Microarray data attached
Answer:
B, D, E are consistent with the data in the microarray
Explanation:
A) False, if you look at the microarray graph, Gene K is red on the right side of the graph (where patients in group II are) and generally more blue on the left side where group I patients are, meaning it is less expressed
B) This is true, look for example at gene F, individual 3 expresses it strongly, whereas individuals 6 and 7 are quite low.
C) No - microarray data looks at the expression of genes, not the genes themselves.
D) Yes, you can see that generally, genes B-Q are highly expressed in group II patients but not group I. In contrast, genes F-U are not expressed in group II but more expressed in group I.
E) Yes, although the data is not always consistent, there are clear patterns in group II patients not present in group I.
Final answer:
The DNA microarray data support conclusions about differences in gene expression levels between cancer patient groups, with color codes indicating relative gene expression. Gene K shows higher expression in Group I if indicated by red, and there is variation within groups. The data do not suggest different sets of genes but rather varying expression levels, which can potentially assign a new patient to a group.
Explanation:
The analysis of DNA microarrays involves comparing patterns of gene expression between different samples, in this case between cancerous cells from different patients. The color codes on a DNA microarray heat map provide valuable insights:
Red indicates a high relative gene expression specific to cancer cells.Blue indicates a low relative gene expression or genes that are more active in non-cancerous tissue.Yellow indicates genes that are expressed at similar levels in both cancerous and non-cancerous tissues.Considering the given scenarios and color codes, let's assess the conclusions that the data support:
Gene K is more highly expressed in cells from patients in Group I than in Group II if the spots corresponding to gene K are predominantly red for Group I and less intensely red or blue for Group II.There is variation in the amount of mRNA present, even among patients in the same group, if the color intensity varies within a group for a particular gene.Patients from Group I and Group II do not carry different sets of genes; instead, the same genes can have different levels of expression in the two groups.A different set of genes is not necessarily transcribed; rather, the same genes may be transcribed at different levels in cells from the two patient groups.Using microarray data, you could likely determine to which group a new patient's cells belong based on the pattern and intensity of gene expression.Professor Stetson is conducting a study on memory with a sample of college students. Before they agree to participate, Professor Stetson tells them about the study and what they will do if they choose to participate. This is an example of:
Answer:
informed consent
Explanation:
When during a study the participants are told in detail about the procedure of the study along with possible risks, benefits and potential consequences and after gaining this knowledge a participant agrees to be a part of the study then it is said to be an informed consent.The medical and research ethics have made it mandatory to get informed consent from the participants before they enrol in any experimental procedure.The informed consent can be given either directly by the participant or if the participant is not in the condition to grant it then the parents or legal guardians can also provide it.Therefore, in the given case Professor Stetson is asking for an informed consent from the participants.For Type A individuals, exposure to stress is especially likely to inhibit organs such as the ________ from removing cholesterol and fat from the blood.
Answer:
For Type A individuals, exposure to stress is especially likely to inhibit organs such as the _liver_ from removing cholesterol and fat from the blood.
Approximately 21% of the human genome is comprised of nucleotides containing C. Given this information, calculate the percentage of the human genome that is comprised of nucleotides containing G, T, and A.
(a) _____% of the human genome is comprised of G.
(b) _____% of the human genome is comprised of T.
(c) _____% of the human genome is comprised of A.
Enter your answer as three numbers separated by commas (example: 12, 26, 32).
Answer:
(a) 21% of the human genome is comprised of G.
(b) 29% of the human genome is comprised of T.
(c) 29% of the human genome is comprised of A.
Explanation:
According to Chargaff's rule, a DNA molecule has the same amount of pyrimidines (T and C) as purines (A and G), and G pairs with C and A pairs with T.
If 21% of the genome contains C, then 21% also contains G.
Therefore 21% + 21% = 42% of the genome is comprised of G+C.
