Answer:
En marina es: sobre pesca, cambio climático, especies invasoras, acuicultura. En terrestres es: inundaciones, calentamiento global, gases de invernadero, radiación reflejada.Explanation:ponme coronaGive reason. Mosquitoes and housefly are placed in the phylum arthropoda
Mosquitoes and Housefly are grouped under the phylum Arthropoda because they both posses jointed legs (or appendages).
Animals that have jointed appendages are called Arthropods (i.e animals with jointed legs). They are triploblastic (i.e they have 3 germ layers) and exhibit bilateral symmetry. They are both six-legged arthropods sub-grouped under the class Hexapoda (coined from the word "hexa" meaning six and "Poda" meaning leg). Their body are divided into three parts; head, thorax and abdomen.
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Suggest how whooping cough spreads from person to person
Answer:
People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease.
Explanation:
Answer:
Explanation:People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease
Which type of seedless plant has a complex leaf arrangement off a vein?
a. java moss
b. club moss
c. ferns
d. horstails
B. Club moss
Explanation:
This is because club moss is an seedless evergreen plants that have scale-like leaves.
Can you plz mark me as brainliest!!!
answer : club moss
explanation: Because they have vascular tissue, seedless vascular plants
are often larger than nonvascular plants. Vascular tissue is spe-
cialized to transport water to all of the cells in a plant.
water can act as either a(n)__or a
Explanation:
Water can act as an acid and a base. As an acid, water donates H+, the hydrogen ion. As a base, water donates OH-, the hydroxide ion
hope this helps you
have a nice day:)
Which of the following would provide the best evidence that species A and species B have a common ancestor?
O A. The limbs of species A perform a function that is similar to the limbs of species B, but they have a different structure.
OB. Species A has limbs, and species B does not have limbs.
O c.
The limbs of species A and species B have a different structure and function.
OD
The limbs of species A have a similar structure to the limbs of species B, but they perform a different function.
Answer:
I believe the answer is D
Explanation:
Answer:
B
Explanation:
The answer is B on Plato. Just took the test and got it right.
A genetically heterogeneous population of rice has a mean in the number of days to maturation of 30. Selection for decreased period of maturation is carried out for one generation. The average period to maturation among the plants selected as parents for the next generation is 25 days. F1 plants mature on average in 27 days. Estimate the narrow sense heritability.
Answer:
h² = 0.6
Explanation:
Before answering the question, we need to know a few concepts.
Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.
Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.
The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,
SD = X₀ - Xs
Heritability in the strict sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.
h² = R/SD
The response to selection (R) refers to the metric value gained from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.
R = h²SD
R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),
R = X₀ - X₁
-------------------------------------------------------------------------------------------------------------
Now that we know these concepts and how to calculate them, we can solve the proposed problem.
Available data:
You are selecting rice´s decreased period of maturation. The population of rice has a mean maturation time of 30 days → X₀ Parental selected average period to maturation is 25 days → Xs F1 plants mature on average in 27 days → X₁ N arrow sense Heritability → h²According to what we sow previously, we need to find out the value of h².
We know that h² = R/SD, so we need to get R and SD first.
R = X₁ - X₀
R = 27 - 30
R = -3
SD = Xs - X₀
SD = 25 - 30
SD = -5
Knowing this, we can calculate h²
h² = R/SD
h² = -3/-5
h² = 0.6
explain how the tissue of the esophagus and tissue of the trachea can be differentiate
Answer:
Trachea: It is the wind pipe — making it a part of the respiratory system
Esophagus: It is the food pipe — making it a part of the digestive system
Trachea: It is shorter, 10–11 centimeters. It connects upper airway to the lungs
Esophagus: It is longer, 25 centimeters. It connects mouth to the stomach
Trachea: It is cartilaginous, made of C-shaped semicircular cartilages. They give it structural stability and prevents it from collapsing
Esophagus: It is muscular. It contracts in a wave-like motion through it’s length to propel food from mouth to stomach a.k.a swallowing
Trachea: It’s opening is protected by Epiglottis, a flap like structure, to prevent food from accidental entering the air passage. It prevents choking
Esophagus: It’s opening is protected by two sphincters. They are muscular rings that constrict to close the esophagus off when food is not being swallowed.
Trachea: It has 2 portions — cervical and thoracic i.e. neck and chest portions.
Esophagus: It has 3 portions — cervical, thoracic and abdominal i.e. neck, chest and stomach portions.
what type of stream valley forms oxbow lakes ?
