Answer:
CUU, CUC, UUG, UUA
Explanation:
Following codons result in leucine amino acid : UUA, UUG, CUU, CUC, CUA and CUG
Following codons result in phenylalanine amino acid : UUU and UUC
CUU : If C is mutated to U, result will be UUU which is phenylalanine codon. U at third position can be mutated to A which will form UUA which us again codon for leucine.CUC : If C at first position is mutated to U, phenlyalanine codon UUC will be formed. C can be mutated to A to form leucine codon UUA.UUG : If G is mutated to U phenylalanine codon UUU will form. U at first position can be mutated to C to form leucine codon CUU.UUA : If A is mutated to U phenylalanine codon UUU will form. U at first position can be mutated to C to form leucine codon CUU.Which bacterial modification allowing the bacteria to survive harsh times, like dessication?
Answer:
The correct answer is endospore and capsule.
Explanation:
Bacteria can do modification in them during harsh conditions which protect them from the harsh time. These modifications are the formation of endospores or capsules.
Endospore is the hard and non-reproductive structure formed outer to the cell wall by mainly bacteria of phylum firmicutes in the absence of nutrients. Endospore helps the bacteria to remain dormant for several years until the right condition comes.
Capsule is a gelatinous layer secreted by bacterial cell which is made up of polysaccharides. It forms outer to cell wall and protects the bacteria from desiccation and engulfment by phagocytes.
Final answer:
Bacteria can survive harsh conditions like desiccation by forming endospores, which are highly resistant structures that protect the bacterial DNA in a dormant state until favorable conditions return.
Explanation:
The bacterial modification that allows bacteria to survive harsh times, such as desiccation, is the formation of endospores. Formed under unfavorable conditions, endospores encapsulate the genetic material of the bacterium and become dormant, allowing the bacterium to withstand extreme conditions including heat, lack of moisture, and exposure to chemicals. They remain in this resistant state until conditions become favorable again, at which point they can re-animate and resume normal cellular functions. Bacterial genera such as Bacillus and Clostridium are well-known for their ability to form endospores.
Other adaptions include changes in cell wall structure, accumulation of energy reserves like starch or oils, and alterations in membrane and protein structure to reduce metabolic activity during harsh conditions. Some bacteria, like Halobacterium, can survive in environments with extremely high salt concentrations, which would be lethal to most other organisms. This extreme halotolerance is an example of how microbial life can adapt to survive and even thrive under conditions that would normally be considered inhospitable.
Calculate the difference in blood pressure between the feet and top of the head.
Answer: this is an incomplete question but pressure is said to decrease with height and increase with depth.
This means pressure at the top of the head is lower than pressure at the feet.
It is taken that blood pressure at the arms is roughly 10% lower than blood pressure at the legs.
Explanation:
The events in the ovarian and uterine cycles are largely controlled by the pituitary gonadotropins and ovarian hormones. Before ovulation are the follicular phase of the ovarian cycle and the menstrual and proliferative phases of the uterine cycle. After ovulation are the luteal phase of the ovarian cycle and the secretory phase of the uterine cycle. How do hormone levels change with the phases and ovulation?
Explanation:
Menstruation is the cyclic changes taking place in women every month which involves the sloughing of the uterine walls.
The menstrual cycle involves the coordination of two-cycle: the ovarian cycle and the uterine cycle.
The menstrual cycle is controlled by the gonadotrophin hormones which are the Luteinizing hormone (LH) and the Follicular stimulating hormone (FSH) and ovarian hormones which are: progesterone and estradiol.
The level of these hormones during each phase are:
1. Follicular phase / proliferative phases- Estradiol level increases/ FSH level declines
2. Ovulation- LH hormone level at the peak, estrogen level at the peak
3. Luteal phase/ secretory phase- Progesterone and estrogen level elevates / LH and FSH level declines
Final answer:
The hormone levels change with the different phases and ovulation. Before ovulation, estrogen levels rise during the follicular phase of the ovarian cycle, while the menstrual phase occurs in the uterine cycle. After ovulation, the ovarian cycle enters its luteal phase and the uterine cycle enters its secretory phase.
Explanation:
The hormone levels in the ovarian and uterine cycles change with the different phases and ovulation. Before ovulation, during the follicular phase of the ovarian cycle, follicles on the surface of the ovary are stimulated to grow by rising levels of follicle-stimulating hormone (FSH). These follicles release estrogen as they grow. At the same time, in the uterine cycle, the menstrual phase occurs where the functional layer of the endometrium in the uterus is shed. After about five days, the estrogen levels rise, and the menstrual cycle enters the proliferative phase, when the endometrium begins to regrow.