The other 58% is comprised of A+T, and because they pair with each other, 29% will correspond to A and 29% will correspond to T,
In garden peas, a single gene controls stem length. The recessive allele (t) produces short stems when homozygous. The dominant allele (T) produces long stems. A short-stemmed plant is crossed with a heterozygous long-stemmed plant. Which of the following represents the expected phenotypes of the offspring and the ratio in which they will occur?
a. 3 long-stemmed plants: 1 short-stemmed plant
b. 1 long-stemmed plant: 1 short-stemmed plant
c. 1 long-stemmed plant : 3 short-stemmed plants
d. Long-stemmed plants only
e. Short-stemmed plants only
Answer:
b. 1 long-stemmed plant: 1 short-stemmed plant
Explanation:
The short-stemmed plant would have genotype(tt) and the heterozygous long-stemmed plant would have the genotype(Tt). So when they will cross with each other half of the offspring will be long-stemmed plants and half will be short-stemmed plant.
T t
t Tt tt
t Tt tt
So the offspring which have Tt genotype will be tall and the offspring which have tt genotype will be short. Therefore the correct answer is b. 1 long-stemmed plant: 1 short-stemmed plant.
Final answer:
When a short-stemmed pea plant (homozygous recessive, tt) is crossed with a heterozygous long-stemmed plant (Tt), the phenotypic ratio of the offspring is b. 1 long-stemmed plant: 1 short-stemmed plant (1:1).
Explanation:
When a short-stemmed plant (which must be homozygous recessive, tt) is crossed with a heterozygous long-stemmed plant (genotype Tt), their offspring can have two possible phenotypes: long-stemmed and short-stemmed. A Punnett square is used to predict the outcome of this cross.
The possible genotypes of the offspring are Tt and tt. The Tt plants will be long-stemmed because the dominant allele (T) masks the effect of the recessive allele (t), and the Tt plants will be short-stemmed. Therefore, the expected phenotypic ratio of the offspring is 1 long-stemmed plant: 1 short-stemmed plant (or 1:1).
The correct answer is b. 1 long-stemmed plant: 1 short-stemmed plant.
Shulte chose several areas in the lower Chesapeake Bay as his experimental sites. At each site, he constructed artificial reefs of two different heights. He also left portions of each site unrestored to serve as points of comparison. Oysters were allowed to colonize the reefs and the unrestored bay bottom safe from harvesting. His hypothesis was that oysters would survive better in the plankton-rich upper waters away from the smothering sediments and hypoxic waters of the bay bottom.A) Reef type is the independent variable.B) Adult oyster density is a dependent variable.C) Spat density is a dependent variable.D) High-relief and low-relief reefs are treatments.
Answer:
All options are correct.
Explanation:
The scientific experiments require proper procedure and experimental set up to perform the experiments. The results and observation was used for the interpretation of the experiment.
The experiments include both the dependent variable, independent variable control and set-up. Here, Shutle used the adult oyster and reef types as dependent and independent variable. Reefs are used as treatments and Oyster density changes and included in the category of density dependent variable.
Thus, all given options are correct.
Restoration efforts by the Chesapeake Bay Foundation aim to increase oyster population density using experimental oyster reefs and disease-resistant strains, though interstate cooperation is required due to upstream pollution.
Oyster restoration efforts in the Chesapeake Bay have been crucial due to the sharp decline in oyster populations. Organizations like the Chesapeake Bay Foundation have focused on increasing oyster population density to improve reproduction rates. Utilizing disease-resistant oyster strains from the Virginia Institute of Marine Science, these restoration initiatives involve constructing experimental oyster reefs to combat problems such as pollution. However, efforts by states like Virginia and Delaware are challenged by pollution from upstream states, necessitating larger inter-state cooperation for a successful restoration. Moreover, understanding the impact of environmental factors on oyster health, such as anthropogenic noise, provides insights that can guide aquaculture practices and coastal development.
John is a newly diagnosed diabetic that contacts the office with complaints of severe nausea and vomiting. What instructions should the APRN provide related to his insulin doses?
Answer:
i think maybe that he should stop using high doses of insulin
Explanation:
it could be the fact that his blood sugar is to low
Every 4 hours, check your blood sugar levels and take regular insulin based on the results.
What is the effect of insulin?A glucometer is typically used to measure blood sugar levels during sliding-scale insulin therapy regimens.
About four times per day, this is performed (every 5 to 6 hours, or before meals and at bedtime). Your blood sugar reading determines how much insulin you receive at mealtime.
The following factors determine whether an APRN is allowed to prescribe prohibited medications to patients: A patient has been given an oral medication with a high first-pass effect.
