This is an example of an Epigenetic effect.
Explanation:
It is the study of heritable changes in the expression of genes, that are not involved in the changes present in DNA sequence. It is regular and naturally occurred, sometimes factors are responsible like age, lifestyle and disease state.
At times it leads to more damaging effect that can cause cancer. So this is also related to various fatal disease. During adulthood epigenetic effect remains stable. It does not occur in mother womb, but during the lifespan. This epigenetic effect can be reversed.
A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. What phenotypes are expected in these offspring? Select all that apply. A. wild type females wild type males B. scute-bristled females scute-bristled males C. ruby-eyed females ruby-eyed D. males scute-bristled, ruby-eyed females E. scute-bristled, ruby-eyed males
Answer:
Expected phenotypes include: Wild type males, wild type females, scute-bristled females, scute-bristled males, ruby-eyed females, ruby-eyed males, and scute-bristled, ruby-eyed males, and scute-bristled, ruby-eyed females
Explanation:
In this sense, it means that all of the options are possible phenotype of expected offspring. It is expected that there will be 425 Scute-bristled, 425 ruby-eyed offspring, 75 wild-type offspring and 75 scute-ruby offspring.
"The expected phenotypes in the offspring from the cross between phenotypically wild F1 females and homozygous double mutant males are:
A. wild type females wild type males
B. scute-bristled females scute-bristled males
C. ruby-eyed females ruby-eyed males
D. males scute-bristled, ruby-eyed females
E. scute-bristled, ruby-eyed males
To understand the expected phenotypes, let's analyze the cross step by step:
1. The initial cross is between scute-bristled females (which we can denote as sc+ rb+) and ruby-eyed males (sc+ rb). The F1 generation will all be heterozygous for both traits, with wild-type phenotypes for both bristles and eyes (sc+/sc rb+/rb).
2. The F1 females (sc+/sc rb+/rb) are then crossed with homozygous double mutant males (sc sc rb rb). Since the males are homozygous for both mutations, they will only contribute the recessive alleles for both traits to their offspring.
3. The possible gametes from the F1 females are:
- sc+ rb+ (wild-type for both traits)
- sc+ rb (wild-type bristles, ruby eyes)
- sc rb+ (scute bristles, wild-type eyes)
- sc rb (scute bristles, ruby eyes)
4. The only possible gamete from the double mutant males is sc rb (scute bristles, ruby eyes).
5. The possible offspring from the cross are then:
- sc+/sc rb+/rb (wild-type females and males)
- sc+/sc rb/rb (wild-type females, ruby-eyed males)
- sc/sc rb+/rb (scute-bristled females, wild-type males)
- sc/sc rb/rb (scute-bristled females, ruby-eyed males)
6. Since the males are homozygous for both mutations, there will be no wild-type or single mutant males in the offspring. All males will be double mutants (scute-bristled, ruby-eyed).
7. The females can be either wild-type, single mutants, or double mutants, depending on which alleles they inherit from their mother.
Therefore, the expected phenotypes are:
A. wild type females wild type males (from sc+ rb+ x sc rb gametes)
B. scute-bristled females scute-bristled males (from sc rb x sc rb gametes)
C. ruby-eyed females ruby-eyed males (from sc+ rb x sc rb gametes)
D. males scute-bristled, ruby-eyed females (from sc rb x sc rb gametes, but this is not a correct description as it implies all males are wild-type, which is not the case)
E. scute-bristled, ruby-eyed males (from sc rb x sc rb gametes)
The correct descriptions for the phenotypes are A, B, C, and E. Option D is incorrectly phrased because it suggests that some males are wild-type, which is not possible given the cross. All males will be scute-bristled and ruby-eyed."
Based on the results of the experiment, which
of the following types of molecules did the
bacteriophages most likely inject into the bacteria
cells?
(A) Simple carbohydrate
(B) Amino acid
(C) DNA
(D) Polypeptide
Answer:
The correct answer is (C) DNA
Explanation:
Bacteriophages are viruses that are known to infect bacteria. Viruses contain the genetic material in a proteinaceous capsid. The genetic material can be DNA or RNA.
During infection bacteriophage transfer the genetic material(DNA) in the bacterial cell and the protein coat remains outside the cell. So as the protein contain sulfur which is present in the protein coat of bacteriophage therefore radioactive sulfur will not be found in the bacterial cell.