After ovulation, the ovarian cycle enters its luteal phase when the ruptured follicle transforms into a structure called a corpus luteum, which produces progesterone and estrogen. In the uterine cycle, the secretory phase begins, during which the endometrial lining prepares for possible implantation of a fertilized egg. The progesterone from the corpus luteum facilitates the regrowth of the uterine lining and inhibits the release of further FSH and luteinizing hormone (LH). The level of estrogen produced by the corpus luteum remains steady for a few days.
With independent assortment, the ratio of phenotypes in the F2 generation of a cross between true-breeding strains (AA bb x aa BB) can be described as 9:3:3:1 when A and B are dominant over a and b. To what phenotype does the "9" in the ratio refer?
a. recessive for both traits
b. dominant for one trait and recessive for the other
c.dominant for the A trait and recessive for the B trait
d. dominant for both traits
Final answer:
The "9" in the 9:3:3:1 ratio refers to offspring that are dominant for both traits, which occurs with a probability of 9/16 when considering two traits with independent assortment.
Explanation:
The "9" in the 9:3:3:1 phenotypic ratio refers to the phenotype that is dominant for both traits. This ratio emerges from the cross between true-breeding strains (AA bb x aa BB) when A and B are dominant alleles over a and b. Independent assortment ensures that alleles for different traits are transmitted to offspring independently of each other. Combining the probabilities for the dominant phenotypes of each trait, both texture and color in this context, we apply the product rule. Since each trait has a 3:1 dominant to recessive ratio, the probability of an offspring being dominant for both traits is (3/4) × (3/4) = 9/16.
The correct option is d. dominant for both traits. The 9 therefore represents the phenotype that is dominant for both traits.
The 9 in the 9:3:3:1 ratio refers to the phenotype that is dominant for both traits. This ratio is derived from the principles of Mendelian genetics, where each trait is determined by a pair of alleles, with one allele inherited from each parent.
In the given cross (AA bb x aa BB), the parents are true-breeding for different traits: one parent is homozygous dominant for trait A (AA) and homozygous recessive for trait B (bb), while the other parent is homozygous recessive for trait A (aa) and homozygous dominant for trait B (BB).
The F1 generation from this cross would all be heterozygous for both traits (Aa Bb), showing the dominant phenotype for both traits due to the presence of at least one dominant allele for each trait.
When the F1 individuals are crossed to produce the F2 generation, the alleles for each trait assort independently. The possible gametes for the F1 individuals are AB, Ab, aB, and ab, each with an equal probability of 1/4. The Punnett square for the F2 generation would look like this:
AB Ab aB ab
-------------------------
AB | AA BB | AA Bb | Aa BB | Aa Bb
-------------------------
Ab | AA Bb | AA bb | Aa Bb | Aa bb
-------------------------
aB | Aa BB | Aa Bb | aa BB | aa Bb
-------------------------
ab | Aa Bb | Aa bb | aa Bb | aa bb
The resulting phenotypic ratio is calculated by counting the number of each phenotypic class:
- Dominant for both traits (AA BB, AA Bb, Aa BB): 4 offspring with the dominant phenotype for both traits.
- Dominant for trait A and recessive for trait B (AA bb, Aa bb): 2 offspring with the dominant phenotype for trait A and recessive for trait B.
- Recessive for trait A and dominant for trait B (aa BB, aa Bb): 2 offspring with the recessive phenotype for trait A and dominant for trait B.
- Recessive for both traits (aa bb): 1 offspring with the recessive phenotype for both traits.
The ratio of these phenotypes is 4:2:2:1, which simplifies to 9:3:3:1 when multiplied by 2 to avoid fractions.
Urbanization often means the loss of green-spaces, causing which of the following problems? Select all that are true.
a. loss of aesthetic and cultural space
b. increased high temperatures (heat-island effect)
c. increased storm-water run-off, flooding, and pollution of waterways
d. increased pollinator and bird populations
Answer:
a. loss of aesthetic and cultural space✔
b. increased high temperatures (heat-island effect)✔
c. increased storm-water run-off, flooding, and pollution of waterways✔
d. increased pollinator and bird populations❌❌❌
Final answer:
Urbanization often leads to the loss of green spaces, causing problems such as loss of aesthetic and cultural space, increased high temperatures (heat-island effect), and increased storm-water run-off, flooding, and pollution of waterways.