Therefore, every 4 hours, check your blood sugar levels and take regular insulin based on the results.
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The success of the program to eradicate smallpox has stimulated experts to pursue what they had not previously considered possible—better control, if not eradication, of the other infections such as measles and yaws.
Answer:
This is an English Language question and not Biology.
Here is full question:
The success of the program to eradicate smallpox has stimulated experts to pursue what they had not previously considered possible -- better control, if not eradication, of the other infections such as measles and yaws.
A. what they had not previously considered possible -- better control, if not eradication, of the other infections such as
B. what they had not previously considered a possibility -- better control, if not eradication, of such infections like
C. something they had not previously considered possible -- better control, if not eradication, of such infections as
D. something not considered a previous possibility -- better control and perhaps eradication, of other infections such as
E. the possibility of what they had not previously considered possible – better control and possibly eradication of infections like
The answer is C
Explanation:
'What they had not previously considered a possibility' is too specific. It might be pointing to just one possibility. But there are two possible possibilities here, 'something' sounds better.
The action in the past, which was the eradication of smallpox, has triggered something, which was the better control or eradication of other diseases.
Previously, the experts had not considered better control possible, but then this action was triggered by the eradication of small pox. Now the experts are considering better control possible because they have been stimulated to pursue it.
Endorphins are found where neurons meet skeletal muscles. less powerful than enkaphalins. pain-controlling chemicals. radically different in function from neurotransmitters. responsible for the fatal reaction that human beings have when bit by a Black Widow spider.
Answer:
Endorphins are "pain-controlling chemicals".
Explanation:
Endorphins are endogenous opioid neuropeptides and peptide hormones in individuals and other animals. They are shaped and deposited in the pituitary gland. The organization of particles as endorphins is founded on their pharmacological action, as different to a exact chemical preparation. Stress and pain are the two most mutual issues foremost to the issue of endorphins. Endorphins cooperate with the opiate receptors in the brain to condense our insight of pain and act likewise to drugs such as morphine and codeine.
Endorphins are the body's natural pain-relieving chemicals, functioning by inhibiting pain signals in the brain. They are not involved in the reactions to a Black Widow spider bite, which is caused by venom, not endorphins.
Endorphins are pain-controlling chemicals similar in structure to opioids such as morphine, codeine, and heroin. They bind to opioid receptors in the brain to reduce pain sensation. Enkephalins, a specific type of endorphin, function by inhibiting the transmission of pain signals at synapses by hyperpolarizing the postsynaptic membrane, effectively reducing the sensation of pain even in the face of intense or chronic pain conditions.
What term is used to describe all the alleles at every locus for all individuals within a population?
Answer: GENE POOL is used to describe all the alleles at every locus for all individuals within a population.
Explanation: The gene pool consists of the sum total of all the genes in a particular population. It includes all the genetic information encoded in these genes.
The Hardy-Weinberg law provides equation to relate genotype frequencies and allele frequencies in a randomly mating population;
p² + 2pq + q² = 1 for two alleles such as A and a, where p and q are probabilities of allele frequencies. That is,
p = homozygous dominant
pq = heterozygous
q = homozygous recessive
The gene pool is commonly used during population studies. Therefore, population genetics is the branch of genetics that describes inheritance on the population level in mathematical terms.
If average speed is calculated using the equation, "distance traveled/elapsed time," what is the average speed of an object that travels 65m in 5s?
Answer:
13m/s
Explanation:
The speed= distance traveled/elapsed time
Therefore speed=65/5
=13m/s
The patient is taking an anticholinergic medication along with an antiparkinson medication. Which statement by the patient indicates more teaching is required?
A. "If it makes me nauseated I can take the medication with crackers."
B. "I can take this medication at bedtime as it makes me sleepy."
C. "No matter what, I cannot take this medication with food."
D. "I can use hard candy to decrease any mouth dryness."
Answer:
The correct answer is A. "If it makes me nauseated I can take the medication with crackers."