Phosphorus is the part of DNA therefore if radioactive phosphorus is found in bacteria then it shows that the viral DNA is present in the infected bacteria. So the correct answer is DNA.
Scientists have modified a clathrin molecule so that it still assembles but forms an open-ended lattice instead of a closed spherical cage. How would this clathrin molecule affect endocytosis in cells?
Answer:
Endocytosis is a cellular process which helps the cell to bring large pr macromolecules inside the cell.
There are many known mechanisms of endocytosis but one of the best-understood mechanism is which involves the molecule called clathrin.
Clathrin is a protein which helps the endocytosis process by assisting the formation of the coated pit in the inner side of the cellular membrane.
When the inner surface when forms a vesicle and pinches off in the cytoplasm, then the clathrin allows the selection of the vesicles cargo and pits to reduce the surface area of the vesicle.
Thus, clathrin plays an important role in endocytosis.
Which of the following is not a hormone involved in water or electrolyte balance? Which of the following is not a hormone involved in water or electrolyte balance? atrial natriuretic peptide thyroxine aldosterone antidiuretic hormone
Answer:
Thyroxine
Explanation:
Thyroxine mainly serves to increase the basal metabolic rate of the body. It increases the rate of ATP production by cellular respiration by increasing the concentrations of enzymes involved in the process. It targets all the body cells to stimulate the glucose breakdown and production of more and more ATP molecules to support growth and development. Thyroxine, along with the other thyroid hormone is also required for the development of the nervous system as it promotes synapse formation, production of myelin and growth of dendrites. However, the hormone has no effect on the water and salt balance of the body.
Aldosterone triggers reabsorption of sodium ions and excretion of potassium ions while anti-diuretic hormone (ADH) stimulates water reabsorption by kidneys. Atrial natriuretic peptide stimulates excretion of sodium ions and thereby loss of more water in urine as it inhibits the secretion of aldosterone.
Assume a thylakoid is somehow punctured so that the interior of the thylakoid is no longer separated from the stroma. This damage will have the most direct effect on which of the following processes?
A) the splitting of water
B) the absorption of light energy by chlorophyll
C) the synthesis
The question is incomplete. The complete question is:
Assume a thylakoid is somehow punctured so that the interior of the thylakoid is no longer separated from the stroma. This damage will have the most direct effect on which of the following processes?
a. the splitting of water
b. the absorption of light energy by chlorophyll
c. the flow of electrons from photosystem II to photosystem I
d. the synthesis of atp
Answer:
d. the synthesis of ATP
Explanation:
During electron transfer from water to PSII and from the reaction center of PSII through an intermediate chain of carriers to PSI and finally to NADP+, protons are pumped from stroma to thylakoid space. This creates a proton centration gradient with higher proton concentration in the thylakoid space than stroma.
Since the thylakoid membrane is impermeable to protons, the downhill movement of protons through a proton channel drives the synthesis of ATP. If the thylakoid membrane is punctured, the proton concentration gradient would not be formed to support the process of ATP synthesis in chloroplasts.
The original source of all genetic variation is __________.
Why do the circulatory systems of land vertebrates have separate circuits to the lungs and to the rest of the body?
Answer:
Because it prevents venous blood from mixing with arterial blood (which is rich in oxygen), in this way the circulation is more efficient.
Explanation:
In land vertebrates, the blood circulation is structured in two independent circuits: the pulmonary circulation, where oxygenation of the blood occurs and the elimination of the carbonic anhydride that it contains, returning back to the heart through its left atrium; and the systemic or major circulation, impelled from the left ventricle, transports the oxygenated blood and the nutrients that it assimilates as it passes through the digestive system, to the tissues of the animal, where it is charged again with anhydride carbonic and other waste substances, returning back to the heart, where it enters through the right atrium. These systems are independent and prevents venous blood (which is poor in oxygen) from mixing with arterial blood. These systems are independent and prevents venous blood (which is poor in oxygen) from mixing with arterial blood.
Land vertebrates have separate circuits to the lungs and the rest of the body in their circulatory systems for efficient oxygen and nutrient distribution. This complex and efficient four-chamberedheart system evolved over time, advantageous for the high metabolic demands of terrestrial life.
