Answer:
Plants, animals, or pathogens that are non-native (or alien) to the ecosystem under consideration and whose introduction causes or is likely to cause harm.”
Explanation:
so, yes
Answer:
No Humans are not ivasive species
Notable examples of invasive plant species include the kudzu vine, Andean pampas grass, and yellow starthistle. Animal examples include the New Zealand mud snail, feral pigs, European rabbits, grey squirrels, domestic cats, carp and ferrets.
A vaccine contains
a) antigens
b) antibodies
c) antibiotics
d) enzyme inhibitors
Answer:
i believe your answer is c)antibiotics
Explanation:
Anatomical barriers play a large role in preventing the entrance of a pathogen into the body. They provide a critical first line of defense. Think about what a potential pathogen like a bacteria or a virus would have to overcome in order to get into your body, or Charlie’s body, and then into the target tissue to untimely make you sick. List several (at least three) of these critical barriers and how they help to eliminate pathogens. Describe how these mechanisms are or could be related to the symptoms you often get when someone falls ill (for example with the cold or a stomach flu).
Answer:Anatomical barriers prohibit any unwanted entrance and colonization of several microbes.
Anatomical barrier are for example the skin,bony encasements and mucous membranes.
Explanation:The skin
The skin is made up of the epidermis and dermis which is dry with a 37 degree Celsius temperature and is also acidic.
The conditions mentioned above are negative for bacterial growth hence they discourage it.
Surface of the skin usually from.the dead , keratinized cells are constantly sloughed off in order to prohibit colonization by microbes
Sweat glands and our hair follicles are also crucial in eliminating bacteria through their toxic lips and lysozyme.
There are also T-lymphocytes underneath the epidermis which fight every foreign microbes in our bodies.
The mucous membranes
Mucus physically traps microbes. They are found in our respiratory tract, the gastrointestinal tract, and the genitourinary tract. Mucus contains lysozyme which destroys bacteria and has secretory IgA which prohibit microbes by trapping them in mucous.
Bony encasements
The example is our skull and thoracic cage that helps us not get affected by an unwanted entry from microbes.
Describe how these mechanisms are or could be related to the symptoms you often get when someone falls ill :
-The cough and sneeze reflex:
As we cough and sneeze the mucus that comes out , goes out with the trapped microbes.
Vomiting and diarrhea: Our body fight infections or any toxins by removing them through the gastrointestinal tract.
The other ways is through actual physical removal of wastes and unwanted products which also includes flushing microbes when we urinate , perspire,cry and and through saliva.
Anatomical barriers such as the skin, chemicals like enzymes, and helpful bacteria play key roles in preventing pathogen entry into the body. Key barriers include mechanical, chemical, and biological defenses.
Here are three critical barriers:
Mechanical Barriers: The skin is an effective physical barrier. It is made out of dead cells that are continually shed, eliminating any stuck microorganisms. Moreover, hairs in the nose trap bigger particles. Synthetic Barriers: Substances like chemicals in sweat and spit kill microbes on body surfaces. For instance, salivary lysozyme prevents infection by breaking down bacterial cell walls. Biological Barriers: Beneficial bacteria on the skin and in the intestines compete with pathogenic bacteria for resources and space, preventing colonies of harmful bacteria.These barriers are a part of the body's natural immune system and can be linked to illness-related symptoms. For instance, when microorganisms infiltrate these obstructions, the body answers with aggravation, prompting redness and expanding.
Scientists use a hierarchical system to classify organisms.
The broadest categories of classification are known as Kingdoms.
Organisms in the Kingdom Monera, or Eubacteria and Archaebacteria, include those who
A. can move independently, such as animals.
B. reproduce by spores, such as mushrooms.
C.
lack a true nucleus, such as bacteria.
D. have a nucleus and chloroplasts, such as algae.
Answer:
lack a true nucleus, such as bacteria.