1. Youthful stream valley
2.mature stream valley
3. Old age stream valley
Answer:
I thik answer is 2 number
The difference between active transport and passive transport is that a. concentration gradients are involved in one and not in the other. b. glycolipids play a role in one and not in the other. c. one requires expenditure of energy by the cell and the other does not. d. ions are transported into and out of the cell by one process and not by the other.
Answer:
D) ions are transported into and out of the cell by one process and not by the other
Traditionally, the classification of fungi has been based on the nature of sexual stages of the life cycle. For Penicillium, however, no sexual stages of the life cycle have been observed. Without evidence from sexual stages, speculate about other possible sources of evidence that scientists may use in classification.
Answer:
Without the evaluation of sexual stages, fungi can be classified according to the type of colony they present, the speed of growth, the formation of pigments and the type of coloration.
Explanation:
Although the observation of sexual stages is extremely efficient for the classification of fungi in the laboratory. This type of analysis is not always possible to be carried out. In that case, scientists need to find other methods that allow for the classification of fungi. These methods are carried out with the help of a microscope, where scientists observe the morphology of the fungi and are able to classify them according to the type of colony they present, the speed of growth, the formation of pigments and the type of coloration.
Histones are essentially identical in sequence/structure in all eukaryotic organisms from yeast to plants to animals. What does this say about the biophysical properties of DNA-packaging and the evolution of eukaryotic organisms
Answer:
It indicates that core histone genes were present in the last common ancestor of yeasts, plants, and animals
Explanation:
Histones are highly basic proteins that can strongly interact with DNA, which is packaged into nucleosomes, the basic structural and functional unit of chromatin. Each nucleosome is composed of approximately 147 base pairs of DNA wrapped around a core of eight histone proteins (two copies of four types of histones H3, H4, H2A, H2B). These core histones are evolutionarily conserved across eukaryotic kingdoms in terms of sequence and structure. Therefore, DNA-packaging into nucleosomes is considered a constraint for the evolution of core histones. Moreover, the presence of conserved core histones in eukaryotic kingdoms (e.g., yeast, plant, and animal kingdoms) is strong evidence that histone-mediated DNA packaging was presumably present in the last common ancestor of eukaryotic genomes.
WHAT IS PHOTOSYNTHESIS?
Answer:
Photosynthesis is the process in which plants make their own food from inorganic substances like carbondioxide and water to organic substances in the presence of sunlight and chlorophyll
Explanation:
Have a nice day
Which type of energy refers to the sum of potential and kinetic energies in the particles of a substance?
Answer:
The answer is Internal energy
KE + PE = IE
Explanation:
The sum of potential and kinetic energy is refers to mechanical energy which is expressed by motion
Answer:C for edge (internal energy)
Explanation:
Part 1 of 1 -
Question 9 of 10
10 Points
When DNA is copied, sometimes there are mistakes. Approximately how often does this
happen?
O A. There aren't any mistakes.
OB. 1 in a billion bases.
OC. 1 in a million bases.
OD. 1 in a trillion bases.
Reset Selection
Answer:
D. 1 in a trillion bases
Explanation:
A mutagen agent can change the genetic information of organisms increasing mutations over the natural level. Mutagens cause changes in the bases, and pairing bases, that compose DNI strands.
A mistake in the process of DNI copy during cell division might cause genetic changes in daughter cells. Defects DNI replication might be inherited if it occurs in germinal cells. But it can also cause many significant epigenetic changes.
Many of these changes can be detected on time by enzymes such as DNI polymerase. This enzyme can correct these mistakes or at least some of them, moving from 3´to 5´direction, and eliminating the mistakes.
The highly effective replication system, together with the action of enzymes, makes it rare to occur a mistake in DNI replication. Generally speaking, the mistaken rates in DNI replication are very low, meaning that only one in a trillion times occurs a mistaken DNI copy.
Glycolysis occurs in
1) mitochondria
2) cytoplasm
3) ER
5) Plastids
what are the effects of under secretion of steroid hormones
Answer:
Anabolic steroids are related to testosterone, the major male hormone. Abusing the hormone can lead to physical and psychological side effects. Problems include breast development and hair loss among men, and facial hair growth, menstrual problems, and a deepened voice in women.
Which life process is vital for the survival of all organisms
Answer:
oxygen
Explanation:
Which structure transports urine to the bladder by peristaltic action?
Answer:
The muscular layer of the ureter consists of longitudinal and circular smooth muscles that create the peristaltic contractions to move the urine into the bladder without the aid of gravity.