Explanation:The circulatory systems of land vertebrates like mammals and birds have separate circuits to the lungs (pulmonary circulation) and the rest of the body (systemic circulation) for efficient oxygen and nutrient distribution. In systemic circulation, oxygenated blood from the heart is circulated to cells throughout the body where oxygen and nutrients are delivered and carbon dioxide and other wastes are collected. In pulmonary circulation, the deoxygenated blood is taken from the heart to the lungs where it is oxygenated before returning to the heart.
In contrast, earlier groups in the evolutionary lineage such as fish and amphibians have simpler heart structures and circulation systems. For example, fish have a two-chambered heart and blood flows unidirectionally through the gills and then the body. Amphibians and reptiles have a three-chambered heart, with some level of blood mixing.
However, the most complex and efficient circulatory system came with land vertebrates having the four-chambered heart that completely separated the oxygen-depleted and oxygen-rich blood further increasing the efficiency of oxygen and nutrient distribution which is particularly advantageous for the high metabolic demands of terrestrial life.
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Tay-sachs disease is caused by a mutation of one nucleotide for a protein that disrupts the activity of an enzyme in the brain. This leads to a toxic level of a substance to build up in neurons in the brain and spinal cord. true or false
Answer:
False
Explanation:
Tay-Sachs disease is an autosomal recessive abnormality that results in a progressive degeneration of the central nervous system, caused by the absence of an enzyme that helps break down fatty substances. These fatty substances, called gangliosides, accumulate at toxic levels in the child's brain and affect the function of nerve cells.
A fellow student brought in a leaf to be examined. The leaf was dark green, thin, had stoma on the lower surface only, and had a total surface area of 10 square meters. Where is the most likely environment where this leaf was growing?
the floor of a deciduous forest
a.a tropical rain forest
b.a large, still pond
c.a dry, sandy region
d.an oasis within a grassland
Answer:
The answer is letter A.
Explanation:
The tropical rain forest is most likely environment where this leaf was growing
Final answer:
The leaf with a large surface area and stomata on the underside is adapted for a tropical rainforest environment, where it maximizes photosynthesis and minimizes water loss under a light-limited canopy.
Explanation:
The leaf described, being dark green, thin, having a large surface area of 10 square meters, and having stomata on the lower surface only, is most likely native to a tropical rainforest. In such an environment, large leaf surface areas are crucial for maximizing photosynthesis under the canopy where light is limited. The stomata, being primarily on the underside, minimize water loss while still allowing for gas exchange away from intense sunlight. Additionally, the thin nature of the leaf is characteristic of plants that do not need to retain large amounts of water, which is consistent with the moist conditions of a tropical rainforest.
Contrastingly, leaves from dry, sandy regions or from plants like cacti have adaptations for minimizing water loss and typically present a smaller surface area. The leaf's characteristics do not align with those one would find in the other environments presented, such as floating on a large, still pond (where we would find wide, flat leaves that can float) or growing in a dry, sandy region (where smaller, thicker leaves with a reduced surface area are advantageous).
A populations carrying capacity
a. may change as environmental conditions change
b. can be accurately calculated using the logistic growth model
c. increases as the per capita growth rate (r) decreases
d. can never be exceeded
Answer:
The correct answer is a. may change as environmental conditions change.
Explanation:
The maximum number of individuals of a species population an ecosystem can sustain for a long time is called its carrying capacity. There are four major factors that influence the carrying capacity of an ecosystem. These are food, water, environmental condition, and space.
Change in environmental condition can affect the carrying capacity for example if the ecosystem is close to human population then pollution can affect the carrying capacity of the ecosystem adversely because many organisms can not survive well in polluted ecosystem. Therefore carrying capacity can decrease for some species population. So the right answer is a.
A mutation occurs in the trp operon DNA of E. coli and results in the change to the two UGG tryptophan codons in the 5′ UTR of the RNA to UAG stop codons. What effect will this mutation be expected to have on the regulation of this mutant trp operon compared to a wild-type operon?
A) In the presence of tryptophan, transcription of the structural genes will be reduced compared with a wild-type operon.
B) In the absence of tryptophan, transcription of the structural genes will be reduced compared with a wild-type operon.
C) In the presence of tryptophan, the repressor will bind to the operator/promoter region with the mutant operon more strongly than with a wild-type operon.
D) In the absence of tryptophan, RNA polymerase will not bind to the operator/promoter region with the mutant operon.
E) In the presence of tryptophan, transcription of the structural genes will be enhanced compared with a wild-type operon.
Answer:
E) In the presence of tryptophan, transcription of the structural genes will be enhanced compared with a wild-type operon.
