………………………………………………………………….
whats with the dots ._.
Inang
mukaangaana
wa
men
CARA
mwen
Stible
ther
Macaman
wa
ut
Answer:
.........................
Explanation:
.................
Answer:
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How does water move around the ocean?
A series of experiments shows that oil content in a diploid grain is influenced by five genes (a through e) with additive alleles. The highest producing strain is 20% oil content; the lowest has close to 0%. A plant of unknown genotype has an oil content of 12%. What is a possible genotype for this plant
Answer:
All the possible genotypes of the plants that have an oil content of 12% include six dominant alleles. Following we are mentioning some of them
AABbCcDdEe, AaBBCcDdEe, AaBbCCDdEe, AaBbCcDDEe, AaBbCcDdEE
Explanation:
Due to technical problems, you will find the complete explanation in the attached files.
The possible genotypes involving a combination of six dominant alleles required for the grain to have an oil content of 12% are:
AABbCcDdEe, AaBBCcDdEe, AaBbCCDdEe, AaBbCcDDEe, AaBbCcDdEE
Additive alleles are alleles which result from polygenic traits.
Polygenic traits are traits which are encoded by several genes resulting in more than two alleles contributing to the phenotype.
The phenotype of additive alleles is continuous due to the many different combinations of the alleles of the genes.
From the data provided:
oil content in a diploid grain is influenced by five genes: a, b, c, d, e The highest producing strain is 20% oil content The lowest has close to 0%Since the grain is diploid, grains that are:
h0m0zygous recessive for the five genes, aabbccddee, will have 0% oil content.h0m0zygous dominant for the five genes, AABBCCDDEE, will have 20% oil content.Each dominant gene will contribute 20% / 5 = 4% of oil content.
Since the grains are diploid, each dominant allele will equally contribute 4% / 2 = 2% to the oil content.
Thus:
0 dominant allele → 0% oil content 1 dominant allele → 2% oil content 2 dominant alleles → 4% oil content 3 dominant alleles → 6% oil content 4 dominant alleles → 8% oil content 5 dominant alleles → 10% oil content 6 dominant alleles → 12% oil contentTherefore, a combination of six dominant alleles are required for the grain to have an oil content of 12%.
All the possible genotype involving a combination of six dominant alleles required for the grain to have an oil content of 12% are given below:
AABbCcDdEe, AaBBCcDdEe, AaBbCCDdEe, AaBbCcDDEe, AaBbCcDdEE.
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"S PRACTICE Exam#1 68Qs.doc
PII CVS PRACTICE Exam#1 68Qs.doc (69 KB)
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30. The largest cells in the blood that leave the bloodstream to become macrophages are th
A. Neutrophils
B. Eosinophils
C. Lymphocytes
D. Monocytes
21
ruthroblastocie fatalie also
Your Answer will be D. Monocytes
[tex] \frak{hope \: it \: helps \: you.....}[/tex]
Answer:
The largest cells in the blood that leave the bloodstream to become macrophages are the
Monocycles
What is transaction between school and post school destination
Answer:
The transition between school to a post school is a difficult phase in one's life because it opens a lot of adjustments whether it is socially, mentally and/or psychologically. Most students find it difficult to transfer because of these following reasons visit these links and see the answers below: brainly.ph/question/2641856, brainly.ph/question/2635171, brainly.ph/question/2627976
There are a bunch of strangers. Not all students are gifted with an extropersonal skills that allows them to make friends easily. Transition makes it difficult for introvert pupils to accept.
The new school is bigger than their previous school. Though in others the size of a place is not bothering their performance, there are individuals who find it difficult to adjust when it comes to the size of their environment. For example a student who studied in a small private school with only one section will find it difficult to adjust moving around a larger environment.
There are different teaching styles coming from the school's faculty. Learning styles are somewhat define by the teacher's teaching styles and strategies. In their previous school, some of the seniors might have already leaked the strategies used by the teacher in their school. But in a post school, some students are 100% newbie wherein they have no idea what is going on inside a class of a single professor or instructor. This is difficult since they do have to observe every single day and learn a new way to follow the lesson's phase.
Friends are not with the same post school. It is said that you can go through anything with your friends but most of the time you will never be with them when you transfer to a post school but still you need to move forward and learn to mix with the public.
Lessons are difficult and new. Most learners are not accustomed in learning quick in a new environment especially when the lesson is more difficult and new.