Answer:
The ureter
Explanation:
The ureter is a long thin tubular structure 10-12 inches long which carries urine produced in the kidney to the bladder. The urine is transported by a process called peristalsis. The ureter actively propels urine from the kidney down into the bladder.
A 36 years old female came to see the doctor with the complaint of amenorrhea (absence of menstrual period) for past three months. She is worried if she is pregnant although she never missed her Contraceptive Pills. She recently noticed some milky discharge from her left breast and abnormal facial hairs. She also said that her period was irregular and less in amount for past one year. She also gained 15 pounds during last year but denies any cold intolerance, dry skin, depression, fatigue or hot flushes.
Her menarche was at 14 and she had regular 21 days cycle till last year when it became irregular and gradually scanty.
Her LMP (last menstrual period) was 3 months back.
She has a 10 years old healthy child (normal vaginal delivery). She has regular Pap test, last test was done 10 months ago which was normal.
Her mother had menopause at age 55.
Her vitals were normal.
Her pregnancy test was done at the office which was Negative.
The doctor reassured and discussed hormonal imbalances with her. She recommended her urine and blood tests for hormone levels and pelvic ultrasound to check her reproductive organs. The doctor advised her to come back with all the test result
1. After reading given case, what will be the most likely diagnosis of this patient?
2. Normal woman does not show and breast discharge, what could be the reason in this patient for breast discharge?
Help me answer to your best of ability
Answer:
The correct answer is -
1. galactorrhea or prolactinoma
2. hormonal imbalance (prolactin imbalance)
Explanation:
The given symptoms experienced by the patient are the absence of menstrual cycles, an increase in weight, some milky discharge from her left breast, and abnormal facial hairs. The pregnancy tests show she is not pregnant and she also does not experience cold intolerance, dry skin, depression, fatigue or hot flushes.
All these suggest that there is a hormonal imbalance in her and due to producing a white substance from the breast it can get that there is too much production of prolactin, which suppresses the effect of estrogen, which is producing by the pituitary gland.
Galactorrhea or prolactinoma are two disorders that affect the production of prolactin that cause milky discharge, it can be caused by overuse of contraceptives or underlying conditions.
2. At what temperature did the prodigiosin-producing genes express in the S. marcescens culture? From the experiment you conducted in this lab, what evidence can you provide to support your claim?
Answer:
The temperature is the key factor for prodigiosin production. It has been shown that , S. marcescens can produce this kind of pigment at about 25 °C, which however could not produce the pigments at elevated temperatures, especially till 37 °C
Explanation:
Considering your knowledge of carbohydrates, evaluate the use of chitin as a component of health foods.
Answer:
Carbohydrates may be defined as energy-rich foods such as sugars and starches. if consumed in excess or not properly metabolized in the body, carbohydrates may lead to obesity and which may also lead a person to severe number of diseases.
Chitins are components of the exoskeletons of foods such as shrimps, crabs and snails, and insects.
Chitin's use as a component in healthy foods is based on its health benefits.
For example it promotes weight loss , prevents obesity, relieving constipation and preventing inflammation associated with refined carbohydrates, cookies and candies
Select the logical fallacy used in this statement: If we ban semiautomatic weapons, it won't stop until handguns and rifles are banned as well, and then the criminals will have all the guns.
Select one:
a.
Slippery Slope
b.
Ad Populum
c.
Ad Hominem
d.
There is no fallacy in this statement
Answer:
A- Slippery Slope
Explanation:
A slippery slope fallacy is when someone claims, without proper evidence, that an action will lead to an often catastrophic consequence via a series of events.
Here, the statement claims that if semi-automatics are banned, so will handguns and rifles, however, doesn't provide any evidence that this is the case.
Lainey is looking for a new apartment and her realtor keeps calling her with new listings. The calls only take a few minutes, but a few minutes here and there are really starting to add up. She's having trouble concentrating on her work. What should Lainey do? a) Tell her realtor she can only receive text messages O b) Limit the time spent on each call O c) Turn off her phone until she is on a break O di Call her realtor back when customers won't see her on the phone
Answer:
c) Turn off her phone until she is on a break
Explanation:
If she does option "a" then her phone will still keep ringing with text message alerts. Option " b" will still be consuming her work time and option "d" can't be because it'll be too late by then so only option c makes sense.
Given that the intracellular concentration of potassium is 150 mEq/L, how would the potassium equilibrium potential be affected if the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L
Answer:
The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.