Explanation:
A mutation can be described as any changes which occur in the DNA of an organism. Some mutations can be useful whereas others have devastating effects.
A stop codon is required to stop the transcription of a gene. If this code becomes mutated, then the process of transcription will continue to happen unless a new stop codon is recognized.
Hence, when tryptophan will be present, the mutations will cause more transcription of the structural genes.
A snail, elodea (aquatic plant), or both were added to the tubes and they were stoppered. Tubes were placed under a grow light for 24 hours and the results recorded. The test tubes were then covered for 24 hours to produce a dark environment and the results recorded. Which statement best explains the change in Test Tube D from the light setup to the dark setup
Answer:
The correct option is "The rate of photosynthesis decreased in the dark, thus the elodea used less CO2. The elodea and snails continued to produce CO2 through cellular respiration, raising the level of CO2 in the tube, causing the solution to become more acidic (yellow)."
Explanation:
The process of photosynthesis is dependent on light energy. As the test- tubes were kept in the dark hence, the rate of photosynthesis was minimal. Less carbon dioxide was used for this process by the elodea.
The process of cellular respiration is not dependent on light. Hence, both the snail and elodea will carry out respiration and release carbon dioxide. Due to this carbon dioxide, the solution will turn acidic.
Answer:
b
Explanation:
You perform an in situ hybridization on Drosophila melanogaster embryos. You apply a probe labeled with a fluorescent stain that is complementary to mRNA transcripts of the gene hunchback. In early stages of embryogenesis you observe fluorescence for the entire anterior half of the embryo only. In embryos at a later stage of development you see the same pattern in the anterior, and also a stripe of fluorescence in the posterior half. What can you conclude about the expression of hunchback from this? Group of answer choices Hunchback moves from the anterior of the embryo to the posterior during development. Hunchback expression is more important in the anterior than in the posterior. Hunchback is never expressed in the posterior half of the embryo. Hunchback is first expressed in the anterior of the embryo and later in a portion of the posterior. Hunchback has been "knocked out" in some portions of the embryo.
Answer:
The morphological features that embryos of various animals share between each other the greater the likelihood they are derived from a common ancestor
Explanation:
The samples that you added to the microplate strip contain many proteins and may or may not contain the disease antigen. What happened to the proteins in the plastic well if the sample contained the antigen?
Answer:
If the sample contain antigen then there will be indirect capture to the protein in plastic well.
Explanation:
The antigen cannot come to the surface of the plate by the antibody which is previously attached to the surface. Antigen is the foreign body that induces immune system and produce antibodies. Antibodies are usually Y-shaped that is produced by the 'B-cells' of the 'immune system' in response to the antigen. Antibody contains paratope that recognizes epitope on antigen and acts like bind which helps to eliminate antigen from the body."Top" carnivores: 1) consume primarily herbivores.
2) rely directly on primary producers for energy.
3) consume primarily carnivores.
4) are more common than secondary consumers.
5) rely on symbiotic bacteria living in their digestive systems to help digest cellulose.
Answer:
3) consume primarily carnivores.
Explanation:
Carnivores are animals that consume other animals to obtain energy and nutrients, this can happen through predation or scavenging. A top carnivore or apex predator is a carnivore at the top of the food chain, this means that it doesn't have predators and it consumes secondary consumers (that are also carnivores).
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Aldosterone is a hormone that is released by the adrenal cortex when your blood pressure is low. It is a fat-soluble molecule and causes water conservation in the kidneys by causing more sodium pumps to be made when it binds with its receptors. Which of the following would be true about his hormone?
1) Aldosterone has receptors on the plasma membrane of kidney cells and activates a signal transduction pathway.
2) Aldosterone has nuclear receptors inside kidney cells and activates gene expression.
Answer:
The correct answer is 2 Aldosterone has nuclear receptors inside the kidney cells and activates gene expression.
Explanation:
Aldosterone is a steroid hormone as a result it is lipophilic in nature.Aldosterone is a mineralocorticoid secreted from adrenal cortex.
As it is lipophilic aldosterone can cross the plasma membrane and binds to the nuclear inside the kidney cell which ultimately result in the genes encoding proteins that stimulate water and Na+ ions conservation by the kidneys.