New policies are too intimidating to break. Some students are struggling when it comes to adjusting to rules whether it is inside the classroom or just a simple campus rule so they feel intimidated and and fearful in breaking one. But it must be remembered that rules keep the society moving forward peacefully.
Explanation:
There are different reason for an individual to fear and feel difficulty adjusting to a post school. But one must remember that humans are created with an adaptability skill that allows us to mix well with a new habitat and environment. Social skills are just part of our adaptability. You can just follow the wind and let the flow take you. Life is a constant change and all you have to do is to embrace it to make yourself belong anywhere you go.
Transaction between school and post school destination is the entering from one stage to another stage of live.
What is transaction between school and post school destination?The transition between school and post school destination means that the youth is ready for next stage of their lives. The perfect transition of the students from high school to college or the adult world is very important.
This is because these years provides the direction to the students for their future so we can conclude that transaction between school and post school destination is the entering from one stage to another stage of live.
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Choose All that apply !!
Help!!
Answer:
1.5
Explanation:
The greatest probability must not exceed 1
[tex] { \boxed{ \sum(P) \leqslant 1}}[/tex]
The information in DNA in the nucleus is used to produce
Answer:
Proteins
Explanation:
DNA courses can be turned into messages that can be used to make proteins
When proteins help transport molecules in and out of the cell, it is called _____.
passive diffusion
facilitated diffusion
active transport
endocytosis and exocytosis
Answer:answer is active transport
Explanation:
Which of the following is not a product of glycolysis?
True
False
Answer:
We need a full question it's a half question
Given that the growing opinion root tip has 16 chromosomes in each cell ( normally). During the cell cycle, while the cell is growing and dividing, specify how many chromosomes will the cell have at the G1 phase? and how many chromosomes will it have after the S phase? Explain your answer with a reason/evidence statement
Answer:
16
It is given that the number of chromosomes present in each onion root tip cell is 16. Hence, the number of chromosomes in the cell will be 16 at G1 phase, after S phase and after M phase
What 2 things change about the molecules of different states of matter?
Answer: speed and proximity
Explanation:
science
Explanation:
heating and evaporation
(c) Predict the growth rate pattern of the E. coli-M bacteria when they are initially added to the growth medium if the mutation does not interfere with glucose utilization.
(d) Provide reasoning to justify your prediction.
The question is incomplete. The complete question is :
Glucose is the preferred carbon source for Escherichia coli bacteria. Researchers isolated E. coli-M, a strain of E. coli with a single mutation that inactivates a key enzyme in one of the three glycolytic pathways E. coli bacteria use to break down glucose. In an initial experiment to study the effect of the mutation on glucose utilization and ATP production by the bacteria, the researchers add 2×104 E. coli-M bacteria to 400 mL of a growth medium that contains glucose. To measure the rate of bacterial growth, they monitor the optical density of the culture as a measure of bacterial concentration over time.
(c) Predict the growth rate pattern of the E. coli-M bacteria when they are initially added to the growth medium if the mutation does not interfere with glucose utilization.
(d) Provide reasoning to justify your prediction.
Solution :
c). It is given that Glucose is preferred as the carbon source for the Escherichia coli bacteria or the E. coli-M bacteria. If the growth medium of the E. coli-M bacteria contains a higher concentration of the acetate, then the growth of the bacteria will inhibit without inhibiting the central metabolism.
d). This is predicted because as E. coli-M bacteria grows, it starts secreting acetate. This is known as overflow mechanism. The regulatory interactions mediated by the acetyl-phosphate that plays an important role in the inhibiting growth by the acetate.
in a cell all organells works together to carry out ?
Answer:
Cell organelles must work together to carry out protein synthesis, utilize proteins within the cell, and transport them out of the cell.
A cross between a plant with red flowers and a plant with white flowers produces offspring with pink flowers. What is this an example of?
Answer:
Example of incomplete dominance
Explanation:
the offspring is different the parent
Nicholas,, a 3 year-old boy, is brought into your gastroenterology practice by his mother. They were referred to you by their pediatrician because Ali has been complaining of severe abdominal pain, especially after eating, for the past few weeks. He has an otherwise normal health history. His pediatrician ordered a fecal occult (hidden) blood test, hematocrit, and urinalysis. The fecal occult blood test is positive and the hematocrit (red blood cell count) is lower than normal for a child his age. Body temperature and urinalysis are normal.