Explanation:
Urbanization often leads to the loss of green spaces, which can cause several problems:
Loss of aesthetic and cultural space: Green spaces such as parks and gardens provide visual beauty and cultural value to urban areas. Their loss can diminish the overall aesthetic appeal and cultural heritage of a city.
Increased high temperatures (heat-island effect): Green spaces help to regulate temperature by providing shade and cooling through evapotranspiration. Without them, urban areas can experience increased temperatures, known as the heat-island effect.
Increased storm-water run-off, flooding, and pollution of waterways: Green spaces help absorb and filter rainwater, reducing storm-water run-off and the risk of flooding. Their loss can lead to increased run-off, which can carry pollutants and cause water pollution.
In an ocean ecosystem, seabirds are predators of fish. What would happen to the balance of an ocean ecosystem if the number of seabirds decreased? A. The population of fish in the ecosystem would increase. B. The population of fish in the ecosystem would decrease. C. The population of fish in the ecosystem would stay the same. D. The population of fish would disappear completely.
Answer: The correct option is A.
Explanation: Seabirds are predators of fish in an ocean ecosystem. When the population of sea birds decreases, the population of fish in the ocean increases. This is because sea birds feed on the fish and fish population is controlled by sea birds whenever seabirds decrease in number, then very low amount of fish is required to feed the sea birds so automatically fish population is increased.
Many muscular/skeletal problems and injuries especially in adults can be attributed to a lack of ___
Answer:
flexibility
Explanation:
Stretching may help prevent injury according to many doctors. Stretching improves your flexibility. "A sports medicine physician will determine your level of flexibility and form an activity or exercise prescription with specific exercises and stretches to improve your flexibility." This shows that your level of flexibility contributes to what activities you can do without injury.
Muscular/skeletal problems and injuries can be attributed to a lack of exercise or physical activity. Regular physical activity and exercise help to strengthen and maintain the health of skeletal muscles.
Explanation:Health: Understanding Musculoskeletal Problems and Injuries
Muscular/skeletal problems and injuries can be attributed to a lack of exercise or physical activity. When a person leads a sedentary life, their skeletal muscles can atrophy due to lack of use, while smooth muscles, which are involuntary muscles found in organs, do not typically experience the same level of atrophy.
Regular physical activity and exercise help to strengthen and maintain the health of skeletal muscles. When muscles are not regularly used and stimulated, they can weaken and lose mass, making them more susceptible to injuries and problems.
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According to Mendel’s law of dominance, which statement best describes the result of a cross between parents with genotypes Rr and Rr?
A) Only RR offspring will have red flowers.
B) Only Rr offspring will have red flowers.
C) All offspring with at least one R will have red flowers.
D) Because of the law of dominance, all offspring will have red flowers.
Answer:
The answer is C: All offspring with at least one R will have red flowers.
Explanation:
Mendel's Law of Dominance states that: “In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. Offspring that are hybrid for a trait will have only the dominant trait in the phenotype.”
This simply means that if there exists two contrasting traits (I.e Rr), one of the traits will always suppress the other, thereby expressing itself. R suppresses r, thus making the offspring with big R have red flowers. The trait is called a dominating trait and the suppressed is called recessive trait.
Of the following lines of flowering plants, which represents the line of evolution that is most closely related to other plant ancestors?
Basal Angiosperms
Answer:
Basal Angiosperms is the correct answer
Explanation:
Basal angiosperms are regarded to be the most primitive group of flowering plants. They are neither eudicots nor monocots but have been around for a very long time and for this time, they have been lumped together with eudicots to form a group known as dicots. They are woody in nature, but produce flowers and seeds.
Discuss the biological importance of each of the following organic compounds in relation to cellular structure and function in plants and animals.
a) Carbohydrates
b) Proteins
c) Lipids
d) Nucleic acids
Explanation:
Carbohydrates: they are the main source of ATP inside the cell and in the membrane, they can participate in cell recognition, adhesion, signaling and also as a physical barrier. Proteins: they form channels or bombs to exchange different molecules into or out of the cell, they also conform most of the enzymes responsible for many metabolic tasks inside the cell. Lipids: they are the main content of the cell's membrane, they form a fluid lipidic bilayer that permeates the cell from its environment and allows the movement of proteins and carbohydrates in it. Nucleic acids: they contain the genetic information of the cellI hope you find this information useful and interesting! Good luck!