Which of the following statements about aldosterone is NOT correct? View Available Hint(s) Which of the following statements about aldosterone is NOT correct? Aldosterone is stimulated by decreased plasma sodium levels and increased plasma potassium levels. Aldosterone increases the number of passive sodium channels in the luminal membrane of the distal tubule and collecting duct, thus aiding sodium reabsorption. Aldosterone is produced in the adrenal cortex. Aldosterone increases sodium reabsorption by increasing the number of Na+-K+ ATPase pumps in the luminal membrane of the proximal tubule.
Answer:
C. Aldosterone increases sodium reabsorption by increasing the number of Na+-K+ ATPase pumps in the luminal membrane of the proximal tubule.
Explanation:
The terminal part of distal convoluted tubules has two different types of cells: principal cells and intercalated cells. These cells are also present throughout the collecting duct. The principal cells serve to reabsorb Na+ and secrete K+ and have receptors for aldosterone and antidiuretic hormone (ADH).
The function of intercalated cells reabsorb HCO3 - and K+ and to secrete H+. Angiotensin II stimulates the adrenal cortex to release aldosterone which in turn stimulates the principal cells in the collecting ducts to reabsorb more Na+ and secrete more K+. The increased reabsorption of Na+ facilitates reabsorption of more water to increase blood volume and blood pressure.
Aldosterone increases sodium reabsorption by increasing the number of Na+-K+ ATPase pumps in the luminal membrane of the proximal tubule.
Explanation:The correct statement about aldosterone is that it increases sodium reabsorption by increasing the number of Na+-K+ ATPase pumps in the luminal membrane of the proximal tubule.
This helps in actively transporting sodium out of the tubule and into the bloodstream.
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What is the most likely consequence of overexpressing either a cdc42 gef or cdc42 gap in migrating cells?
Answer:
The Cdc42 guanosine triphosphatase is essential for cell polarization in several organisms and in vitro for the organization of polarized epithelial cysts. A long-standing question concerns the identity of the guanine nucleotide exchange factor (GEF) that controls this process. Using Madin-Darby canine kidney cells grown in Matrigel, we screened 70 GEFs by RNA interference. Of these, six positives were identified that caused a multilumen phenotype, including Tuba, a Cdc42-specific GEF localized below the apical cortex. Loss of Tuba abolishes Cdc42 enrichment at the apical cortex. Normal lumen formation is rescued by human Tuba or active Cdc42 but not by a GEF-negative Tuba mutant. Silencing Cdc42 causes a similar phenotype, including multilumen formation and reduced atypical protein kinase C (aPKC) activity. Lumen disorganization after depletion of Tuba or Cdc42 or inhibition of aPKC is caused by defective spindle orientation. Together, our findings implicate Tuba as a key activator of the Cdc42 GTPase during epithelial ductal morphogenesis, which in turn activates apical aPKC to ensure that spindles orient parallel to the lateral plane.
The human insulin gene contains a number of sequences that are removed in the processing of the mRNA transcript. In spite of the fact that bacterial cells cannot excise these sequences from mRNA transcripts, explain how a gene like this can be cloned into a bacterial cell and produce insulin.
Answer:
In order to produce human insulin in a bacterium you need to isolate the mature mRNA (after the excision of the introns), produce a DNA copy using the enzyme retrotranscriptase, clone the gene in an expression vector and transform bacterial cells with this genetic construct to produce the insulin.
Explanation:
Eukaryotic cells perform a process called splicing, in which introns (the non-coding fragments of the gene). are excised from the mRNA Prokaryotic cells (i.e. bacteria) don’t perform this process. So, in order to produce a recombinant protein in a bacterium, insulin in this case, the best strategy should be to produce a complementary DNA (cDNA) of the mature mRNA (AFTER the splicing process, the excising of the introns).
All animals are________ , organisms with cells that are relatively large, complex, and contain membrane-enclosed organelles such as the nucleus.
Answer:
Eukaryotes
Explanation:
All the animals are eukaryotes because they are made up of eukaryotic cells. Eukaryotic cells are different from prokaryotic cell as they are larger and complex than prokaryotic cell. Eukaryotic cell contain a membrane-bound nucleus in which the genetic material of all eukaryotes are assembled.
Eukaryotic cells also have other membrane-bound organelles like mitochondria, Golgi apparatus, endoplasmic reticulum, etc. As nucleus is absent in prokaryotes, therefore, transcription and translation takes place simultaneously in prokaryotes. Therefore the correct answer is eukaryotes.
