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.
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
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.
100 POINTS!!!!!!
The theories surrounding the formation of our solar system are based on many
biased opinions
incorrect facts
non-testable data
scientific investigations
Answer:
"scientific investigations."
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:
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.
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
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.
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.
To know more about the potassium equilibrium refer to the link :
https://brainly.com/question/2742049
Lectins often bind their ligands via multiple weak interactions. bind their ligands with relatively low specificity. prevent viruses from binding to their target cells. are carbohydrates that bind to receptor proteins.
Answer:
The correct answer is - B.often bind their ligands via multiple weak interactions.
Explanation:
Lectins are specific types of proteins that identify and bind to specific carbohydrates present on the cell surfaces. They have an essential role in interactions and communication between various cells for identification and recognition.
Binding sites of lectins on the surface of one cell bind to the Carbohydrates on the surface of another cell. A lectin usually has two or more binding sites for carbohydrate units.
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
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.
Give 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.
Learn more:
https://brainly.com/question/2244172?referrer=searchResults
For each of the following structures, first indicate its function in the fetus; then, note its fate (what happens to it or what it is converted to after birth).
a. Umbilical artery
b. Umbilical vein
c. Ductus venosus
d. Ductus arterious
e. Foramen ovale
Answer:
1. Functions:
a. Umbilical artery >> carries deoxygenated blood from the fetus to the placenta
b. Umbilical vein >> transports oxygenated blood from the placenta to the fetus
c. Ductus venosus >> allows oxygenated blood from the placenta to bypass the liver
d. Ductus arterious >> allows most of the blood from the right ventricle to bypass the fetus's non-functioning lungs
e. Foramen ovale >> oxygenated blood from the umbilical vein to bypass the pulmonary circulation
2. After the bird:
1. Umbilical artery >> medial umbilical ligament
2. Umbilical vein >> round ligament of the liver
3. Ductus venosus >> ligamentum venosum
4. Ductus arteriosus >> ligamentum arteriosum
5. Foramen ovale >> fossa ovalis
Explanation:
The umbilical artery is a paired artery localized in the abdominal and pelvic regions, which carries deoxygenated blood from the fetus to the placenta through the umbilical cord. The medial umbilical ligament is the obliterated part of the umbilical artery that arises from the internal iliac arteries. In utero, the umbilical arteries carry waste products back to the placenta, whereas the umbilical vein carries oxygenated blood from the placenta to the fetus. The round ligament of the liver (also known as ligamentum teres hepatis) is a remnant of the umbilical vein that exists in the embryonic stage, it connects the left lobe of the liver to the umbilicus. The ductus venosus is a slender shunt that allows oxygenated blood from the placenta to bypass the liver, it connects the intra-hepatic portion of the umbilical vein to the inferior vena cava. The ligamentum venosum is an extrahepatic, slender, and fibrous remnant of the fetal ductus venosus that travels between the left portal vein and the inferior vena cava. The ductus arteriosus is a fetal artery that connects the aorta to the pulmonary artery. The ligamentum arteriosum is a nonfunctional vestige of the ductus arteriosus, it is attached to the superior surface of the pulmonary trunk. The foramen ovale is an oval-shaped, small, opening in the wall (septum) between the two upper chambers of the heart. The fossa ovalis is a vestige stricture of the foramen ovale of the embryonic heart, which forms a depression in the right atrium of the heart.
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.
Which is a compound that allows plants to get nitrogen from the nitrogen cycle?
Answer:
Plants can use ammonia as a nitrogen source. After ammonium fixation, the ammonia and ammonium that is formed will be transferred further, during the nitrification process. Aerobic bacteria use oxygen to convert these compounds.
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
Glycolysis occurs in
1) mitochondria
2) cytoplasm
3) ER
5) Plastids
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.
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
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.
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:
56:25
If blood is in short supply, which blood type would be the most beneficial to have on hand if someone needed a blood transfusion?
O+
O–
AB+
AB–
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:)
Match the following description with the appropriate type of respiration:
a. occurs in the mitochondria
b. resting muscles depend on this type of respiration
c. lactic acid builds up in muscle fibers
d. rapidly produces ATP for short time periods
e. produces large quantities of ATP but takes longer to synthesize
1. anaerobic respiration
2. aerobic respiration
Answer:
The correct answer is -
1. c, and d.