Explanation:
The anticholinergic antiparkinson medication administrated via oral is absorbed in the duodenum and the first portion of the jejunum. The stomach acts as a reservoir and the entrance to the circulatory system, from the intestine, is mediated by active transport. In this step, the medicine competes with other amino acids and ramified chains for its absorption and when they are taken together, they affect medication absorption. Among other considerations, administration of the medicament separated at least by 30 minutes of meals is a useful habit to optimize its kinetic.
The incorrect statement indicating more teaching is required is "No matter what, I cannot take this medication with food." This suggests a misunderstanding, as taking medications with food can often alleviate side effects like nausea.
Explanation:The question is assessing the patient's understanding of how to manage the common side effects of anticholinergic and antiparkinson medications. Looking at the options provided, the answer to which statement indicates that more teaching is required is C. "No matter what, I cannot take this medication with food." This statement is incorrect as taking medication with food can be an effective way to minimize gastrointestinal side effects, such as nausea. Many medications can be taken with or without food, and the healthcare provider or pharmacist can provide personalized advice based on the specific medication.
Options A, "If it makes me nauseated I can take the medication with crackers," and B, "I can take this medication at bedtime as it makes me sleepy," are generally correct. These statements suggest the patient understands how to mitigate some common side effects such as nausea and drowsiness. Option D, "I can use hard candy to decrease any mouth dryness," is also correct because anticholinergic medications can cause dry mouth, and sucking on hard candy can help alleviate this symptom.
In conclusion, the patient expressing a rigid stance against taking medication with food demonstrates a lack of understanding and indicates that more teaching about managing medication side effects is necessary.
An outer membrane is seen in: all bacteria but not archaea all archaea but not bacteria all gram - A.bacteria all gram B. bacteria C.all prokaryotes but no eukaryotes
Answer:
All gram (-) ve bacteria
Explanation:
The structure of bacteria demonstrates that the bacteria possess a cell wall which is composed of teichoic acid and peptidoglycan.
The cell wall thickness varies in the bacteria which is tested by the Gram staining technique. The bacteria with thick cell wall shows positive color or stain color so are called gram-positive bacteria and the bacteria with thin cell wall do not show the color and thus are called gram-negative bacteria.
However, the gram-negative bacteria also possess a layer of lipopolysaccharide and lipoproteins called LPS layer or outer layer surrounding the cell wall which is absent in the gram-positive and archaea bacteria.
Thus, All gram (-) ve bacteria is the correct answer.
Under the Physical Exam and Autopsy provision, how many times can an insurer have the insured examined, at its own expense, while a claim is pending?
Answer:
Unlimited.
Explanation:
Health insurance provide medical benefits and covers the medical expense of the client. The health insurance can be provided by both the government and private sectors.
The client pays some fixed amount of money to the insurer for the continuity of the medical benefits. The insurer can unlimited or finite time can examine even when the claim is not completed.
Thus, the answer is unlimited.
Under the Physical Exam and Autopsy provision, an insurer can have the insured examined multiple times at its own expense while a claim is pending. This provision is designed to prevent fraud and confirm the validity of a claim.
Explanation:The Physical Exam and Autopsy provision in an insurance policy typically gives the insurer the right to examine the insured when deemed necessary, as many times as they want, while a claim is pending. This means that even multiple examinations can be arranged by the insurer at its own expense. The key idea behind this provision is to prevent fraud and confirm the validity of the claim.
For example, if an insured person claims disability pay, the insurer may request an independent medical examination by a doctor of their choice. If discrepancies appear or if the condition is expected to improve over time, repeated examinations might be required.
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Timothy was born without testes. With respect to hormone production and sexual behavior, which of the following is the most likely outcome?
A. Timothy will have some testosterone, but he will probably be unable to achieve an erection.
B. Timothy will lack testosterone, but he will probably be able to achieve an erection.
C. Timothy will have some testosterone and will probably be able to achieve an erection.
D. Timothy will lack testosterone and will probably be unable to achieve an erection
Without testes, Timothy would not produce significant amounts of testosterone, the hormone responsible for male sexual characteristics and the ability to achieve an erection. Therefore, it is most likely that he would lack testosterone and be unable to achieve an erection.
Explanation:The testes are the major source of testosterone in males. Without testes, a male would not produce significant amounts of testosterone. Testosterone plays a vital role in the development and maintenance of male sexual characteristics, as well as in the process of getting an erection. Without testosterone, it is likely that Timothy will be unable to achieve an erection. Therefore, the correct answer is D. Timothy will lack testosterone and will probably be unable to achieve an erection.