Answer:eukaryotic
Explanation:
Albuterol and epinephrine both have bronchodilation properties that improve the amount of oxygen that a person can inhale and absorb. However, Albuterol is administered only for asthma, whereas epinephrine is administered for both asthma and anaphylaxis. Why is epinephrine, and not Albuterol, the first choice for anaphylaxis?
Answer:
The correct answer is Albuterol is not a vasoconstrictor.
Explanation:
Anaphylaxis is a dangerous allergic reaction caused by any chemical contained in a medicine or protein antigens that can cause death by suffocation if not treated quickly.
When the immune system is faced with a substance to which the body is allergic, it automatically releases chemicals that cause the allergy. One of these chemicals released is histamine.
In the case of anaphylaxis, the blood pressure drops suddenly, the airways close and extreme vasodilation occurs. Vasodilation is caused by histamine and what it produces is that the blood vessels dilate, and produce an outflow of fluid. If this does not stop immediately, the patient may suffer bronchoconstriction where their bronchial tubes narrow due to histamine.
Albuterol is not vasoconstrictor, that is, it is not able to close these vessels to prevent the passage of fluid, therefore the most appropriate in these cases is to use epinephrine.
This medicine is vasoconstrictor, which will generate a relaxation in the muscles of the airways and narrow the blood vessels.
Final answer:
Epinephrine is favored for anaphylaxis due to its ability to act on both alpha and beta-adrenergic receptors, raise blood pressure, and provide anti-inflammatory effects, which are essential in treating the severe symptoms of anaphylactic shock.
Explanation:
Epinephrine is the first choice for treating anaphylaxis because it exerts effects through both α- and β-adrenergic receptors. While both epinephrine and albuterol are used for their bronchodilation properties to improve the amount of oxygen a person can inhale, epinephrine also raises blood pressure and has anti-inflammatory effects on the immune system, which are crucial in counteracting severe allergic reactions and the quick drop in blood pressure seen in anaphylaxis. On the other hand, albuterol is typically prescribed for asthma as it primarily acts on the β2-adrenergic receptors to relax the bronchial muscles, and it lacks the broader systemic effects of epinephrine that are necessary to manage anaphylactic shock.
Anaphylaxis involves systemic mast cell degranulation causing a rapid decrease in blood pressure and bronchial smooth muscle contraction, potentially leading to fatal anaphylactic shock. Since epinephrine raises blood pressure and relaxes the bronchial smooth muscle, it is life-saving in these situations. The rapid administration of epinephrine is crucial, and patients with known severe allergies are advised to carry automatic epinephrine injectors.
The first step leading to angiotensin II production is the secretion of what by the kidneys? Multiple Choice Calcitriol Angiotensin converting enzyme Angiotensin I Angiotensinogen Renin
Answer:
The correct answer is "Renin".
Explanation:
The secretion of the enzyme renin by the kidney is one mechanism used by the kidneys for blood pressure regulation. The secretion of renin is the first step leading to angiotensin II production, in what is known as the renin–angiotensin system (RAS). Renin catalyzes the conversion of angiotensinogen to angiotensin I, which is subsequently converted to angiotensin II by the angiotensin-converting enzyme (ACE).
The first step leading to angiotensin II production is the secretion of renin by the kidneys, which then acts on angiotensinogen to produce angiotensin I.
Explanation:The first step leading to angiotensin II production is the secretion of renin by the kidneys. When there is a low fluid (blood) volume or low sodium concentration, the kidneys produce renin through the juxtaglomerular complex. Renin then converts angiotensinogen, a protein produced by the liver, into angiotensin I. Following this initial step, angiotensin converting enzyme (ACE), primarily found in the lungs, plays a critical role by converting angiotensin I into the physiologically active form, angiotensin II, which helps regulate blood pressure and fluid balance by constricting blood vessels and triggering various hormonal responses, including aldosterone and ADH release.
Francesca s immune system has begun to turn against normal body tissues, which is referred to as __________." selective optimization primary aging an autoimmune response compression of morbidity
Answer:drfdrgfdhbrf
Explanation:
Which balance-stabilization exercise is when an individual lifts one leg while maintaining the optimal alignment including level hips and shoulders, and then lifts a medicine ball in a diagonal pattern until the medicine ball is overhead?
Single leg lift and chop balance stabilization exercise is when an individual lifts one leg while maintaining the optimal alignment including level hips and shoulders, and then lifts a medicine ball in a diagonal pattern until the medicine ball is overhead.
