Answer:
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Explanation:ASjkbxzjb
SCamzhb
what best describes technology's use in science?
Technology, since the 19th century, has very much helped the development of Scientific theories. It has helped Scientists use telescopes to look into germs or zoom into the stars, it has also helped Chemists determine the weight of, say, a mashed up apple with soda.
If all 15 presynaptic neurons stimulate the post-synaptic neuron at the same time, will an action potential be produced?
Answer:
Yes
Explanation:
If all 15 presynaptic neurons stimulate the post-synaptic neuron at the same time, the action potential will be produced. In a chemical synapse, an action potential in the presynaptic neuron is responsible for the release of a chemical messenger known as neurotransmitter. After this, the neurotransmitter diffuses across the synapse and binds to the postsynaptic cell so we can say that potential can be produced.
Albinism, lack of pigmentation in humans, results from an autosomal recessive gene. Two parents with normal pigmentation have an albino child. What is the probability that their next three children will be wild type
Answer:
50
Explanation:
because this disorder occurs when the alleles are recessive meaning the alleles Will be heterozygous
In one study, 108 regions of the genome were identified to be significantly associated with schizophrenia. Below is a Manhattan plot showing the results from this study. Assuming that none of the causative variants are linked and there is a single causative variant per peak region, how many regions of chromosome 6 with causative variants were identified in this study
Answer: hello your question is incomplete attached below is the missing part of the complete question
The top of the highest peak corresponds to a region of the genome called the MHC locus. This means that of all the variants identified by this study, ____________.
the variant in MHC is observed the most frequently in the affected cohort.
the variant in MHC can cause schizophrenia in a Mendelian fashion.
the variant in MHC is the least likely to be a false positive result.
the variant in MHC has the largest odds ratio.
answer:
i) 4
ii) the variant in MHC is the least likely to be a false positive result. ( C )
Explanation:
The number of regions of chromosome 6 with causative variants identified is 4
While the variant in MHC is the least likely to be a false positive result. this is because, it is the highest peak in of all the causative variants identified in this study (i.e. closely related to the disease) and been the highest peak will make it least likely to return a false positive result.
The membrane potential that occurs when neurotransmitters bind to their receptors is called _______.
Answer:
action potential
Explanation:
Transmission of a signal within a neuron (from dendrite to axon terminal) is carried by a brief reversal of the resting membrane potential called an action potential. When neurotransmitter molecules bind to receptors located on a neuron's dendrites, ion channels open.
Answer:
I guess action potential is the correct one
Absolute zero temperature is not the zero energy temperature.explain
The absolute zero is the temperature at which molecular motion ceases i.e translational kinetic of molecules become zero, but the molecules possess the potential energy, according to the Kinetic theory of gases. Thus, the absolute zero is not the zero energy temperature.
Outermost layer of virion fullfills which of the following functions maturation biosynthesis release attachment
urine is made of nutrients and water
Answer:
what's the question over here?!
what is defferenciated cell and and undeferenciated cell.with example
Answer:
Differentiated cells refer to the specialized cells in multicellular organisms modified to carry out a particular function, such as transporting a certain substance or executing a specific task, while undifferentiated cells refer to the stem cells capable of giving rise to indefinitely more cells of the same type,.
example of differentiated cell=zygote to embryo then a comlex foetusexample of undifferentiated cell= embryonic stem cell.hope it helpsstay safe healthy and happy.Further studies of the genes in the region around the MHC locus identified C4 as a likely causative gene. A particular allele of C4 was significantly associated with schizophrenia. In addition, it was found that individuals with high expression of this allele of C4 had elevated risk of developing schizophrenia. Which of the following variants would be most likely to increase risk of schizophrenia from this allele?