The biological importance components:
The carbohydrates are the main ATP and participate in cell recognition. Proteins are forms of channels that exchange different molecules. Lipids are the main content of the cell membrane. Nucleic acids consist of genetic information from cellsWhat are Carbohydrates ?These are the main source of ATP in the cells and membrane, they can participate in cell recognition, adhesion, signaling, and also as a physical barrier.
While Proteins are channels or bombs to exchange different molecules into or out of the cells. Lipids: they are the main content of the cell's membrane, fluid layers. Nucleic acids consist of the genetic matter of the cell.
Find out more information about the compounds.
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Hydrothermal processes are most likely involved in the formation of which set of resources?
Answer:
mineral resources like the Zinc, copper, gold and silver.
Explanation:
The hydrothermal process usually includes the movement s of the subsurface hot waters and the upwelling of the magma from the earth mantel. This process leads to the formation of deposits such as those of the iron, zinc and gold and silver. What is the genus of plant that is easily identified by having no true leaves, dichotomous branching as well as lobed sporangia (usually yellow in color)
-None of these
-Psilotum
-Cardiospermum
-Ginkgo
-Equisetum -Marchantia
Answer:
Psilotum.
Explanation:
Psilotum is also known as whisk fern, it is a fern like vascular plants genus. This plant is found in tropical areas, and more temperate areas. Plants of this genus lack true leaves, and roots (they are anchored by creeping rhizomes).
The stem of Psilotum contains many branches with paired enations (like small leaves) but they do not have vascular tissues. By the fusion of three sporangia, above the enations synangia formed, and produced the spores.
The fluctuation of the water level in the small arm of the water seal with respirations is called:_________
Answer: Tidaling
Explanation:
It is rise and fall of water in water-seal chamber, used to determine the degree of re expansion of the lungs. It reduction shows the lungs reexpands.
However, with respiratory efforts it is normal.
If Avery had observed transformation using only the extracts containing degraded DNA, degraded RNA, and degraded protein, but NOT the extract containing degraded polysaccharides, he would have concluded that
Answer:
Polysaccharides are the genetic material.
Explanation:
Avery did not observe transformation using the extracts containing degraded DNA. On the other hand, extracts with degraded RNA, proteins, and polysaccharides exhibited transformation. Therefore, he concluded that DNA is the genetic material responsible for transformation. If he would have observed the process of transformation using extracts containing degraded DNA but not with degraded polysaccharides, he might have concluded that "polysaccharides were the genetic material responsible for the process of transformation."
Synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the__________.
Synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the Liver.
What is the Cori cycle?The Cori cycle refers to the metabolic pathway in which lactate is produced by the fermentation in the muscles and moves to the liver and is converted to glucose which then returned to the muscles and is converted back to lactate.
According to the context of this question, in animals, lactate formed in the muscles is successfully recycled to glucose in the liver. Lactate produced in the muscles is transported from the muscles to the liver where it is reoxidized by liver LDH to pyruvate.
Through the process of gluconeogenesis, pyruvate is finally converted into glucose in the liver.
Therefore, the synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the Liver.
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Final answer:
Glucose synthesis from pyruvate in the Cori cycle primarily occurs in the liver, where pyruvate undergoes gluconeogenesis to form glucose.
Explanation:
Synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the liver. The liver plays a crucial role in this metabolic cycle by converting lactate, which is produced by anaerobic glycolysis in the muscles during periods of intense activity, back to pyruvate. This pyruvate is then synthesized into glucose through a process called gluconeogenesis. The newly formed glucose is released into the bloodstream, providing energy for muscle cells and other tissues, thereby completing the Cori cycle.
(a) Addition of malate to an extract of muscle cells stimulates oxygen consumption by the extract. Explain why malate stimulates oxygen consumption.
(b) Malonate is an inhibitor of succinate dehydrogenase. If malonate is added to a suspension of bacteria that are catabolizing glucose then oxygen consumption is inhibited. Explain why malonate inhibits oxygen consumption.
Answer:
Oxygen consumption is a measure of the activity of the first two stages of cellular respiration: glycolysis and the Kreb's cycle. The addition of malate stimulates the citric acid cycle and thus stimulate respiration.
Explanation:
The added malate serves a catalytic role, because it is regenerated in the later part of the citric acid cycle. With higher concentrations of oxaloacetate or malate, higher flux of acetyl CoA will be utilized into the citric acid cycle increasing the oxygen consumption much greater levels.