Explanation:
The Kingdom Monera, which is now split into the kingdoms Eubacteria and Archaebacteria, includes organisms who lack a true nucleus, such as bacteria. In Monerans, the genetic material is loose within the cell.
Which bead phenotypes avoided predation best? Why do you think this phenotype performed better than the other?
Phenotype of the predators themselves avoid predation, which include their strong physical body structure, desire to make others prey.
Explanation:
Phenotype are the set of characteristics which are said to be seen in an organism. These are the observable kind of characteristics. Phenotype includes in them the physical form, structure or the morphology of the organism.
It also includes in the biochemical, physio social properties of the organism. The phenotype of the organisms which themselves are the predators are the ones responsible for avoiding their own predation.
A geneticist crossed pure breeding black mice (BB) with pure breeding brown mice (bb). All the mice in the F1 (first) generation had black coats. When these mice were crossed, they yielded 961 black-coated mice, and 317 brown-coated mice. The Punnett square below describes this cross. The numbers 1 to 4 are shown instead of the genotypes for the offspring.
Which of the genotypes, if any, is MOST likely to occur in the second generation that is shown in the Punnett square above? (2 points)
Group of answer choices
Bb
The Punnet square that shows each of the boxes is shown in the image attached.
What is the Punnet square?
When it comes to forecasting the inheritance of traits governed by basic Mendelian genetics—in which alleles obey the laws of dominance, recessiveness, and segregation—the Punnett square is especially helpful. The patterns that emerge in the squares offer information about the likelihood that the offspring will have distinct genotypes.
In genetics, a Punnett square is a graphical representation that is used to forecast the genotype distribution of an offspring and the possible results of a genetic cross between two individuals.
Final answer:
The most likely genotype to occur in the second generation of a cross between Bb and Bb mice is the Bb genotype, as it is expected in a 3:1 ratio of dominant to recessive phenotypes.
Explanation:
The geneticist crossed pure breeding black mice with genotype BB (dominant black coat color) with pure breeding brown mice with genotype bb (recessive brown coat color). All the offspring of the F1 generation had black coats, indicating that black is the dominant trait. When these F1 mice, which are all heterozygous Bb, were crossed, the result in the F2 generation followed a Mendelian 3:1 ratio, where approximately 75% of the mice had black coats and 25% had brown coats.
The observed numbers of 961 black-coated mice and 317 brown-coated mice are close to that 3:1 ratio, demonstrating that the Bb genotype is indeed the most likely to occur in the second generation of this cross.
Help me please :
1) in a systemic circulatory system, blood flows from the heart to the lungs and returns to the heart. (true/false)
2) coagulation of the bloodis an effect from the action of receiving a compatible blood group. (true/false)
Answer:
1.true
2.false
Explanation:
the only time blood coagulates is when an incompatible blood group comes in contact with it due to the effect of antibodies
A 26-year-old woman in the clinic today for her annual physical examination. As you take her medical history, it becomes apparent that she is quite concerned about her risk for developing breast cancer. Her mother and a maternal aunt both developed breast cancer in their late 40s. She has heard that genetic testing can be done to see whether she carries "breast cancer genes."
***Assuming J.F. is talking about the BRCA1 and BRCA2 genes, what is the role that these genes have in the genetic mechanisms of breast cancer? (select all that apply)
a. BRCA1 and BRCA2 are tumor suppressor genes.
b. BRCA1 and BRCA2 are tumor promoter genes.
c. BRCA1 and BRCA2 are oncogenes.
d. BRCA1 and BRCA2 are proto-oncogenes.
Answer: the correct option is A (BRCA1 and BRCA2 are tumor suppressor genes).
Explanation:
BRCA1 and BRCA2 genes( breast cancer genes) are both tumor suppressor genes in every human expressed in breast cells and other tissue cells and it helps in
- repairing damaged DNA,
- regulate cellular growth and
- suppress the development of tumors . BRCA1 is located in chromosome 17 while BRCA2 is located in chromosome 13.