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________ are particularly polarizable. ________ are particularly polarizable. Small polar molecules Large nonpolar molecules Large polar molecules Small nonpolar molecules Large molecules, regardless of their polarity,
Answer:
____Large molecules, regardless of their polarity____ are particularly polarizable.
Explanation:
All the large molecules are called polar molecules because they have many bonds including ionic bond, Van der Waals forces or hydrogen bond and flexibility in conformation. This property also increase the boiling point of large molecules.
Dispersion forces, also known as London dispersion forces or Van der Waals forces, are the primary intermolecular forces between all molecules. Larger molecules, regardless of their polarity, are particularly polarizable and have stronger dispersion forces. Small nonpolar molecules have weaker dispersion forces.
Explanation:Dispersion forces are the primary intermolecular forces between all molecules, including polar and nonpolar molecules. These forces are also known as London dispersion forces or Van der Waals forces. Larger molecules, regardless of their polarity, are particularly polarizable and have stronger dispersion forces. Small nonpolar molecules, on the other hand, have weaker dispersion forces.
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An 18-month-old has fanning of the toes and dorsiflexion of the big toe seen on physical exam. Based on this finding the nurse should:
Answer: Fanning of the toes and dosriflexion of the big toe is an abnormal response called the babinski reflex. Its a neurologic relfex. Newborn babies are not neurologically mature and therefore show a Babinski response. Many young infants do this, too, and it is perfectly normal. However, in time during infancy the Babinski response should vanish completely.
Explanation:
Positive Babinski sign happens when stroking the sole of the foot, the big toe bends up and back to the top of the foot and the other toes fan out. This can mean that you may have an underlying nervous system or brain condition that's causing your reflexes to react abnormal. This abnormal response is a problem of the CNS normally the pyrmidal tract. In this case the baby is 18 months old and is not mature neurologically so this response is perfectly okay.
But if it persists after the age of 2 years or as an adult it should be considered as a problem.
Scientists have modified a clathrin molecule so that it still assembles but forms an open-ended lattice instead of a closed spherical cage. How would this clathrin molecule affect endocytosis in cells? Choose one: A Vesicles would be larger, increasing the cargo endocytosed. B. All movement of molecules into and out of the cell would cease. C. Vesicles cannot form properly without a clathrin cage, thus inhibiting endocytosis. D. Endocytosis would be unaffected, since adaptors and receptors can still interact.
Answer:
Vesicles cannot form properly without a clathrin cage, thus inhibiting endocytosis.
Explanation:
Clathrin is important for the vesicle fusion. These vesicles are important for the transport and secretion of the proteins from one cell compartment to the next compartment.
The clathrin proteins helps in the vesicle formation and form the close cage in the vesicle. This shape is important for the vesicles formation. The mutation in clathrin will disrupt the shape of the vesicle from the cage to the liner structure. This disrupts the endocytosis process and the protein transport will also blocked.
Thus, the correct answer is option (C).
A gene has a base sequence of GTC. Due to mutation, the base sequence changes to GTG. Answer the follow questions
a) what are the bases of mRNA coded for this section of DNA, before the mutation?
c) what amino acid is coded by this section before the mutation?
d) what amino acid is coded by this sequence after mutation?
Answer:
i only have a few of the answers
Explanation:
a) What are the bases of mRNA coded for by this section of DNA, before the mutation? Hint: In RNA,
A pair with U. (1 point) CAG.
b) What are the bases of mRNA coded for by this section of DNA, after the mutation? Hint: In RNA, A
pairs with U. (1 point) CAC.
c) What amino acid is coded for by this sequence before the mutation? Hint: Use the codon table. (1 point) Glutamine.
d) What amino acid is coded for by this sequence after the mutation? (1 point) Histidine.
Calculate the sample standard deviation and sample variance for the following frequency distribution of heart rates for a sample of American adults. If necessary, round to one more decimal place than the largest number of decimal places given in the data.