The inhibition of succinate oxidation by malonate is a well known phe-
nomenon. Since the oxidation of succinate to fumarate is an integral
part of the Krebs cycle of oxidations, it has been generally assumed that
the inhibitory effect of malonate upon the oxidation of any member of
the cycle is the result of the inhibition of the succinate to fumarate step. malonate inhibits oxidations in the cycle by at least two mechanisms: in addition to the inhibition resulting from a block of succinate oxidation, malonate inhibits oxidation by another mechanism that is believed to involve combination with magnesium ions.
Calculating allele frequencies in a gene pool1. To calculate the frequency of the brown allele, count the number of BROWN ALLELES and divide by the total number of alleles in this population2. In this beetle population, the number of brown alleles is 83. In this beetle population, the total number of alleles for color is 204. The frequency of the brown allele in this beetle population is .45. The frequency of the green allele in this beetle population is .6
Answer:
1. To calculate the frequency of the brown allele, count the number of BROWN ALLELES and divide by the total number of alleles in this population
2. In this beetle population, the number of brown alleles is 8
3. In this beetle population, the total number of alleles for color is 20
4. The frequency of the brown allele in this beetle population is .4
5. The frequency of the green allele in this beetle population is .6
Reducing equivalents are key to the process of cellular respiration and ultimately ATP production. Illustrate the direct inter-linkage of the glycolytic pathway to cellular respiration, concluding with the general illustrative mechanism and description to the production of ATP. (Hint: draw the reason that you sometimes get 30 ATP out of glycolysis instead of 32.
→The correct question should be 'draw the reason that you sometimes get 30 ATPs out of cellular respiration instead of 32'
Answer :Glucose is high energy-rich ccompound. Therefore to initiate the breakdown down of bonds in glucose for phosphorylation of glucose; to obtain Pyruvate for ATPs synthesis 2ATPs must be supplied...
The 2 ATPs were ‘borrowed’ – to convert
1. Glucose to fructose phosphate,
2. And later to fructose bisphosphate. This is phosphorylation of glucose.
The end product of this reaction is PYRUVATE. THE NET ATPs produced is 4ATPs per molecule of glucose, but the total gained ATPs is 2, because the 2ATPs 'borrowed' from the phosphorylation must be paid back. This is substrate level phosphorylation and it takes place in the cytoplasm without OXYGEN.
Link reaction is the process of conversion of pyruvate from glycolysis to Acetyl CoA .It is the reaction pathway which linked glycolysis with citric acid cycle, in cellular respiration in the presence of oxygen. If no oxygen is available, link reaction does not take place; rather the pyruvate moves to fermentation
It involves decarboxylation, and dehydrogenation (with NADH) of 3Cpyruvate, and reaction with Coenzyme A to form 2C-acetyl CoA,
The 2C-acetyl Coenzyme A reacted with 4c oxaloacetate in the mitochondrial matrix to form 6C citrate. This involves decarboxylation and dehydrogenation with NADH.
The citric cycle continues on till 2ATPs are produces with each glucose molecule; until the 4 molecule oxaloacetate is reproduced. 4CO2, 6NADH, 2FADH2 and 2ATPs molecules per molecule of glucose are produced.
The NADH and FADH2 fetched Hydrogen into the mitochondrial matrix where it is split to
Protons and electrons.
The electrons are transferred by protein carries in the matrix. As these electrons moved through proteins carriers of different energy levels ENEGY is released;and used to pump protons from the matrix across the intramembrane to the cytosol. The proton gradient drives protons down the concentration gradient through the hydrogen channel of ATPase synthase. The enzyme tapped the energy from the protons to synthesise ATP from ADP.
This is oxidative phosphorylation.it produces 28ATPs
The balanced sheet of ATP per molecule of glucose =
→4ATPS from glycolysis ( 2ATPs used ).
→2ATPs from Citric Acid
→28 ATPs from Oxidative phosphorylation.
⇒Total =34 - (2ATPs) used =30ATPs.
Explanation:
50 POINTS PLZ HELP ASAP
Use the image to answer the question.
Question:
Explain how heat transfer is occurring through convection, radiation AND conduction in this picture. (3 points)
Answer:
The Heat is being transferred through the spoon via conduction and convection while whatever is hot in the mug is being transferred via radiation into the air
Explanation:
Answer and Explanation:
A cup of tea/coffee transfers heat via conduction, convection and radiation.
Conduction:
Heat transfer from a coffee to the spoon is in direct contact
Inner part of the tea cup against hot liquid will transfer heat to the outer part and saucer.
Convection:
Movement of molecules within the coffee. Hot molecules of the tea at the middle of the drink will be eventually make their way to the outer part of the drink near the tea cup, where they will loose their heat.
