Answer:
Norepinephrine
Explanation:
Nerve cells and circulating hormones are the one responsible for vasoconstriction of the vessel walls, as they do not enter the tunica media of the blood vessel, the nerves do not synapse directly on the smooth muscle cells. Instead, they release the neurotransmitter, norepinephrine, which diffuses into the media and acts on the nearby smooth muscle cells, which result in contraction of the entire muscle cell layer and thus reducing the radius of the vessel lumen.
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.
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.
The DNA polymerase used in the polymerase chain reaction (PCR) was initially isolated from the bacterium Thermophilus aquaticus that lives in hot springs like Old Faithful in Yellowstone Park. Now called Taq polymerase, this enzyme can be produced synthetically and ordered from a catalog. What best explains why Taq polymerase must be used in PCR rather than human DNA polymerase?
Choose one:
A. Human DNA polymerase would need to be extracted from humans, which would be unethical.
B. Taq polymerase adds DNA bases more quickly than human DNA polymerase.
C. Taq polymerase is much easier to synthesize than human DNA polymerase.
D. Taq polymerase can withstand the high temperature needed to separate the DNA strands.
Answer:
The correct answer will be option-D
Explanation:
Polymerase chain reaction or PCR is the technique which creates the copy or replicates the DNA from a sample. The PCR technique is based on the DNA replication mechanism which employs the DNA polymerase enzymes and primers.
Since PCR is a technique therefore, to break the hydrogen bonds of the DNA, the sample is heated at a temperature above 94° C in which the natural DNA Polymerase will not withstand. So, an equivalent enzyme called Taq polymerase enzyme is used.
Since the Taq polymerase enzyme is extracted from a Thermophilic bacteria known as Thermus aquaticus therefore, the temperature can withstand the high temperature which is provided during the PCR.
Thus, option-D is the correct answer.
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.
The nurse educates a patient treated for psychosis with perphenazine (Trilafon) regarding the measures to be taken to manage and prevent side effects. The patient makes certain statements to reiterate the teaching. What statement of the patient indicates a need for further teaching?
Answer:
Explanation:
If the patient makes any statement that includes the wrong measures or actions that should be taken to prevent side effects then it is clear that they need further teaching. On the basis of treatment for psychosis with perphenazine then such wrong statement may be something like the following: "If I have a dry mouth then I need to take lozenges or eat hard candy for a period of 2 weeks"
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:
Which major branch of the left coronary artery curves to the left within the coronary sulcus, giving rise to one or more diagonal branches as it curves toward the posterior surface of the heart?
Answer: The circumflex artery
Explanation: the circumflex artery curves to the left around the heart inside the coronary sulcus, there by forming one or more left marginal arteries. the marginal arteries are also called obtuse marinal artery.
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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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.
Which statement is the best physical description of a gene? A chromosome is a segment of a gene. Genes carry many chromosomes. A gene is a specific segment of DNA on a chromosome. A gene consists of all of the DNA on a chromosome.
Answer:
The correct option is C. A gene is a specific segment of DNA on a chromosome
Explanation:
Option A is false because a gene is a segment on a chromosome.
Option B is false because a chromosome carries many genes.
Option D is false because a gene carries a segment of DNA, not all of the DNA.
Option C is correct because there are many genes present on a single chromosome. Each gene carries a segment of DNA. The position of a gene on the chromosome is known as its loci.
Answer:
c,
Explanation:
a gene is a specific sediment of DNA on a chromosome
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.
What is the difference in the structure of amylose and cellulose
Final answer:
Amylose and cellulose differ in their glycosidic linkages and overall structure. Amylose has alpha-1,4-glycosidic linkages and a helical structure, while cellulose has beta-1,4-glycosidic linkages and a more extended structure. Cellulose is insoluble in water and has mechanical strength, while amylose can form a colored product with iodine.
Explanation:
Amylose and cellulose are both linear polymers of glucose units, but they differ in the glycosidic linkages between the glucose units. Amylose has alpha-1,4-glycosidic linkages, while cellulose has beta-1,4-glycosidic linkages.
Cellulose has a more extended structure than amylose due to the beta-1,4-linkages, allowing for extensive hydrogen bonding between adjacent chains. This close packing results in cellulose being insoluble in water and having mechanical strength.
On the other hand, amylose has a helical structure formed by the alpha-1,4-linkages, and it can form a colored product when bound to iodine.