2. a, b, and e.
Explanation:
Anaerobic respiration is the short and quick way of producing energy, however, it produces less amount of energy than aerobic respiration and produces lactic acid as a byproduct. It takes place in the cytoplasm of the cell and working or acting muscles depend on such respiration.
Aerobic respiration is the main and important respiration process that takes place in mitochondria and it takes time to produce ATPs that are much more than anaerobic respiration. Resting muscles get energy by this type of respiration.
A sample from a stock of a bacterial colony in liquid media was diluted by a factor of 106, and 2 ml of this dilution was spread on a Petri dish of solidified media. 56 colonies were observed. What was the concentration of bacteria of the initial stock?
Answer:
28 × 10⁶ colonies/ml
Explanation:
Let C be the concentration of bacterial in the initial stock. When it is diluted by a factor of 10⁶, the new concentration is C' = C/10⁶.
When 2 ml of this concentration is spread on a Petri dish of solidified media, 56 colonies were produced. The number of colonies, n after spreading the 2 ml of C' is C' × 2 ml = 2C' = 2C/10⁶.
So, n = 2C/10⁶.
Since the number of colonies after spreading on a Petri dish of solidified media is 56, n = 56 colonies.
So, 2C/10⁶ = 56
Making C subject of the formula, we have
C = 56 × 10⁶/2
C = 28 × 10⁶ colonies/ml
So, the initial concentration of bacteria is 28 × 10⁶ colonies/ml
Issued in 1974, 45 CFR 46 raised to regulatory status:
A) The 1974 National Research Act
B) The Nuremberg Code
C) Kefauver-Harris Drug Amendments to the Federal Food, Drug & Cosmetics Act
D) US Public Health Service Policy
Hello!
The answer is D, The U.S Public Health Service Policy. Its main purpose was providing protection for human subjects for research work which was conducted by federal agencies. You can read more about it on HHS.gov, as it is a very interesting regulation.
I hope this helps! :)
Issued in 1974, 45 CFR 46 raised to regulatory status US Public Health Service Policy. Option D
What is the regulatory status?The restrictions known as the Common Rule for the protection of human subjects in research done by or supported by federal agencies are outlined in Title 45 of the Code of Federal restrictions, Part 46. These laws define moral principles and requirements for the protection of research subjects who are being used as human subjects.
The US Public Health Service (PHS) published the policy that would eventually become 45 CFR 46 in 1974. This policy established standards for the examination and approval of research involving human subjects as well as the legal foundation for the protection of those individuals. Informed consent, weighing risks and benefits, and the creation of Institutional Review Boards (IRBs) to regulate research techniques were all adopted.
Learn more about regulatory status:https://brainly.com/question/30479939
#SPJ6
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.
Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email
Answer:
d) Send a detailed email
Explanation:
Send a detailed email is the best way to deliver an update to her because in the email he can send all the information in detail form which can satisfy his owner. He can't call or message because it takes too much time to provide information so email is the best way to provide information. Booking a one-hour meeting is not worth it and the reason for this is that there is no big presentation which takes one hour of description one email is enough for it.
The photic zone Select one: a. has the most nutrients closer to land. b. is an area with sufficient light for photosynthesis. c. has an abundance of photosynthetic organisms. d. is very shallow. e. All of the answer choices are correct.
the answer for this would be E
The photic zone has the most nutrients closer to land, an area with sufficient light for photosynthesis, has an abundance of photosynthetic organisms, and is very shallow, Thus, the correct option for this question is E, i.e. all of the following.
Where is the photic zone present in the aquatic ecosystem?The photic zone present in the surface layer of the aquatic ecosystem significantly receives sunlight for photosynthesis. It is the topmost layer that receives sunlight. Hence, this zone is also known as the Sunlight zone.
According to the question, due to the sufficient availability of sunlight, this area performs a good rate of photosynthesis and occupies an abundance of photosynthetic organisms. Phytoplanktons are the characteristic members of this region. As this region has sufficient availability of light, the concentration of nutrients is highly rich.
Therefore, according to the photic zone, the correct option for this question is E, i.e. all of the following.
To learn more about the Photic zone, refer to the link:
https://brainly.com/question/14576008
#SPJ2