Explanation:It is also a strength exercise which requires total body movement.People perform this exercise to enhance their general fitness. While performing single leg lift and chop, it is important to understand how it is performed rather than how many times.
The Chop and Lift is a great exercise to help correct movement patterns, increase whole body integration and eliminate compensation patterns as well as provide an effective training stimulus to develop power, strength, stamina and stability (control).
The events listed below generally take place during meiosis.I. Synapsis occurs. II. Crossing-over is completed. III. Condensation of chromosomes begins. IV. Separation of homologous chromosomes begins.Which of the following is the correct sequence of these events?
Answer:
III. Condensation of chromosomes begins.
I. Synapsis occurs.
II. Crossing-over is completed.
IV. Separation of homologous chromosomes begins
Explanation:
The prophase I of meiosis I begins with the condensation of chromosomes. The process of compaction makes the individual chromosomes visible and the stage is called leptotene.
Leptotene is followed by zygotene of prophase I during which the homologous chromosomes are paired together. The process is mediated by the synaptonemal complex. As the homologous chromosomes are paired, each pair is visible as a tetrad as each of the chromosomes of a pair has two sister chromatids.
The exchange of part of chromatids occurs during crossing over. Crossing over is the event of the pachytene stage of prophase I. After crossing over, the homologous chromosomes begin to separate from each other during diplotene but stay paired at the points of crossing over. These points are called chiasmata. Diakinesis of prophase I is marked by the dissolution of chiasmata.
Cell division can be of two types mitotic or meiotic division. Meiosis is a reduction division in which chromosomes number gets halved in parental cells to produce four gametes.
This type of division occurs in the reproductive cells of an organism. For example sperm cells and eggs.
The correct order of the events are:
III. Condensation of chromosomes begins.
I. Synapsis occurs.
II. Crossing-over is completed.
IV. Separation of homologous chromosomes begins.
Explanation for the correct order:
Meiosis occurs in two stages: meiosis 1 and meiosis 2. In meiosis, there is a separation of homologous chromosomes and the cell reduces to the haploid stage.The first stage of meiosis 1 is prophase 1, in which DNA and protein condense to form chromosomes. The homologous chromosome then forms synapses and the paired chromosome are called bivalents. In the pachytene stage of prophase 1 crossing over takes place in the part of chromatids. At the diplotene stage of prophase separation of homologous chromosomes begins and at the end of the stage called diakinesis, the chiasmata dissolve.Therefore the correct order of events in meiosis is divided into stages leptotene, zygotene, pachytene, diplotene, and diakinesis.
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Identify the state of DNA (just DNA, in a plasmid, or in a virus) that is transferred from one organism to another in transformation. Explain Griffith’s experiment including if rough and smooth strains of S. pneumonia are pathogenic or nonpathogenic and how the rough strain was changed into a smooth strain of the bacteria?
Answer: The correct answer is : The state of the DNA that is transferred from one organism to another is Naked DNA.
Griffith's experiment went like this:
* Rough strain
* do not cap
Smooth strain
* capsule
* virulent (pathogenic) avirulent
Through DNA-mediated transformation, bacteria enter the DNA
Smooth strain (capsule) -> killed the mouse
Rough strain (no capsule) -> no effect on mouse
Heat killed smooth strain -> no effect on mouse
Heat killed smooth strain + living rough strain mixed -> killed the mouse
An important challenge to traditional (pre-1860) ideas about species was the observation that seemingly dissimilar organisms, such as hummingbirds, humans, and whales, have similar skeletal structures. This most directly suggested to biologists that
only the best-adapted organisms can survive
advantageous changes can be passed along to offspring
most evolution occurs rapidly following a mass extinction
dissimilar organisms might have evolved from a distant, common ancestor
all of the above
Answer:
The correct answer will be option- dissimilar organisms might have evolved from a distant, common ancestor
Explanation:
Evolution is a field of biology which studies the history of survival of the organism on Earth or the history of life but not the origin of life on earth. The evolutionary biologists thus collect the evidence in the form of fossils which provides a clue to understanding the survival of species in past times.
Before the concept of natural selection provided by Charles Darwin, Evolutionary Biologists were puzzled with their same skeletal structural specimens collected but of the dissimilar organisms as it lacks the links between these species.
It was the theory of evolution which suggested the mechanism of divergent evolution which solved this puzzle.