These genes in relation to genetic mechanisms of breast cancer, can be tested to know the possibility of the patient being susceptible to breast cancer in future or not. Mutation in either BRCA1 or BRCA2 or even in both genes makes the woman have a higher risk of developing breast,
Identify the best way to determine if two organisms are closely related
The best way is to compare the DNA of the two species. The more nitrogenous bases and amino acids two species have in common, the more closely related they are.
Which nerve is involved with carpal tunnel syndrome?
answer :Carpal tunnel syndrome is caused by pressure on the median nerve as it travels through the carpal tunnel.
Symptoms: Paresthesia; Hypoesthesia; Pain
Jeff's father has Huntington disease, which is caused by a rare autosomal dominant allele (H ). Two-thirds of people with the dominant Huntington allele show no symptoms by age 50, and Jeff is 50 and has no symptoms. Jeff's wife is is pregnant. What is that chance that Jeff's child will eventually show symptoms of Huntington disease?
a. 1/8
b. 1/4
c. 1/5
d. 1/6
e. 1/12
f. 1/24
g. 2/3
Which one?
Answer:
g. 2/3
Explanation:
Huntington disease is a autosomal dominant disorder. This disorder can be inherited and there is 50% chance that a child may get infected from his father's disorder. Jeff is 50 and has not shown any disorder symptoms. There are 2/3 people who do not show any symptoms before age of 50. Therefore there is 2/3 chance that Jeff will eventually show symptoms of Huntington disorder.
The chance that Jeff's child will eventually show symptoms of Huntington disease is 50%. The correct option that represents this probability is not provided in the given selections (a-g), but based on the information provided, 1/2 is the accurate probability.
Huntington's disease is an autosomal dominant disorder, which means that a single copy of the dominant allele (H) is sufficient for the expression of the disease. If Jeff has the disease, then he must have at least one copy of the dominant allele. Assuming that Jeff's wife does not have the Huntington allele (is homozygous recessive, hh), the Punnett square shows that each child has a 50% chance (Hh) of inheriting the allele and therefore the disease. The fact that Jeff has not displayed symptoms by age 50 is not factored into the genetic risk for their children, as the disorder can manifest symptoms later in life, and Jeff might still possess and pass on the allele. Given that two-thirds of people with the allele show no symptoms by age 50, it seems Jeff is among this group, but his potential transmission of the allele to offspring remains unaffected by this statistic.
You are working in an outpatient clinic when a mother brings in her 20-year-old daughter, C.J., who has type 1 diabetes mellitus (DM) and has just returned from a trip to Mexico. She has had a 3-day fever and diarrhea with nausea and vomiting. She has been unable to eat and has tolerated only sips of fluid. Because she was unable to eat, she did not take her insulin as directed. You note C.J. is unsteady, so you take her to the examining room in a wheelchair. While assisting her onto the examination table, you note her skin is warm and flushed. Her respirations are deep and rapid, and her breath is fruity and sweet smelling. C.J. is drowsy and unable to answer your questions. Her mother states, "She keeps telling me she's so thirsty, but she can't keep anything down."
Required:
List four pieces of additional information you need to elicit from C.J.'s mother.
Answer:
The additional questions are listed in the explanation
Explanation:
1. For how long has C.J. been drowsy
2. What have C.J.'s blood glucose levels been running?
3. How much fluid has C.J. been able to keep down over the past 3 days?
4. How often has C.J. been vomiting over the past 3 days?
In the case of C.J., a 20-year-old with type 1 diabetes mellitus presenting with signs of possible diabetic ketoacidosis, it is critical to obtain information on the duration and control of her diabetes, changes to her insulin regimen, the timeline of her symptoms, and any past DKA episodes or hospitalizations.