You perform a surface abdominal exam, which appears normal, but Ali complains of pain in the lower right abdominal quadrant during palpitation. You order a CT enterography procedure, which uses a contrast dye to make an X-ray of the small intestine. The CT shows a diverticulum. After confirmation with a technetium scan, you diagnose Nicholas withMeckel’s diverticulum.
Treatment for Meckel’s diverticulum includes surgical removal of the diverticulum.
What do you mean by diverticulum?
Diverticulum is a small pouch or sac that forms in the wall of the digestive tract. It can occur in any part of the digestive tract, including the esophagus, stomach, small intestine, and large intestine. Diverticula can cause symptoms such as abdominal pain, constipation, and diarrhea.
You recommend that Nicholas have a laparoscopic removal of the diverticulum, a minimally invasive procedure that involves making small incisions in the abdomen and inserting a camera and surgical tools to remove the diverticulum. You explain to Nicholas’ mother the benefits of the procedure and the potential risks associated with it.
After the procedure, you recommend that Nicholas follow up with his pediatrician in two weeks. You also suggest that Nicholas and his mother modify his diet to include more fiber and fluids to help ease his symptoms and prevent the diverticulum from returning. You also suggest that Nicholas take an antacid to help reduce the abdominal pain.
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#SPJ1
CODIS is divided in to Indexes. (Offenders, Forensic Unknowns, and Relatives of Missing Persons etc.) Is law enforcement allowed to search all of the profiles in all indexes against each other? Why or why not?
Question 2 options:
1)
Yes, if the data is in the database then it is lawful to search it anyway that helps the case.
2)
No, but only because it is not helpful to the case.
3)
Yes; but only if you have the permission of the victim in the case.
4)
No; it is against a federal law.
5)
Yes; but only in cases when the suspect is a male.
Answer:
1
Explanation:
they are always allowed to
Describe the neural and renal mechanisms that control blood pressure.
Answer:
The autonomic nervous system is the main neural regulator of circulation and blood pressure in the short term and beat by beat and exerts its function through various reflexes that regulate vasomotor tone, heart rate and cardiac output. At the renal level, the renin–angiotensin–aldosterone system is possibly the most important in the maintenance of arterial homeostasis.
Explanation:
Blood pressure is regulated by a series of interrelated autonomic systems and humoral reflexes, which continually adjust the determining elements of the system (heart rate, stroke volume, total peripheral resistance and circulating volume).The effective circulating volume is controlled by a series of reflex systems, which obtain information about the perfusion pressure (baroreceptors in the carotid bulb and aortic arch), plasma osmolarity (hypothalamus) and urinary sodium (distal tubule).The kidney has its own self-regulatory mechanisms. The reduction in renal blood flow is detected at the level of the mesangial cells of the juxtaglomerular apparatus, starting the renin-angiotensin system. The increase in angiotensin II produces on the one hand local vasoconstriction, and on the other hand stimulates the production of aldosterone by the adrenal cortex with the consequent tubular reabsorption of sodium and water.Antidiuretic hormone or vasopressin (released from the hypothalamus by stimulation of arterial baroreceptors and also by stimulation of angiotensin II) also acts at the renal level, which acts as a powerful and water-saving vasoconstrictor in the distal tubule.
As the most junior member of a lab, you are tasked with generating cell lines that accumulate DNA damage to investigate how random mutations affect transformation of cells into cancer cells. You decide to mutate proteins in the p53 pathway. Which three proteins would you mutate
Answer:
In p53 pathways, 3 proteins which are mutate and cause cancer that proteins are:
1.BRCA2
2. p53 in Rb genes binds with deoxyribonucleic acid that stimulates another gene to produce a protein known as p21 that interacts with cdk2. When p21 is combined with cdk2 cell will not pass through to the next stage of cell division. This is Li-Fraumeni syndrome (when p53 causes cancer from parents to offspring)
3.BRCA1
The parent that transmits mutation in one of these genes has a 50% chance of inheriting the mutation. The effects of mutations in BRCA2 and BRCA1 are seen when the person's second copy of the gene is normal.
Can somebody help me?