Radiation:
Thermal radiation is the loss of heat from a solid object in to atmosphere via the release of electromagnetic radiation. So hot outer edge of the teacup loss the energy to its surroundings via thermal radiation.
Reproduction by mitosis duplicates:A. chromosomesB. genesC. DNAD. all of the above
I do not understand why this was deleted, but Reproduction by mitosis duplicates all of the above, because all of these are in the gene being duplicated, the chromosomes, the gene itself, and the DNA.
Hope this helps, if not, comment below please!!!!
Reproduction by mitosis duplicates by chromosomes, genes and DNA which makes the answer option D
Reproduction by mitosis duplicates?Reproduction by mitosis results in the duplication of chromosomes, genes, and DNA. Mitosis is a type of cell division that ensures each daughter cell receives an identical set of genetic information to the parent cell. During mitosis, the chromosomes are replicated, and the genetic material, including genes and DNA, is duplicated to be equally distributed between the two daughter cells. This process allows for the growth, development, and tissue repair in multicellular organisms and ensures the genetic continuity of the cell's genome.
The answer to this problem is D. All of the above
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Sympathetic nerves may leave the spinal cord at which vertebra?
a. first coccyx
b. first thoracic
c. third lumbar
d. second cervical
Answer:
b. first thoracic
Explanation:
Sympathetic nervous system is one of the divisions of the autonomic nervous system. The other divisions are parasympathetic nervous system and enteric nervous system.
Autonomic nervous system is composed of preganglionic neurons and postganglionic neurons.
Similarly, the sympathetic nervous system is composed of postganglionic and preganglionic neurons.
The cell bodies of preganglionic neurons of sympathetic nervous system is found in intermediolateral cell columns of spinal cord segments T-T12 and L1-2 or 3. Therefore, sympathetic nervous system is also called thoracolumbar division.
The cell bodies of parasympathetic nervous system are found in the brainstem nuclei which exit via cranial nerves 3, 7, 9, 10 and also in the sacral spinal cord segements S2-4, therefore parasympathetic nervous system is also known as craniosacral division.
QUESTION:
b. first thoracic
This is the correct answer choice as sympathetic nervous system cell bodies are found in intermediolateral cell column of T1-12 and L1-2 or 3 spinal cord segments. Hence, a sympathetic nerve is most likely to leave via first thoracic vertebra.
Third lumbar is less likely as sympathetic nervous system does not extend to this level in every individual.
Sympathetic nerves leave the spinal cord at the first thoracic vertebra, playing a critical role in the regulation of the body's involuntary mechanisms.
Explanation:The sympathetic nerves, which are a key part of the autonomic nervous system that regulates body's unconscious actions, typically leave the spinal cord at the first thoracic vertebra. Therefore, the correct answer to your question is b: first thoracic. It's important to note that while they start at this vertebra, sympathetic nerves extend throughout the body to help regulate various involuntary mechanisms, such as heart rate, blood pressure, and digestion.
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In Labrador retriever dogs, coat color is determined by the interaction of two genes (pigment and deposition of pigment). This is called epistasis. Coat color can be black (B) or brown (b) and deposition of pigment into the hair shaft is deposited (E) or not deposited (e). If a retriever has the genotype where they have _ _ ee, then they will have a yellow coat regardless of what color they inherit (e.g., Bbee or bbee gives a Yellow lab).
1. Calculate the phenotypic ratios of a cross between a male and a female Labrador retrievers that are both heterozygous for coat color and heterozygous for deposition of pigment.
What is the genotype of both the male and female parents?
What are the possible gametes of both the male and female parents?
What is the phenotypic ratio of the offspring?
Answer 1,2,3,4:
As both are heterozygous therefore Male genotype will be (BbEe) and the genotype of female will be (BbEe). While both will have the phenotype black.
Parent Genotype
Male X Female
BbEe X BbEe
Possible gametes from Parents
Male X Female
BbEe X BbEe
(BE) (Be) (bE) (be) X (BE) (Be) (bE) (be)
...
F1 generation will have following genetypic ratio
Black Labrador = 9 = B_E_
Brown Labrador = 3 = bbE_
Yellow Labrador = 3 = B_ee
Yellow Labrador = 1 = bbee
Furthermore, the phenotypic ration will be
Black Labrador = 9
Brown Labrador = 3
Yellow Labrador = 4
1.Genotype of Parents:
Both the male and female parents have the genotype BbEe, being heterozygous for coat color and deposition of pigment.