Explanation:
Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.
Nernst equation:
E = 58 millivolts/z. [Log₁₀ (C-out/C- in)
Where,
• E = Equilibrium potential
• 58 millivolts/z = Constant
• z = Ion charge + positive or negative symbol
• C-out = Ion concentration out of the cell
• C-In = Ion concentration inside the cell
By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.
Now let us see the provided values,
• z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+
• C-out = Ion concentration out of the cell ⇒ 5 mEq/L
• C-In = Ion concentration inside the cell ⇒ 150 mEq/L
E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)
E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)
E = 58 millivolts (Log₁₀ 30)
E = 58 millivolts (1.477)
E = 85.67 millivolts
85.7 mV is the absolute value of equilibrium potential.
E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)
E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)
E = 58 millivolts (Log₁₀ 42.85)
E = 58 millivolts (1.63)
E = 94.65 millivolts
94.7 mV is the absolute value of equilibrium potential.
If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.
The potassium equilibrium ability might increase, which means that greater K+ might be leaving the cell.
Let us expect that the handiest ion transported via the cell membrane is K+. We want to apply the Nernst equation to recognize the future of the ion.
Nernst equation:
[tex]E = 58 millivolts/z. [Log₁0 (C-out/ -in)[/tex]
Where,
E = Equilibrium ability.58 millivolts/z =Constant.z=lon charge + advantageous or terrible symbol. C - out = Ion awareness out of the cell. C-ln= ion awareness in the cell.For convenience, withinside the Nerts equation, the larger awareness fee corresponds to the numerator and the smaller awareness fee to the denominator. Now allow us to see the supplied values,
[tex]z=lon charge + effective or terrible ⇒+1 ⇒ K+\\C - out = lon awareness out of the cell ⇒5 mEq/L\\C-ln= lon awareness withinside the cell ⇒150 mEq/LE = 58 millivolts/+ 1.[Log 10 ( 50mEq / L / 5mEq/L)\\E = 58millivolts (Log30)\\E = 58 millivolts (1.477)[/tex][tex]E = 85.67 millivolts\\85.7 mV is absolutely the fee of equilibrium capability.\\E = 58 millivolts/z. [Log10 (lon in/lon out)\\E =58 millivolts/+1. [Log 10 (a 100and 50 mEq/L / 3.five mEq/L)\\E =58 millivolts (Log10 42.85)\\E = 58 millivolts (1.63)\\E = 94.65 millivolts94.7 mV is absolutely the fee of equilibrium capability.[/tex]
What happens if the extracellular attention of potassium is modified from 5.0 to 3.5?If the extracellular attention of potassium is modified from 5.0 to 3.5mEq/L, there can be a growth withinside the membrane capability from 85.7 to 94.7 mV. The growth withinside the equilibrium capability will bring about extra potassium diffusing out of the cell, turning the cell indoors much less high quality than before.
Thus it is clear from this that the potassium equilibrium potential is affected.
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Macronutrients is the most readily available for energy production
Answer:
The correct answer is - true.
Explanation:
All the macronutrients carbohydrates, fats, and protein all three produce energy. Carbohydrates are the most preferred source of energy for the human body as it is the macronutrient that your system most requires.
The human body easily breaks down most carbohydrates and provides a significant amount of energy. The energy-providing process is called cellular respiration starts with glucose as a substrate. Glucose is the simplest carbohydrate.
What is the biggest part of the brain?
Answer:
CerebrumExplanation:
The cerebrum is divided into two hemispheres or halves and is the biggest portion of the brain. The cerebrum is in charge of voluntary movement, speech, intellect, memory, emotion, and sensory processing, among other things.
OAmalOHopeO
what is The Catalys?
Answer:
A catalyst is a chemical substance that speeds up the rate of a chemical reaction at any given conditions.
The body regulates the amount of hormones are released by using feedback loops. A __ feedback loop increases the response whereas a __ feedback loop decreases the response.
Positive feedback loop increases the response whereas a negative feedback loop decreases the response.
What is positive feedback?Positive feedback is the amplification of a body's response to a stimulus. For example, in childbirth, when the head of the fetus pushes up against the cervix (1) it stimulates a nerve impulse from the cervix to the brain (2).
A feedback mechanism resulting in the inhibition or the slowing down of a process.
Examples of processes that utilise negative feedback loops include homeostatic systems, such as thermoregulation (if body temperature changes, mechanisms are induced to restore normal levels), blood sugar regulation (insulin lowers blood glucose when levels are high ,glucagon raises blood glucose when levels are low).