You are dealing with a case of a 20-year-old patient, C.J., who has type 1 diabetes mellitus (DM) and is presenting with fever, diarrhea, nausea, vomiting, and signs of dehydration and potential diabetic ketoacidosis (DKA), given the fruity, sweet-smelling breath (a sign of ketosis). In order to provide the best care for C.J., there are several pieces of additional information that would be crucial to elicit from her mother:
The duration of diabetes and the control level of C.J.'s diabetes prior to this incident.Any recent changes to C.J.'s insulin regimen, including the doses and types of insulin she uses.Details of her symptoms timeline, especially the onset of the fever, vomiting, and diarrhea, to better understand the progression of her current condition.Any prior episodes of diabetic ketoacidosis (DKA) or hospitalizations related to her diabetes, to assess her risk and previous management strategies.What statement best describes the conflict in this excerpt?
Read this excerpt from "Eleven".
Not mine, not mine, not mine, but Mrs. Price is already
turning to page thirty-two, and math problem number four. I
don't know why, but all of a sudden I'm feeling sick inside,
like the part of me that's three wants to come out of my
eyes, only I squeeze them shut tight and bite down on my
teeth real hard and try to remember today I am eleven,
eleven.
It is internal because Rachel is impatient about her
birthday
It is internal because Rachel fights against the urge to cry.
It is external because Mrs. Price is a challenging teacher.
It is external because Mrs. Price does not call on Rachel.
Mark this and retum
Save and Exit
Next
Submit
What’s this
Answer:
internal cos she figts against the urge to cry
Explanation:
Answer:
I think the answer is c
Explanation:
I read the story
What is the only commercial product derived from an orchid?
Aspirin
Saffron
Vanilla
Turmeric
Answer:
Saffron
Explanation:
Vanilla is the only commercial product derived from an orchid. It comes from a special orchid plant native to Central America, and due to a complex curing process, genuine vanilla extract is of higher quality and flavor than synthetic correct answer is vanillin.
The only commercial product derived from an orchid, as mentioned in the question, is vanilla. The vanilla beans are sourced from an orchid-type plant that is native to Central America. These beans undergo a complex curing process, which contributes to the premium price of natural vanilla. Vanilla is widely used in flavoring products and is known for its unique taste and aroma.
Imitation vanilla extracts use a synthetic compound called vanillin, which is made to mimic the flavor of genuine vanilla. However, pure vanilla extract is considered superior in terms of quality and flavor over the artificial counterpart. Due to the intensity of artificial vanilla, it has to be used more sparingly compared to the real thing.
Which of the following statements are true about the impact of competition and environmental filtering on community structure. A. Competition shapes a species fundamental niche B. Phylogenetic clustering and increased niche overlap occur when environmental filtering dominates C. Phylogenetic overdispersion and decreased niche overlap occurs when competition dominates D. Both competition and environmental filtering shape community structure
:
D. Both competition and environmental filtering shape community structure.
Explanation:
Based on competition theory, two organisms occupying the same ecological niche can only coexist provided there is spatial or conditioning niche partitioning.This prevent competitive exclusion , ensuring survival of the two species in their niche. Since the two species developed greater variability in (over-dispersed traits) to resist the selection pressure,(competition).Many overdisperesed traits is present in the communities.
However environmental filtering, is concerned with selection of species with natural ability or triats to survive in a given communities,Thus if two species of organism meet the same ecologocal requirements in a community, environmental filtering, fitters them and ensure their survival ahead of others who do not meet the requirements, and could not survive.
Thus the simultaneous operation of theses two opposing selection pressures, structured communities ensuring coexistence and favoured ability of organism to survive.
The lab you work in has discovered a previously unidentified extracellular signal molecule called QGF, a 75,000-dalton protein. You add purified QGF to different types of cells to determine its effect on these cells. When you add QGF to heart muscle cells, you observe an increase in cell contraction. When you add it to fibroblasts, they undergo cell division. When you add it to nerve cells, they die. When you add it to glial cells, you do not see any effect on cell division or survival. Given these observations, which of the following statements is most likely to be true? And why?