Thus, the selected option is the correct answer.
A nurse observes a few small, yellow nodules on the cervix of a client during the speculum exam. They are not painful or odorous, and a thin, clear discharge is present. The nurse recognizes that these are most indicative of what type of condition
Answer:
A nurse observes a few small, yellow nodules on the cervix of a client during the speculum exam. They are not painful or odorous, and a thin, clear discharge is present. The nurse recognizes that these are most indicative of nabothian cysts.
Explanation:
Nabothian cysts or nabothian follicles are also called mucinous retention cysts or epithelial cysts. It is a mucus-filled cyst on the surface of the cervix. Many women have multiple cysts they are common, benign and considered a normal feature of the adult cervix. They may be translucent or opaque, whitish to yellow, and range from a few millimeters to 3 to 4 cm in diameter. They are most often caused when stratified squamous epithelium of the ectocervix which is the nearest portion to the vagina that grows over the simple columnar epithelium of the endocervix which is the nearest portion to the uterus.
There are no serious complications or threat to your health with nabothian cysts.
answer correctly / explain a lil im timed
Q.1A monomer is to a polymer, like
a. one skittle is to a bag of skittles
b. a bucket of fried chicken is to a family reunion.
c. a classroom of students is to one student
d. four white tailed deer are to a heard of hippos
Q.2 Identify the products in this reaction:
water + carbon dioxide ----> glucose + oxide
a. glucose and oxide
b. carbon dioxide and glucose
c. water, glucose, and oxide
d. water and carbon dioxide
Q.3 Which of the following is a plausible or scientifically acceptable general principle offered to explain phenomena?
a. theory
b. scientific law
c. hypothesis
d. fact
Q.4 The smaller subunits that make up nucleic acids
a. phosphorous
b. nucleotides
c. carbon
d. disaccharides
Q.5
What are the building blocks of carbohydrates?
a. Amino Acids.
b. Monosaccharides.
c. Glycerol and fatty acids.
d. Nucleotides.
Q.6 parkle Bunny Unicorn measured the pH of her rainbow candy. She finds it to be 3.7. What can she infer from this data?
a. Her candy is very acidic
b. Her candy is neutral
c. Her candy is very basic
d. Her candy should be pulled off the shelves of every store making Sparkle Bunny Unicorn a sad mystical creature.
Q.7
Which of the following is not an example of adhesion?
a. water sticking to itself forming larger drops
b. a water strider walking on water
c. water traveling up a tree
d. water sticking to a straw
Answer AND Explanation:
a classroom of students is to one studentglucose and oxidetheorynucleotidesMonosaccharides.Her candy is very acidic water sticking to itself forming larger dropsWhat is the probability that a cross between a true-breeding pea plant with spherical seeds and a true-breeding pea plant with wrinkled seeds will produce f1 progeny with spherical seeds?
Answer:
1 or 100%
Explanation:
Let the allele for seed shape be represented by S.
True breeding spherical shape seeded pea plant will have SS as genotype
True breeding pea plant with wrinkled seeds will have ss as genotype
Crossing the two: SS x ss
Progeny: Ss, Ss, Ss, Ss
Based on the assumption that spherical shape allele is dominant over wrinkle shape allele, all the F1 offspring will have spherical seeds.
Hence, the probability is 1 or 100%
The probability of a cross between a true-breeding pea plant with spherical seeds and a true-breeding pea plant with wrinkled seeds producing F1 progeny with spherical seeds is 100% because the spherical seed trait is dominant over the wrinkled seed trait according to Mendel's principle of dominance.
Explanation:The probability of a cross between a true-breeding pea plant with spherical (round) seeds and a true-breeding pea plant with wrinkled seeds resulting in F1 progeny with spherical seeds is 100%. This is based on Gregor Mendel's principle of dominance. According to Mendel, one trait can be dominant to another, meaning the offspring (F1 generation) from a cross between true-breeding parents each showing a different trait will all show the dominant trait. In the case of pea plants, the trait for round seeds is the dominant trait over the trait for wrinkled seeds. Therefore, all F1 progeny will display round seeds.
However, when these F1 plants undergo self-fertilization, in the next generation (F2), the trait for wrinkled seeds reappears in approximately 25% of the offspring. This is due to what is now known as Mendel's principle of segregation. In other words, genes for different traits separate in the formation of sex cells.
Learn more about Genetics here:https://brainly.com/question/30459739
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