Answer:
Let's complete the question using these data
Heart Rates in Beats per Minute
Class Frequency
61-66 2
67-72 11
73-78 11
79-84 3
85-90 8
Sample variance = 56.6924
Sample std dev = 7.5
Explanation:
First, let's find the average
The average
(2 x 63.5 + 11 x 69.5 + 11 x 75.5 + 3 x 81.5 + 8 x 87.5)
___________________________
(2+11+11+3+8)
2666.5
= _______
35
= 76.18571
variance = 1927.54/34
= 56.6923529
Sample variance = 56.6924
std dev = sqrt( variance)
= sqrt(56.6924)
= 7.529
Sample std dev = 7.5
To calculate the sample standard deviation and sample variance, find the mean of the data, calculate the squared deviations from the mean multiplied by their respective frequencies, sum up the squared deviations, divide by the sum of the frequencies minus one to get the sample variance, and take the square root of the sample variance to obtain the sample standard deviation.
Explanation:To calculate the sample standard deviation and sample variance for the given frequency distribution, we need to first calculate the mean of the data. Then, we will calculate the deviations of each observation from the mean, square those deviations, and multiply them by their respective frequencies. Next, we sum up the squared deviations and divide it by the sum of the frequencies minus one to calculate the sample variance. Finally, we take the square root of the sample variance to obtain the sample standard deviation.
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Kiara loves to play outdoors and enjoys hiking and watching birds in her backyard. Her father notices this interest and decides to sign her up for a local Nature Club that includes birdwatching and plant identification on local short hikes. What kind of gene-environment interaction is this?
Answer:
"Evocative" type of gene - interaction is this.
Explanation:
It is the correlation that occurs only in case an individual behavior that evokes an environmental response. It is also the association between an individual genetically predisposed behavior and others reaction towards that behavior. The environment gene interaction is an very complex interaction between the individual. In the gene interaction a series of the functional association among the genes. In the Interaction the non allelic gene effect of one will subsequently affect the another gene. Which result in the formation of suppression of the effect when together they produce a new character.
A researcher has introduced molecules of human myoglobin and human hemoglobin into a solution containing oxygen. Which of the following statements is true of human hemoglobin and myoglobin?
(Complete questions below).
ANSWER
→When introduced into an oxygen-containing solution, a single molecule of human myoglobin will bind to more oxygen molecules than a human hemoglobin molecule will.
Explanation:
Myoglobin(MB) and haemoglobin (HB) are oxygen binding pigments in mammals while Hb transports oxygen and other gases in the blood streams,Myoglobin stores maximum oxygen in the muscles. Thus this explains while it is possible to seize breath with no effects for few seconds.The oxygen storing capacity of myoglobin made it a higher binding oxygen pigments than haemoglobin. Hb does not store oxygen but transports it in the body.
Structurally,.Heamoglobin is made up of 4-Heme group; while myoglobin has 1 Heme group.The strong affinity and storage capacity for oxygen is due to non- bonding of distal HIistidine group (His 64) with iron atom.(iron ii) which gives enough flexibility for interactive reactions of Histidine with oxygen for maximum oxygen binding and storage,(the proximal histindine group(His-63) is bonded directly to the iron atom therefore no flexibity for bonding and stroge with oxygen..
Therefore 1 molecule of MB will bind to more oxygen molecules(MB02) , than 1 Hb molecule.,(Hb08) even at low concentration.
Complete Question.
A researcher has introduced molecules of human myoglobin and human hemoglobin into a solution containing oxygen. Which of the following statements is true of human hemoglobin and myoglobin?
Because myoglobin does not interact with oxygen molecules, only human hemoglobin will bind to oxygen molecules in this solution.
When introduced into an oxygen-containing solution, a single molecule of human myoglobin will bind to more oxygen molecules than a human hemoglobin molecule will.
When introduced into an oxygen-containing solution, human hemoglobin will bind oxygen molecules more tightly than human myoglobin.
Compared to human hemoglobin, human myoglobin will likely bind to oxygen molecules more quickly when introduced into an oxygen-containing solution.
When introduced into an oxygen-containing solution, human myoglobin and hemoglobin will behave similarly and will bind to the same number of oxygen molecules (with the same affinity).
alpha helix and beta sheet secondary structure is located mainly in the interior of folded proteins whereas irregular loops occur on the outside_________________.