(a) Because it acts on so many diverse cell types, QGF probably diffuses across the plasma membrane into the cytoplasm of these cells.
(b) Glial cells do not have a receptor for QGF.
(c) QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells to produce the different responses observed.
(d) Heart muscle cells, fibroblasts, and nerve cells must all have the same receptor for QGF.
Option A is not possible because a 75 kDa protein is too large to diffuse through the plasma membrane.
Option B also cannot be assumed without more information. Even though the addition of QGF does not affect the cell division or survival of glial cells, there may be other signaling pathways that are triggered.
Option D is also not possible because the same receptor would give rise to the same signaling cascade.
Option C is correct - QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells to produce the different responses observed.
Assuming it was the same cascading pathway, the same responses would be observed. However, this is not the case (heart muscle cells - cell contraction; fibroblasts - cell division; nerve cells - death; glial cells - no effect). Therefore, the different responses are due to different pathways.
Final answer:
Given the diverse effects of QGF on different cell types, the correct statement is that QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells, due to the specificity of cell surface receptors and the signaling cascades they initiate.
Explanation:
The observation that a previously unidentified extracellular signal molecule, QGF, has different effects on various cell types (increasing contraction in heart muscle cells, inducing division in fibroblasts, causing death in nerve cells, and having no effect on glial cells), suggests specificity in the way cells respond to QGF. The most accurate statement considering these observations is that QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells to produce the different responses observed. This specificity is likely because cells have protein receptors on their plasma membranes that, upon binding to a signal molecule like QGF, initiate a cascade of intracellular signals that result in diverse cellular responses. This complexity underscores the concept that the same extracellular signal can lead to vastly different outcomes depending on the cell type, due to the presence of specific receptors and the unique signaling pathways activated within each cell type.
Option (c), "QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells to produce the different responses observed" is the correct option. This option best explains the varied effects of QGF based on established principles of cell signaling involving receptor-ligand interactions and the activation of specific intracellular pathways. Such specificity in response mechanisms highlights the role of cell surface receptors and the tailored intracellular signaling cascades that they initiate, which can lead to a range of cellular outcomes including division, death, or functional modulation.
For Mendelian traits, ALL members of the F1 generation are expected to resemble the parent with the dominant trait. For cytoplasmic traits, the members of the F1 generation are expected to resemble the mother, with potential for variation due to the genetics of the cells from which the eggs developed.True/False
Answer:
True
Explanation:
Mendel's law of dominance states that in a hetero-zygote, one trait will conceal the presence of another trait for the same characteristic. Rather than both gene contributing to a phenotype, the dominant allele will be expressed exclusively
In case of cytoplasmic inheritance, distinct maternal effects are observed. There is more contribution of cytoplasm to the zygote by female parent than male parent and we know generally that ovum contributes more cytoplasm to the zygote than sperm.
We are all born as naturalists eager to explore the world through our senses.
True
Fals
Answer:
That is false because some people are blind, or deaf,
Explanation:
The corpus callosum
A. consists of a broad band of gray matter.
B. . is found at the base of the transverse fissure.
C. s a band of commissural fibers that connects the right cerebral hemisphere to the left cerebral hemisphere.
D. connects the frontal lobe to the occipital lobe.
E. connects the frontal lobe to the parietal lobe.
Answer:
The corpus callosum C. is a band of commissural fibers that connects the right cerebral hemisphere to the left cerebral hemisphere
Explanation:
The corpus callosum is the largest commissural tract in the human brain, it is not a broad brand of gray matter. It consists of about 200-300 million axons that connect the two cerebral hemispheres. So, as we can see, options D and E are incorrect because it does not connects the frontal lobe to the occipital lobe or to the parietal lobe. It is not at the base of the transverse fissure, as says option B, it is beneath the cerebral cortex in the brain.