Amylose has α-1,4-glycosidic linkages leading to a helical structure, while cellulose has β-1,4-glycosidic linkages resulting in a straight chain that enables tight packing and significant mechanical strength.
The difference in the structure of amylose and cellulose lies in the type of glycosidic linkages that bind the glucose units together. In amylose, the glucose units are connected through α-1,4-glycosidic linkages, which promote a helical structure. On the other hand, cellulose is characterized by β-1,4-glycosidic linkages, which result in a straight, extended polymer chain. This structure of cellulose encourages extensive hydrogen bonding between adjacent chains, making them pack closely into fibers, giving cellulose its great mechanical strength and making it insoluble in water.
Due to the different linkages, amylose, and cellulose exhibit distinct physical properties. Amylose can interact with iodine to form a colored complex, which allows it to be identified in starch. In contrast, the lack of a helical structure in cellulose means it cannot bind to iodine to form a colored product.
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.
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.
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.
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 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.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:
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
What 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.
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Autonomic nervous system sympathetic and parasympathetic
Answer:
The answer is the subdivisions of the Autonomic nervous system.
Explanation:
The sympathetic and parasympathetic are the subdivisions of the Autonomic Nervous System. Their actions are usually antagonistic to each other.
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 explains how an oxygen-rich atmosphere was possible?
a) Rates of oxygen consumption were lower than rates of photosynthesis.
b) Photosynthetic organisms have existed on earth longer than heterotrophs.
c) All of these choices are correct.
d) Rates of carbon fixation and respiration have always been equal.
e) The number of plants has always outnumbered the animals.
Answer:
a
Explanation:
the rate of oxygen consumption were lower than rates of photosynthesis. Photosynthesis occurs at a faster rate than oxygen consumption, more oxygen present in atmosphere than carbon dioxide
Place the following in the correct order _____ Actin and myosin form links _____ Acetylcholine crosses the gap at the neuromuscular junction _____ Calcium is released into the sarcoplasm _____ Myosin cross-bridges pull actin filaments inward.
Answer:
The correct oder is given below:
Explanation:
The oder of the events during muscle conratction as as follows:
Acetylcholine crosses the gap at the neuromuscular junction.Calcium is released into the sarcoplasm. Actin and myosin form links. Myosin cross-bridges pull actin filaments inward.A nurse teaches the parents of a 7-year-old girl who has been prescribed long-term phenytoin therapy about care pertinent to this medication. Which statement indicates that the teaching has been effective?
a) "We give the medication between meals."
b) "We'll call the clinic if her urine turns pink."
c) "She's eating high-calorie foods, and we encourage fluids, too."
d) "We'll have her massage her gums and floss her teeth frequently."
Answer: I think c) "She's eating high-calorie foods, and we encourage fluids, too." is the correct answer.
Explanation:
"A technician is reading the plates from a sputum culture. On the sheep blood agar (SBA), she sees flat spreading colonies with a metallic sheen. On cetrimide agar, she sees a fluorescent green color in the media with clear colonies. On MacConkey, she sees medium clear colonies that have a "fruity or grape-like odor." What is the most likely organism?"
Answer:
P. aeruginosa
Explanation:
P. aeruginosa is a gram-negative bacteria that belongs to the family Pseudomonadaceae.
From the given question the following points lead us to conclude that the colony that will be growing would be of P. aeruginosa :
1. Flat spreading colonies with a metallic sheen on SBA - P. aeruginosa is known to produce smooth colonies with flat edges.
2. Fluorescent green color in the media with clear colonies on cetrimide agar - P. aeruginosa is known to produce pyoverdin which is a fluorescent pigment under low iron conditions.
3. Medium clear colonies that have a "fruity or grape-like odor" on MacConkey Agar - P. aeruginosa has a sweet fruity odor which is its characteristic odor because of the production of trimethylamine.
Thus, from all these characteristics one can conclude that the organism in the culture is P. aeruginosa.
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."
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.
A new species of plant is discovered that closely resembles a known species. Upon further examination, it is discovered that the new species has a tetraploid number of 4n = 12, while the known species has a diploid number of 2n = 6.Which of the following statements provides a reasonable explanation for the evolution of this new species of plant?
Answer:
Explanation:
An error in cell division occurred in the known species, leading to the development of an autopolyploid tetraploid offspring that evolved into the new species.