant genotype would be AA and a heterozygous genotype would be Aa. Only in the homozygous recessive genotype does the recessive trait get expressed because there is no dominant allele to dominate over the recessive allele.
Infectious agents could be organisms or viruses. You observe a case of conjunctivitis. Thisillness could be caused by an organism (Chlamydia trachomatis) or by Adenovirus.Both of these etiologic agents are obligate intracellular pathogens. Explain why one is a virus the other one is not by applying the criteria suggested by AndreLwof . Discuss how the two pathogens differ at each point covered in le
Answer:
Explanation:
Conjunctivitis can be caused by either a bacteria or a virus or even sometimes due to allergic reactions.
According to Andrelwof, viruses are classified based
i. on nature of DNA or RNA,
ii. nature of nucleocapsid ,
iii. symmetry of nucleocapsid,
iv. diameter of nucleocapsid with helical symmetry
So Chlamydia trachomatis is a bacteria, they have a nucleocapsid. They are just a single cell. But Adenovirus is a virus, that has a head and tail, contains nucleocapsid.
Chlamydia is Gram negative obligate intracellular pathogen. They can spread through direct or indirect contact through fomites, hand, contaminated towels. They affect both the eyes and symptoms include itching, irritation, discharge, swelling of eyelids, photophobia, and pain.
Laboratory diagnosis can be done by direct detection of inclusion bodies with Giemsa staining of conjunctival smears.
Adenoviruses are the most important and most frequent cause of follicular epidemic keratoconjunctivitis . The major mode of transmission is by direct inoculation by fingers. symptoms include itching, tearing, burning and foreign body sensation as well as photophobia.
the best method for diagnosis is the quantitative real-time polymerase chain reaction (PCR) and Enzyme immunoassays which are a cheaper and faster option with a high sensitivity.
In a laboratory experiment, C. elegans and B. thuringiensis were cultured individually (control) or together (experimental) for 150 days. Under optimal conditions, the generation time of C. elegans is approximately 3.5 days and the generation time of B. thuringiensis is approximately 25 minutes. At the end of the experiment, the change in virulence of B. thuringiensis and the change in resistance of C. elegans to BT toxin were determined. (a) Calculate the maximum number of generations that is possible in individual cultures of C. elegans AND the maximum number of generations that is possible in individual cultures of B. thuringiensis in 150 days.
Given 150 days, the maximum number of generations for C. elegans is approximately 4,114 generations, and for B. thuringiensis is approximately 8,640 generations, calculated based on their respective generation times.
Explanation:To calculate the maximum number of generations for C. elegans and B. thuringiensis, we will need the total time frame of 150 days and their generation times. The generation time for C. elegans is about 3.5 days, and for B. thuringiensis, it’s approximately 25 minutes.
C. elegans: First, convert the total experiment time from days to the same time unit for generation time. There are 150 days * 24 hours/day * 60 minutes/hour = 216,000 minutes. Then, divide this total available time by the organism's generational cycle: 216,000 minutes / (3.5 days/generation * 24 hours/day * 60 minutes/hour) = about 4,114 generations.
B. thuringiensis: Similarly, we calculate the number of generations: 216,000 minutes / (25 minutes/generation) = approximately 8,640 generations.
The reason behind this large difference in number of generations is the stark difference in generation times of these organisms. The comparatively shorter generation time of B. thuringiensis allows it to produce more generations within the same timeframe.
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