Answer:
Sorry but this isn't clear. What I understood was something related to meosis or something. Not sure tho. Plz post it clear so that you do get answer
Explanation:
why do we find tea to be tasteless after we have eaten sweet or candies
Explanation:
The answer is that when you eat the sweet candies and chocolates the buds will remain sweet so that's why the tea become tasteless. The buds only relive the sweetness by staying for few minutes or eat something else.
I hope this answers to your question
If a bacterial isolate show intermediate to moderate resistance to an antimicrobial, how might this drug still be successfully used in the treatment of this microbe
Answer:
Due to its effectiveness.
Explanation:
This drug still be successfully used in the treatment of this microbe because of its effectiveness on the microbes. All the microbes did not got resistance against the antimicrobial chemical and they are not full proof from the antibiotics so that's why the antimicrobial are used in the treatment of microbes. If all the microbes attain resistance against antimicrobial drug so it can't any affect on the microbes.
Give three reasons why you might have a situation in which bacterial colonies were found on the first section of a streak plate, but not on sections two and three.
Hi. You have not shown the sections this question refers to, nor have you provided more information about those sections. This makes it difficult for your question to be answered. However, I will try to help you as best I can.
Generally speaking, we can consider that there was no growth of bacterial colonies in sections 2 and 3, because, for some reason, the bacteria was not inoculated in sections 2 and 3. What could also have happened, is that the bacteria in the sections 2 and 3 were eliminated, but the bacteria from section 1 managed to survive and form colonies.
Another possibility is that after inoculating the bacteria in section 1, you didn't handle the bacteria correctly in the other sections, leaving that bacteria to dry out and die.
Finally, the culture media in sections 2 and 3 could be inappropriate for bacteria to develop and form colonies.
ecologically, termites are classified as a carnivore b, detritivores c, insectivore d, herbivore
Answer:
b. detritivores
Explanation:
Detritivores (also known as detrivores, detritophages, detritus feeders, or detritus eaters) are heterotrophs that obtain nutrients by consuming detritus (decomposing plant and animal parts as well as faeces).
The closest living ancestors of birds are crocodilians. What provides morphological evidence for this relationship?
a) both are homeothermic
b) both have extensive air sac system
c) both have a 4-chambered heart
d) both have differentiated teeth
e) none of the above
Answer:
c) both have a 4-chambered heart
Explanation:
The heart that consists of four chambers where the blood flows. In this, the blood entered into the right atrium and passes via the right ventricle. Now the right ventricle pumps the blood into the lungs in which it is oxygenated. Now the oxygenated blood would be bring back to the heart via the pulmonary veins that entered into the left atrium
So as per the given situation, the option c is correct
What part of the cell does: Photosynthesis? Respiration?
10 POINTSSSSSSSSS!!!!!!!!!!!!!!!!!!
Answer:
13. A group of Roaches moving in a group
14. Stepping on the Roaches to kill them
Answer: First image shows 3 groups with 14 Roaches in each group moving towards an human.
Second image shows the human stepping on the middle group of roaches.
Explanation: Roaches are nasty. Get an exterminator, ya nasty.
You have a novel virus that can kill mice when injected into them. This novel virus can transform a non-virulent virus, causing it to become virulent. You want to understand what the transforming factor is made of. You heat kill the novel virus and treat different samples of the heat killed sample with either an RNase, a DNase, or a Proteinase. Of the novel virus uses double-stranded RNA as its genetic material, which treated sample will no longer be capable of transforming the non-virulent strain
Answer:
The sample treated with RNase
Explanation:
According to this question, a novel virus was found which was capable of killing mice when injected into them and capable of making a nonvirulent virus to be virulent by transforming them.
After heat killing the novel virus, different samples were treated with either an RNase, a DNase, or a Proteinase. If the novel virus uses double-stranded RNA as its genetic material, the sample treated with RNase will no longer be capable of transforming the non-virulent strain. This is because RNase is a catalytic enzyme that degrades RNA, hence, the RNA will no longer be present to transform.
hola como estan soy nueva
Answer:
Buena gracias
Explanation:
Assuming the oligopeptide alfahelicks forms one continous a helix the glutamate residue is hydrogen bonded to:___.
a. leucine.
b. cysteine.
c. serine.
d. lysine
e. isoleucine.
Answer:
B. Glutamate is always bonded to Cysteine when dealing with DNA.
Explanation:
Hope that's what you meant.
If this has to do with RNA or mRNA then still g pairs with c.
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