2.Possible Gametes:
Male (BbEe): BE, Be, bE, be
Female (BbEe): BE, Be, bE, be
3.Phenotypic Ratio of Offspring:
Black with pigment (BBEE, BBEe, BbEE, BbEe)
Brown without pigment (bbEE, bbEe)
Yellow (BbEe, BBEe, bbEe, bbEE).
Genotype of Parents:
Both the male and female parents are heterozygous for coat color and heterozygous for deposition of pigment, denoted as BbEe.
Possible Gametes:
Male (BbEe): BE, Be, bE, be
Female (BbEe): BE, Be, bE, be
Phenotypic Ratio of Offspring:
When the alleles are combined, the possible genotypes of the offspring are:
BBEE (Black with pigment)
BbEE (Black with pigment)
BBEe (Black with pigment)
BbEe (Black with pigment)
bbEE (Brown without pigment)
bbEe (Brown without pigment)
BBEe (Brown without pigment)
BbEe (Brown without pigment)
The presence of "ee" ensures a yellow coat regardless of the color inherited. The phenotypic ratio is 4:2:4, representing the combinations of black with pigment, brown without pigment, and yellow, respectively.
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In humans, the AMY1 gene produces the enzyme amylase in cells of the salivary glands. Amylase breaks down starch (a polysaccharide) into the sugar maltose (a disaccharide). People from cultures with diets high in starch produce more amylase than people from cultures with diets low in starch because of a mutation in the AMY1 gene. Explain in two to three sentences why the frequency of this AMY1 mutation would have increased in frequency in populations with a high starch diet.
Answer:
Natural selection
Explanation:
AMY1 produces amylase, a protein that breaks down starch from food into maltose, which is the first step in properly digesting starch.
A mutation in the AMY1 gene that causes it to produce more amylase, would mean that individuals carrying that mutation would have more amylase in their saliva, and would be better at breaking down the starch.
If you can more efficiently break down the starch in your diet, it will be a better source of energy for you. This would have been particularly important thousands of years ago, when food security was poorer. If you lived in a region or were part of a culture where starch was an important part of your diet, you might have been better nourished if you could properly break down these starches.
Over time, natural selection would be at work: the individuals carrying the AMY1 mutation would have improved fitness, because of their ability to digest starch and get all the nutrition and energy from it (i.e. maybe they were less likely to get sick, to starve etc.). Therefore, very slowly, this mutation would become more frequent in those regions where it is beneficial.
The AMY1 mutation likely became more prevalent in high-starch diet cultures because it offers a survival advantage, allowing for more efficient starch digestion and nutrient absorption.
Explanation:The frequency of the AMY1 mutation that increases amylase production would have increased in populations with high-starch diets due to the principles of natural selection and evolution. Essentially, individuals with this mutation could more efficiently digest starch resulting in greater nutrient absorption and, consequently, a potential survival and reproductive advantage. Over many generations, these benefits would increase frequency of this AMY1 mutation within the population, reflecting the adage 'survival of the fittest'.
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A 35-year-old male, Mr. NX, presents to your clinic today with complaints of back pain and "just not feeling good." Regarding his back, he states that his back pain is a chronic condition that he has suffered with for about the last 10 years. He has not suffered any specific injury to his back. He denies weakness of the lower extremities, denies bowel or bladder changes or dysfunction, and denies radiation of pain to the lower extremities and no numbness or tingling of the lower extremities. He describes the pain as a constant dull ache and tightness across the low back.He states he started a workout program about 3 weeks ago. He states he is working out with a friend who is a body builder. He states his friend suggested taking Creatine to help build muscle and Coenzyme Q10 as an antioxidant so he started those medications at the same time he began working out. He states he also takes Kava Kava for his anxiety and garlic to help lower his blood pressure.1. His historical diagnoses, currently under control, are:A. Type II diabetes since age 27B. High blood pressureC. Recurrent DVTs2. His prescribed medications include:A. Glyburide 3 mg daily with breakfastB. Lisinopril 20 mg dailyC. Coumadin 5 mg daily
Answer:
Recurrent DVT
Coumadin 5mg daily
Explanation:
DVT is called Deep venous thrombosis.. It is characterized by blood clots in the legs which can dislodge to any part of the body and cause blockage of blood vessels causing recurrent pain..