Answer:
of folded proteins
Explanation:
In Secondary structures; the irregular loops seems to be positioned on the outside of folded proteins while the Alpha and beta sheets is categorically positioned at the centre or interior of the protein.
The loops are positioned on the outside of folded proteins in order for the peptide bonds to be open to form H bonds with water.
As a result of stress, the anterior pituitary releases ________, which stimulates release of hormones from the adrenal cortex that retain sodium and water, increase blood sugar, and begin breaking down fats.
A) thyroidstimulatinghormone
B) growthhormone
C) ACTH
D) ADH
Answer: The answer is ACTH (adrenocorticotropic hormone)
Explanation:
ACTH, known as Adrenocorticotropic hormone, is a hormone secreted by the Anterior Pituitary gland. Its function is primarily to stimulates the adrenal cortex gland, which in turns release its own hormones generally known as Corticosteriods such as
- Aldosterone
- deoxycorticosterone etc
These corticosteriods exert several functions ranging from
- transport of electrolytes and distribution of water in tissues to
- metabolism of carbohydrates to release blood sugars; and break down of proteins and fats
Under stress, the anterior pituitary gland releases ACTH, which promotes the adrenal cortex to produce cortisol to maintain bodily functions, including retaining sodium and water, increasing blood sugar, and initiating fat breakdown.
Explanation:When the body is under stress, the anterior pituitary gland reacts by releasing ACTH (Adrenocorticotropic hormone). This hormone then stimulates the adrenal cortex which in turn produces cortisol. The release of cortisol into the bloodstream results in several reactions such as retaining sodium and water to increase blood volume, increasing blood sugar for energy, and the breaking down of fats for additional sources of energy. Therefore, the correct answer is C) ACTH.
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When ejecting a protein sample into a well of the SDS-PAGE gel with a pipetter, why should you only push down on the pushbutton to the first stop and not the second stop?Select one:
a. Pushing down to the second stop will eject more sample than required.
b. The pushbutton should only be pushed down to the second stop when loading the sample into the pipetter.c. Pushing down to the second stop when loading the samples will not affectthe SDS-PAGE results.d. Pushing down to the second stop could denature the protein sample.e. Pushing down to the second stop will result in ejection of air from the pipet tip, which could blow the content intended for that particular well to the surrounding wells
Answer:
.e. Pushing down to the second stop will result in ejection of air from the pipette tip, which could blow the content intended for that particular well to the surrounding wells
Explanation:
The pipette has two stops the first and the second stop. The first stop is the one that changes in other words the amount that is set. The second stop is for extra push to remove all the liquid from the tip. When the pipette is pushed to the second stop some air is ejected.
This is the main reason when loading samples on wells during SDS-PAGE ,we stop at the first stop or else the contents will go to the neighboring wells.
A heat-killed, phosphorescent (light-emitting) strain of bacteria is mixed with a living, non-phosphorescent strain. Further observations of the mixture show that some of the living cells are now phosphorescent. Which of the following observations would provide the best evidence that the ability to phosphoresce is a heritable trait?
Answer:
B. DNA passed from the heat-killed strain to the living strain.
Explanation:
This is similar to Grifitth's transformation experiment which uses 2 different strains of Streptococcus pneumoniae.
Pieces of DNA from the heat-killed strain must have been taken up by the living strain and this pieces became incorporated into the genome of the living strain. Th DNA pieces contained the phosphoresce gene and this gene was passed on to the living strain in the process of taking up the DNA.
The correct option is B.
A student is observing a cell under the microscope. It is observed to have supercoiled DNA with histones. Which of the following would also be observed by the student?
a.A single circular chromosome
b.A nucleus
c.Free-floating nuclear material
d.No organelles
Answer:
The correct answer is b. A nucleus
Explanation:
Prokaryotes are the organism that does not have a nucleus and their genetic material is present in the cytoplasm. Prokaryotes do not have histone protein. Eukaryotes are the organisms who have membrane-bound nucleus which contains DNA(genetic material) which can found supercoiled around histone protein.
So as histone protein is found in the eukaryotic nucleus which helps in the supercoiling of DNA, therefore, the student who is observing the cell which has supercoiled DNA with histone is eukaryotic cell and the organelle is the nucleus. Therefore the correct answer is b. nucleus.