a. a variant that decreases binding of an activating transcription factor to the C4 promoter
b. a 5’ UTR variant that destabilizes the C4 mRNA
c. a duplication of the C4 gene
d. a nonsense coding variant
Answer:
c. a duplication of the C4 gene
Explanation:
Schizophrenia is a disease that affects the brain, which is characterized by an imbalance of neurotransmitters and the presence of damaged brain tissues. In humans, the Complement component 4 (C4) protein is found at the neurons (i.e., cell bodies, dendrites and axons), which are the most common brain cells, and neuronal synapses. Moreover, gene duplication can increase gene expression when repressive mechanisms (e.g., DNA methylation at promoter sites) capable of silencing the expression of the duplicated gene are lacking. Recently, it has been discovered that a chromosome rearrangement associated with a tandem duplication of a genomic region that contains a C4 gene may contribute to schizophrenia risk by increasing the expression of this gene. In this example, a, b and d options are associated with a decrease in expression of the C4 gene, either by decreasing its transcription (options a and d) or by posttranscriptional mechanisms (option c).
You are a well-trained Cytogenetic Technologist. You work as a Genetic Counselor at a medical institution. You are meeting a nervous couple in your office, who recently found out that they are going to be first-time parents. They have come to you to discuss the chances of having a child with birth defect or genetic disorder. To ease their nerves, you greet them with a smile in your white coat, guide them to their seats, and slowly go through the options of prenatal tests available for them.
KARTOTYPE TEST METHODS
First you introduce to the parents the importance of karyotype test, by describing its domy to detect two broad categories of abnormalities in chromosomes, (1)_____and (2)_____. You provide a few examples of genetic disorders that can be caused by chromosomal abnormalities, such as (3)____and (4)____. Some parents would like to know what is the main cause of these genetic disorders. Without going into too much detail, you explain to them that chromosomal abnormalities are usually caused by (5)____in meiosis I and melosis II, when the homologous chromosomes or sister chromatids fail to separate during meiotic cell division
After the parents have decided to conduct a fetal karyotype test, you collect a small sample of chorion cells through chorionic villi sampling (CVS). Chorion is a membrane around the embryo that eventually fuses with the amnion. The chorion cells are grown in culture because chromosomes are most visible at (6)___stage of cell division. You use a technique called banding to create distinguishable landmarks on the chromosomes. There are several types of staining techniques you can use for karyotyping, some advanced techniques like FISH (fluorescent in situ hybridization) uses fluorescent probes to add colors onto different DNA sequences on the chromosome. You choose the most common type of staining techniques called G- banding to label the fetus chromosomes with Glemsa dye.
After taking photos of the chromosomes from a microscope, you sort the (7)____pairs of autosomes in the following order, from (8)____to (9)____, and keep the sex chromosomes last. The sorted photographed image of chromosomes is called (10)____. Now, you are ready to examine the karyotype of the fetus.
Answer:
1 monosomy
2 trisomy
3 Down syndrome
4 Turner syndrome
5 non-disjunction
6 metaphase
7 twenty-two
8 largest (chromosome 1)
9 smallest (chromosome 22)
10 karyotype
Explanation:
Non-disjunction during meiosis I or II occur when homologous chromosomes or sister chromatids refuse to separate. This makes the resulting gametes to have too many or too few chromosome numbers in their genomes.
A gamete with too few chromosome number that participates in fertilization with a normal gamete will result in a zygote with abnormally less chromosome number. This situation is referred to as monosomy. Whereas, the fusion of a gamete with too many chromosome number with a normal gamete will result in a zygote with too many chromosome, a situation known as trisomy.
The manifestation of monosomy and trisomy in humans results in Turner and Down syndromes respectively.
Humans have 23 pairs of chromosomes out of which a pair determines maleness or femaleness - the sex chromosome. Other chromosomes are referred to as autosomes. Hence, there are 22 autosome pairs and 1 pair of sex chromosome.
Chromosomes are studied by karyotyping, a process that involves photographing chromosomes at the metaphase stage of the cell cycle, cutting them out, and then arranging them in decreasing order of size.
Ariella is a 12-year-old whose BMI-for-age is above the 95th percentile. What conclusion could be drawn? a. She is at risk for undernutrition b. Her body fat percentage is in an unhealthy range. c. She is going through a growth spurt. d. She is at risk for future obesity.
Answer:
She is at risk for future obesity
Explanation:
The correct conclusion could be drawn is she is at risk for future obesity. BMI is a body mass index that is derived from the mass and height of the person.