Coumadin is a drug used in treating and preventing DVT
After the first exposure to an antigen, a ________ stimulates growth and multiplication of antigen-reaction cells. After the first exposure to an antigen, a ________ stimulates growth and multiplication of antigen-reaction cells. phagocytic immune response secondary innate immune response hyperactive cytotoxic response primary adaptive immune response
Answer:
Primary adaptive response
Explanation:
After the first exposure to an antigen, a primary adaptive response stimulates growth and multiplication of antigen-reaction cells. phagocytic immune response secondary innate immune response hyperactive cytotoxic response primary adaptive immune response
A client's head protrudes during a pulling assessment. Which of the following muscles should be stretched?
Answer:
Option C
Explanation:
Please see the attachment
DNA codes for the sequence of amino acids in the primary structure of a protein, but not for sugars or lipids. This is because?A. only proteins are involved in living metabolic reactions. B. sugars and lipids code for their own replication. C. sugars and lipids are ever present in the living environment and are not used in living structures. D. other hereditary molecules code for sugars and lipids. E. proteins are the main structural and functional components of cells.
Answer:
E. proteins are the main structural and functional components of cells.
Explanation:
DNA sequence codes for proteins since proteins are the main structural and functional components of cells. Proteins are structural components of cell membranes. The DNA packaging proteins pack the DNA into chromatin. The immune system distinguishes the self from non-self by identification of specific peptides present on the cell surfaces.
Hemoglobin is the oxygen carrier protein of RBCs. Immunoglobulins are one of the most abundant proteins of blood serum and are responsible for adaptive immune response. Enzymes are proteins with catalytic functions. Myosin and actin are the major structural and functional proteins of muscle cells. Proteins with different primary structures perform these diverse functions.
On his trip to the laguna atascosa wildlife refuge, mr Ortiz wants to analyze the overall health of alligator pond. Which of the tests should he perform?
Answer and Explanation:
To determine the overall health of the alligator pond, Mr. Ortiz should carry out the following tests:
pH level testpH is extremely important in this type of research as a reduction of pH could lead to a higher availability of toxic metals that may be absorbed by aquatic organisms such as alligators and fish species.
pH is acceptable in a range of 6.5 to 8.2.
Nitrates and Ammonia (NH3) levelsOversaturation of nitrates and ammonia could lead to the death of organisms that are crucial for the maintainance of a healthy pond and for the balance of the food chain.
Moreover, it could lead to algal blooms that result in a lower concentration of oxygen available for other organisms.
Over 0.2 per L could be lethal to some fish species that are also important for the health of the alligator pond.
Less than 0.1 indicates a healthy alligator pond.
Bacterial coliform testThis test determines the amount of bacteria that comes from waste (animal or human), which could lead to mild to serious bacterial infections and diseases.
If it is a healthy pond, the results should indicate less than 10 fecal coliform bacteria/100 mL.
Populations evolve for many reasons. Suppose there is a population of plants that have either purple flowers or white flowers, and the allele for purple flowers is dominant. This means that plants with two purple alleles have purple flowers. Plants with one purple allele and one white allele also have purple flowers. Only plants with two white alleles have white flowers. For each event or condition described below, answer the following questions.
a. Which mechanism of evolution is at work?b. How does this event affect the populations gene pool? Do the frequencies of the two alleles change, and if so, how?
Answer:
Natural selection by directional selection, was the driving force of evolution at work. The purple flowers parents were able to withstand the selective pressure and therefore emerged from completions with other flowers from previous generation, thus their dominance follow directional selection. This was the trait inherited by their offspring, the dominant purple
The gene pool of purple flower will be large; because of dominance in expression with each successive generation in both homozygote and heterozygote state, and therefore higher frequency of expression compared to the white flowers.
The frequency of allele depends on the rate expression of the allele in the population. Thus, the dominant purple flowers have high frequency of expression compared to the recessive white flowers which can only express itself in homozygote recessive state
During intramembranous ossification, the developing bone grows outward from the ossification center in small struts called _
Answer:
During intramembranous ossification, the developing bone grows outward from the ossification center in small struts called spicules.
Explanation:
The intramembrane ossification is the one that preferentially produces flat bones and, as the name implies, takes place within a connective tissue membrane. In this process, some of the mesenchymal cells that form the connective tissue membranes are transformed into osteoblasts constituting an ossification center around which bone is formed. Consequently, a certain amount of flat bones develop that are characterized by the presence of bone spicules. These spicules radiate from the primary ossification centers to the periphery. As growth continues during fetal and postnatal life, these bones increase in volume by apposition of layers on their outer surface and by osteoclastic resorption from the inside of each bone.