Answer:
Option C
Explanation:
The Corpus callosum is also referred to as callosal commissure and is a wide, thick nerve tract made up of a flat bundle of commissural fibers that divides the cerebral cortex lobe into the right and the left hemispheres. It allows for communication/transfer of information between the two hemispheres. It is made up of white matter tissue
what is operant behaviour
Answer:
Operant behavior (which goes along with operant conditioning) refers to behavior that "operates" on the environment or is controllable by the individual. Operant behavior is done because it produces some type of consequence.
• What part of a phosphodiglyceride is hydrophobic?
The hydrophobic part of a phosphodiglyceride consists of two fatty acid tails that repel water. The phosphate head is the hydrophilic part that attracts water, allowing these molecules to form cell membranes with distinct interior and exterior environments.
In a phosphodiglyceride, the hydrophobic part is comprised of two fatty acid tails. These tails are long chains of hydrocarbons that are nonpolar and do not interact well with water. On the other hand, there is a hydrophilic part, which is the phosphate group attached to a glycerol backbone. The phosphate head is polar and interacts readily with water. This dual nature allows phosphodiglycerides to form cell membranes where the hydrophobic tails face inward away from water and the hydrophilic heads face outward towards the aqueous environment.
All healthy, highly productive ecosystems provide which of these essential ecosystem services that involve gases in the atmosphere?
Answer:
because
Explanation:
It is an Incomplete question and the complete question is
All healthy, highly productive ecosystems provide which of these essential ecosystem services that involve gases in the atmosphere?
A. increasing carbon dioxide levels in the atmosphere by the action of plants and animals.
B. removing carbon dioxide from the atmosphere and storing carbon in the tissues of organisms.
C. removing water vapor from the atmosphere and using it to support plant growth.
D. removing methane from atmosphere and storing it's carbon and hydrogen in the tissues of organisms.
Answer:
The answer is removing carbon dioxide from the atmosphere and storing carbon in the tissues of organisms.
Explanation:
The water and carbon dioxide are byproducts and in which the photosynthesis removes CO2 from the atmosphere and replaces it with O2. In case of respiration it takes O2 from the atmosphere and places it with CO2.
To remove the CO2 it is directly put out of the air and burried it under ground saline aquifers this process is known as carbon capture and sequestration.
Sequestration is a process in which CO2 is removed from atmosphere and stored in the soil carbon pool. This process is mainly done through photosynthesis.
Healthy ecosystems provide essential ecosystem services involving gases in the atmosphere, such as the replenishment of oxygen through photosynthesis.
Explanation:Healthy, highly productive ecosystems provide essential ecosystem services that involve gases in the atmosphere. One of these services is the replenishment of oxygen through the process of photosynthesis. Green plants convert carbon dioxide and water into glucose and oxygen using the energy of light. This process not only supports human respiration but also contributes to the overall composition of the Earth's atmosphere.
Learn more about Ecosystem services involving gases here:https://brainly.com/question/15669128
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Two cells in the same organism differ only in the number of chloroplasts they contain. The first cell has multiple chloroplasts, and the second cell has very few. What would most likely characterize these cells?
A. There would be no difference between the functioning of the cells because the chloroplasts are not essential cell structures.
B. The second cell would become larger because it would have fewer chloroplasts regulating its size and shape.
C. The second cell would shrink because it would not be able to store water and maintain cell shape.
D. The second cell would not be able to produce as much food because it could not capture sunlight.
Answer: D
Explanation: there right i just need the points
The correct answer is D. The second cell would not be able to produce as much food because it would not be able to capture sunlight.
What is the reason for chloroplasts being necessary ?Chloroplasts are essential for photosynthesis, in which plants and other photosynthetic organisms use sunlight to convert carbon dioxide and water into oxygen and sugars.
Without chloroplasts, the second cell would not be able to capture sunlight and convert it into energy, and therefore would not be able to produce as much food as the first cell.