Some of the chemicals that make up living organism have only few structures (like DNA) other have many different structures (like proteins). Why might this be the case?
Answer:Proteins have many different functions, while DNA has limited functions.
Explanation:
Chemicals that make up the living organisms consists of more protein than DNA. This is because proteins have different functions while the function of DNA are limited. Excluding water and fat, the human body is made up almost entirely of protein. Most of the structural framework of the body such as bones, muscle, organs, skin and nails are made up of protein.
Protein are responsible to do most of the work in cells by acting as enzymes, hormones and carriers. Proteins are largely required for the structure, function, and regulation of the body's tissues and organs.
Why do individuals who suffer from brainstem injuries have difficulty with their autonomic functions?
Answer: The brainstem controls autonomic functions through cranial nerves arising from it, along with the medulla oblongata, therefore an injury to the area can cause loss of these functions.
Explanation:
The brainstem comprises of the midbrain, and the pons and medulla of the hindbrain. It is in direct continuation with the spinal cord. Ten cranial nerves arise from this part, and a number of tracts pass through this region. The medulla of the brainstem are particularly involved in maintaining heart rate, breathing and blood pressure. In case of a brainstem injury, the cranial nerves responsible for autonomic functions could get damaged leading to dysfunction and/or the areas of medulla consisting of centres controlling heart rate, blood pressure or other autonomic functions, if damaged could cause difficulties.
Answer:
Because the brainstem regulates autonomic functions through cranial nerves arising from it.
Explanation:
The brainstem is the posterior part of the brain that is continuous with the spinal cord. The brainstem has three parts which includes; midbrain, pons and medulla oblongata.
The midbrain play major part in wakefulness and regulation of homeostasis. It is associated with vision, hearing, motor control, sleep and wake cycles, alertness, and temperature regulation.
The pons is situated between the medulla oblongata and the midbrain. It transmit signal information via nerves to the cerebrum through the medulla and to the cerebellum. It also sends sensory signals to the thalamus via sensory nerves. The pons performs both motor and sensory functions. Sensory functions like hearing, equilibrium, taste, and facial sensations such as touch and pain. Motor roles like eye movement, facial expressions, chewing, swallowing, urination, and the secretion of saliva and tears.
The medulla oblongata controls autonomic functions and connects the spinal cord to higher levels of the brain. It also regulate several basic functions of the autonomic nervous system like breathing, heart and blood vessels, digestion, vomiting, coughing, sneezing, and swallowing.
Individuals that suffers Injury to the brainstem, particularly to the medulla oblongata, will have difficulty with these autonomic functions.
The study of life in the universe is a multidisciplinary field of scientific research, involving scientists with training in many different areas. This scientific discipline has been given the name ____. (The American Physical Society has created a division with this name and NASA has given it the same name as the one adopted by your textbook.)
(a) exobiology (b) bioastronomy (c) astrobiology (d) biochemistry (e) biophysics
c
Answer: The answer is astrobiology
Explanation:
Astrobiology is a new scientific discipline employing scientists in the field of biochemistry, biophysics, chemistry, geology and many others to study the existence of life activities anywhere in the universe, including the Earth and other planets.
Similar to other vertebrate animals, humans possess retroviruses that exist in two forms: as normal genetic elements in their chromosomal DNA (endogenous retroviruses) and as horizontally-transmitted infectious RNA-containing viruses which are transmitted from human-to-human (exogenous retroviruses, e.g. HIV and human T cell leukemia virus, HTLV). Endogenous retroviruses in animals and humans probably evolved from transposable elements, some of them gaining the ability to package themselves in a virion structure, leave the cell and infect another cell.
Answer:
True
Explanation:
Retrovirus is a virus containing an RNA as a genetic material which infects the host cell by integrating its RNA into the genome of the host cell.
The retrovirus uses their reverse transcriptase enzyme to synthesise DNA from the RNA and then insert the DNA into a host DNA.
The studies on the genome of the human have shown that about 5-8% of the human genome is comprised of these retrovirus called endogenous retroviruses which act as transposable elements and some retroviruses use humans as the host to replicate and vector to transfer from human to a human called exogenous retroviruses.
Thus, true is the correct answer.