How BMI is calculated?The formula is BMI = kg/m2
Where kg is a person’s weight in kilograms and m² their height in meters squared.
BMI categoriesUnderweight = <18.5Normal weight = 18.5–24.9Overweight = 25–29.9Obesity = BMI of 30 or greaterThus, it is concluded that Ariella is at a risk for future obesity according to her BMI percentile.
Learn more about BMI here:
https://brainly.com/question/2064513
#SPJ2
which is more vulnerable to disturbances, a simple food web with only a few species or a more complex one
Answer:
few species
Explanation:
in a complex one im not sure as to how the question measures complexity but a complex one may have more options and more things to adapt to
What organisms help break down dead leaves in an ecosystem
Answer:
Decomposers are vital organisms that grow by breaking down dead and decaying matter. Some of these are scavengers - macro-organisms that feed on dead or decaying matter, e.g. flies, cockroaches, earthworms. Others are decomposers, generally microscopic bacteria and fungi, that break down wastes.
Explanation:
Select all of the characteristic monosaccharide features that are absent in this molecule: H O H CO о -C-H H-C -O CH H-C- -O H H-C-H H
Answer:
Carboxyl functional group.
Explanation:
Carboxyl functional group is the feature or characteristic which us absent in the monosaccharide. The carboxyl group is an organic, functional group comprising of a carbon atom which is double-bonded to an oxygen atom and have singly bond with a hydroxyl group. In other words we can say that it is a carbonyl group (C=O) that has a hydroxyl group (O-H) which is attached to the carbon atom.
The energy source for active transport is ________ , while the force driving facilitated diffusion is ________.
Answer:
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Explanation:
In most unsaturated fatty acids found in animal fats, vegetable oils, and biological membranes, the cis isomer predominates. true false
Answer:
The correct answer is - true.
Explanation:
Most unsaturated fats are oils and normally are biological membranes or plant-based such as vegetables. These fats mostly have cis isomer in them. Unsaturated fats help to lower blood cholesterol levels.
However, few oils show saturated fats but these are few in number, the majority are made up of unsaturated fats that have double bonds.
Even after taking into account all of the variants discovered by this study, there are still likely to be other variants that you have not discovered. The affected and unaffected cohorts are roughly the same size and are drawn from a similar population with similar ancestry. All of the following modifications to the study are likely to allow identification of new variants associated with the disease, EXCEPT ____________.
Question Completion with Options:
a. increasing the size of your cohorts.
b. increasing the diversity of your cohorts.
c. increasing the size of your affected cohort.
d. increasing the diversity of your affected cohort.
Answer:
All of the following modifications to the study are likely to allow identification of new variants associated with the disease, EXCEPT
a. increasing the size of your cohorts.
Explanation:
Option A is chosen because increasing the size of the cohorts will be equally divided between the affected and unaffected cohorts since the study discovered that these two classes are roughly the same size. Therefore, new variants of the disease may not likely be identified and isolated unless the other modification options are followed instead of Option A.
What are physical and chemical properties
4: Åcil fast bacteria are
a. Neisseria
b. Staphylococci
e !vlycobacteria
d. All of the above
。
Answer:
c. Mycobacteria
Explanation:
The mycobacteria is an acid fast bacteria. So, option (c) is the correct answer.
_____________ is a post-translational modification that occurs in the endoplasmic reticulum or Golgi apparatus to make proteins functional.
Answer:
Lipidation is a post-translational modification that occurs in the endoplasmic reticulum or Golgi apparatus to make proteins functional.
we should conserve environment give reason
Answer:
for healthy living and long life
Can the use of badger control (including culling) be justified in the
prevention of the spread of bovine tuberculosis (
Answer:
no. badger vaccination is better
Explanation:
spreading mostly happens cattle to cattle
"killing badgers will not solve the problem. Badgers are not the primary cause of the spread of bTB in cattle: the primary route of infection is cattle-to-cattle contact[2]. The Government's badger cull is flying in the face of science. It should be putting more resources into speeding up the development of an effective cattle vaccine, amongst other measures"
wildlife life trusts
Aspirin may still irritate the stomach as a side effect. What functional group in the molecule is responsible for this effect
Answer:
The functional group that irritates the stomach is known as the salicylic acid molecule.