A is incorrect because chloroplasts are necessary for photosynthesis and are therefore important cell structures. B is incorrect because the number of chloroplasts does not directly affect cell size.
C is incorrect because the number of chloroplasts has nothing to do with a cell's ability to retain water and maintain its shape. The photosynthesis is all done in the chloroplasts where the leaves are having the cells where the photosynthetic units.
Learn more about Chloroplasts at:
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True/False: Mark the following statements as true (T) or false (F). If the statement is false, correct it to make it a true statement. a. Gametes result from two rounds of cell division. b. Meiosis results in diploid cells. c. Homologous chromosomes overlap at points called chiasmata or crossover. d. The DNA replicates prior to the start of meiosis II.
Answer:
a. Gametes result from two rounds of cell division. True
b. Meiosis results in diploid cells. False
c. Homologous chromosomes overlap at points called chiasmata or crossover. True
d. The DNA replicates prior to the start of meiosis II. False
Explanation:
False statements corrected to make them true:
b. Meiosis results in haploid cells
d. The DNA does not replicates prior to the start of meiosis II.
Mules, the offspring of donkeys and horses, are commonly used as pack animals. Despite the existence of mules, horse
and donkeys are considered separate species,
Which discovery would change that thinking and define horses and donkeys as the same species?
O
O
O
O
if mules looked more like donkeys
if mules looked more like horses
if mules could produce offspring
if mules were sterile
Answer:
If mules could produce offspring.
Explanation:
Even though horse and donkey produce mule, they are still regarded as different species because the mule produced cannot produce an offspring of its own.
The only why they can be considered as same species is if the mule can also reproduce.
You conduct a mark-recapture survey of eastern fence lizards (Sceloporus undulates) in a 2-hectare plot in eastern Arizona. You mark 250 lizards and are able to recapture 75, of which 24 are marked. What is the estimated size of the population of fence lizards in this plot? Calculate the estimated population size for the plot. Use the following formula: M/N = R/C where M is the number of initially marked individuals, N is the population size, R is the number of recaptured marked individuals, and C is the total number of individuals recaptured.
a. 80 b. 7.2 c. 781.3 d. 250 e. 24
Answer:
Option C
Explanation:
Given -
Total number of initially marked lizards (M) [tex]= 250[/tex]
Total number of recaptured lizards (C) [tex]= 75[/tex]
Total number of recaptured marked lizards (R) [tex]= 24[/tex]
As we know that -
[tex]\frac{M}{N} = \frac{R}{C}[/tex]
We have to find "N" . "N" is the total population size of lizards in a 2-hectare plot in eastern Arizona.
Thus, on rearranging the above equation, we get -
[tex]N = \frac{M * C}{R}[/tex]
Substituting the given values in above equation, we get -
[tex]N = \frac{250 * 75}{24} \\N = 781.25\\N = 781.3[/tex]
Hence, option C is correct
Where is the green sample of water from
Answer:
Add a Picture
Explanation:
Add a picture please so I can answer it, otherwise im just going to have to guess its Moss, Algei or Bacteria lol
Which process will be immediately inhibited if cytochrome c is unable to interact with cytochrome c oxidase?
a. transfer of electrons from cytochrome c to coenzyme Q
b. transfer of electrons to H 2 O H2O
c. transfer of electrons from cytochrome c to O2
d. transfer of electrons from coenzyme Q to cytochrome c
Answer:
the correct answer is C
Explanation:
What cytochrome C does is change electrons to oxygen to reduce it to two water molecules
Answer:
c. transfer of electrons from cytochrome c to O2
Explanation:
Actually in the electron transport chain cytochrome c oxidase is the final enzyme. From each 4 cytochrome c molecule It receives electron , transfer them dioxygen to form 2 molecules of water. 4 protons are used in order to make 2 water molecules and 4 protons are translocated across the membrane which create electrochemical potential.Hence electrochemical potential is generated and is utilized by ATP synthase to form ATP.