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How are oxygen and carbon dioxide exchanged between the alveoli and the
capillaries?
A. Endocytosis
B. Osmosis
C. Simple diffusion
5
D. Active transport
Answer:
B. Osmosis
Explanation:
Osmosis is the process in which oxygen and carbon dioxide exchanged occur between the alveoli and the capillaries because the oxygen enters the body and the carbondioxide gas leaves the cell through a semi-permeable membrane and we know that Osmosis is a process in which smaller molecules moves from higher concentration to lower concentration through semi-permeable membrane of the cell.
Match the terms in column B to the descriptions in column A.
Column A:
1. Connects the larynx to the main bronchi
2. Includes terminal and respiratory as subtypes
3. Food passageway posterior to the trachea
4. Covers the glottis during swallowing of food
5. Contains the vocal cords
6. Indentation on the lung where the lung root structures enter and exit
7. Pleural layer lining the walls of the thorax
8. Site from which oxygen enters the pulmonary blood
9. Connects the middle ear to the nasopharynx
10. Pleural layer in contact with the surface of the lung
11. Increases air turbulence in the nasal cavity
12. Separates the oral cavity from the nasal cavity
Column B:
a. alveolus
b. bronchiole
c. conchae
d. epiglottis
e. esophagus
f. hilum
g. larynx
h. palate
i. pharyngotympanic tube
j. parietal pleura
k. trachea
l. visceral pleura
Answer:
1. Connects the larynx to the main bronchi k. trachea
2. Includes terminal and respiratory as subtypes b. bronchiole
3. Food passageway posterior to the trachea. e. esophagus
4. Covers the glottis during swallowing of food d. epiglottis
5. Contains the vocal cords g. larynx
6. Indentation on the lung where the lung root structures enter and exit f. hilum
7. Pleural layer lining the walls of the thorax j. parietal pleura
8. Site from which oxygen enters the pulmonary blood a. alveolus
9. Connects the middle ear to the nasopharynx. i. pharyngotympanic tube
10. Pleural layer in contact with the surface of the lung l. visceral pleura
11. Increases air turbulence in the nasal cavity c. conchae
12. Separates the oral cavity from the nasal cavity h. palate
Explanation:
1. The trachea is between the main bronchi and the larynx. It has semicircular rings of cartilage. The function of this organ is to conduct the air from the larynx to the primary/main bronchi.
2. After the primary (main), secondary, and tertiary bronchi, the terminal and the respiratory bronchioles come. They also conduct the air towards the alveoli so that the oxygen can enter the blood. The diameter of the bronchi terminal is smaller than the bronchi, and the diameter of the respiratory bronchi is smaller than the previous ones.
3 and 4. The esophagus is not part of the respiratory system. It is a tube that belongs to the digestive system since food has to pass through it to go to the stomach. The esophagus is posterior to the trachea, and the epiglottis closes the entrance to the larynx when we swallow to stop food from going to the lungs.
5. The vocal folds are in the larynx. This one is between the trachea and the pharynx. There are two types of vocal folds, the true and the false vocal folds. Both of them vibrate when air passes through them, allowing us to speak and make different tones.
6. The hilum is the lungs section where the bronchus, the pulmonary artery, and the pulmonary vein enter the lung. It is an indention that is in the middle part of the lungs.
7 and 10. The pleura has two sides, the parietal one and the visceral one. The first one is in contact with the lungs and the second one with the thorax's walls. Between them, there is a space called the pleural cavity. The cavity has fluid that allows the movement of the two pleurae. As a result, the lungs can move and fill with air.
8. After the respiratory bronchioles, we have the alveolar duct. The alveolar ducts lead to alveolar sacs. The alveolar sacs has the alveolus. They are thin walls that are in contact with capillaries. When the air is there, the oxygen passes through the thin walls. Then it goes through the capillaries' walls and into the blood.
9. The pharyngotympanic tube is also known as the Eustachian tube. It connects the middle ear to the nasopharynx. Its function is to regulate the pressure in the ear.
11. The conchae are in the nasal cavity. They are three projections in the nasal cavity, the inferior one, the middle one, and the superior. Their function is to increase the surface of the nasal cavity so that more air can enter with every inspiration. As they are projections, they modify the laminar airflow producing a turbulent flow.
12. The palate is between the nasal cavity and the oral cavity. It has two parts, the soft palate, and the hard palate. The palate helps in the production of certain sounds and divides the nasal cavity from the mouth.