Hence answer is c. transfer of electrons from cytochrome c to O2.
When mosquitoes are abundant, purple martins flock to the area and feed exclusively on them. When mosquito populations are not large, purple martins are similarly scarce and feed on other insects. This is an example of A) density-independent regulation. B) density-dependent regulation. C) ecosystem carrying capacity. D) community carrying capacity.
Answer:
Density-dependent regulation
Explanation:
density-dependent processes takes place when the density of a population population regulate growth rates.
Some factors that affect density-dependent factors could be biotic (biological in nature), such as;disease, predation, intraspecific competition, accumulation of waste, and interspecific compitition.
For instance, in interspecific competition or intraspecific compitition, reproductive rates of the individuals will definitely reduced,which results to reduce in population’s growth rate.When the prey density is low, the mortality of its predator is increased.
Therefore,when mosquitoes are abundant and purple martins flock to the area and feed exclusively on them, since the mosquito populations are not large, purple martins are similarly scarce and feed on other insects, this is reffered to as Density-dependent regulation.
A mutation creates a dominant negative allele of a particular gene. The normal allele of the gene encodes a protein that forms a trimer within the cell. If one or more of the subunits has the mutant structure, the entire trimeric protein is inactive. In a heterozygous cell, if the proteins of both alleles are present at the same levels, what percent of the trimers present in the cell will be inactive
Answer:
87.5% of the trimers in the cell will be inactive.
Explanation:
A monomer is a single subunit. Look at an easy comparison. If a monomer is a ball. For the protein to be functional, it needs to be a trimer .A trimer = 3 monomers joined / bound together with bonds. Therefore, if a monomer is a ball, we need 3 balls, we attach it together and that is the primer . if one ball becomes missing, that protein won't work .
In chemistry, a trimer is regarded as three molecules or ions of the same substance (that is, same type) which are associated together .
Double negative : Each gene has 2 alleles. So, for a positive protein, both allele needs to be same, let's say + means on. So for a x protein , if it's x+/x+ it means the protein is homozygous and active. If for some mutation, it's x+/x- , it's a heterozygous protein. Double negative means when it is x-/x- and this means the protein has lost is function . It's not functional at all.
The heterozygous cell will form one or more subunits has the mutant structure when the gene is expressed. From the question if there is a mutation in dominant negative allele, it will cause the inactivation of one trimer resulting in 87.5% inactive trimers for a heterozygous individual.
There are 87.5 % percent of the trimers present in the cell which will be inactive for a heterozygous individual, If at least one of the subunits has the mutant structure.
The probability for each individual subunit to be a mutant is 1/2
Gene codes for a protein that forms trimer so the percentage of the trimers present in the heterozygous individual is third power of 1/2 =1/8
1/8=0.125 will be active.
1- 0.125 will be inactive.
=0.875 will be inactive.
0.875 × 100= 87.5%
Therefore 87.5% will be inactive.
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Which factor most directly influenced the evolution of the diverse types of 1 point
beaks of the finches shown in the diagram?*
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11
5
1
DEME
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LEHELE
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Illum FBiH
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Elle
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O
A) predation by humans
a
.
TITLE
nem
LE!
LE
allille
alle
TELEEEEEEEEEE
till
THEHE
SELETTE
O
B) available food sources
ces
BE
Sallalla
DEBEEEEE
Till
20 TL
Viime
En
W
C) oceanic storms
SW
ELLE
O
WWWWW
D) lack of available niches
==
We
WWW
es va
WWWWWWWW
W WWWWWWWWW
WWW WWWWW
=
war
WWWWW
WWWWWWWWW
Www WW
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=
1 point
State the name of one species of finch from the diagram shown that is
..
Answer:
Um I think this person was dying when typing this agaga
Explanation:
They said crazy things