Explanation:
what is the function of bile
Answer:
Bile breaks down fats into fatty acid during digestion.
Answer:
Bile helps with digestion. It breaks down fat into fatty acids. Which can be taken into rhe body by the digestive tract. Bile contains:mostly cholesterol
In 1985 a biologist counted 750 pine trees in a 250 hectare forest. Using similar counting techniques, the biologist counted 1,250 pine trees in 1990 and 1,500 pines in 1995.
What was the average change of the size of the population from 1985 to 1995?
What was the density of pine trees each year that they were counted?
What was the average change of density from 1985 to 1995?
Answer:
Suppose that we have a given function f(x)
The average rate of change of the function between two values x₁ and x₂ is given by:
[tex]r = \frac{f(x_2) - f(x_1)}{x_2 - x_1}[/tex]
a) We want to find the average (rate) of change on the size of population from 1985 to 1995.
We have that:
f(1985) = 750
f(1995) = 1500
Then we have:
[tex]r = \frac{1500 - 750}{1995 - 1985} = 750/10 = 75[/tex]
This means that the population of trees increases, in average, at a rate of 75 trees per year.
b) What is the density of trees each year that they were counted?
This will be equal to the quotient between the number of trees and the area.
1985: number of trees = 750 pines
area = 250 ha
Then the density is:
D(1985) = (750 pines)/(250 ha) = 3 pines/ha
So 1985, there were 3 pines per hectare.
1990: number of trees = 1250 pines
area = 250 ha
Then the density is:
D(1990) = (1250 pines)/(250 ha) = 5 pines/ha
1995: number of trees = 1500 pines
area = 250 ha
The density is:
D(1995) = (1500 pines)/(250 ha) = 6 pines/ha
3) now we want to get the average change between 1985 and 1995 in the density, this will be:
[tex]r = \frac{D(1995) - D(1885)}{1995 - 1985} = \frac{6 pines/ha - 3pines/ha}{10} = 0.3 pines/ha[/tex]
So, on average, each year the number of pines per hectare increases by 0.3
According to a famous article by Max Kleiber, the scaling of the metabolic rate or energy consumption for mammals Pmetab (measured in kcal/day) with the body mass Mb (measured in kilograms) is Pmetab=70M0.75b.
a. What is the specific metabolic rate, i.e. the metabolic rate per kilogram of body mass for the following animals?a mouse whose mass is 30g. Give your answer to three significant figures.
b. What is the specific metabolic rate, i.e. the metabolic rate per kilogram of body mass for the following animals?a cow whose mass is 300kg. Give your answer to three significant figures.
Answer:
Explanation:
Given that:
The scaling of the specific metabolic rate is:
[tex]=\dfrac{P_{metab}}{m}[/tex]
where;
[tex]P_{metab} = 70M_b^{0.75}[/tex]
[tex]=\dfrac{ 70M_b^{0.75}}{m}[/tex]
(a) For a mouse whose mass = 30 g = 0.030 kg
The specific metabolic rate [tex]=\dfrac{ 70*(0.030)^{0.75}}{(0.030)}[/tex]
= 168.197 kg
≅ 168 kg to 3 significant fig.
(a) For a cow whose mass = 300 kg
The specific metabolic rate [tex]=\dfrac{ 70*(300)^{0.75}}{(300)}[/tex]
= 16.8197 kg
≅ 16.8 kg to 3 significant fig.
Explain how the results from testing Barley High Lysine (BHL) are related to the primary, secondary and tertiary structure of protein?
Answer:
Enhanced derivatives of barley chymotrypsin inhibitor-2 are known as Barley high lysine (BHL) proteins. BHL has tryptophan, threonine, and isoleucine in a folded structure similar as the tertiary structure of protein.
Derivatives of BHL are digestible in gastric juice and other fluid. Denaturation of these proteins shows proteins in gastric fluid minus pepsin which is looks as the secondary structure of protein. It has four soluble protein groups –
a) Albumins
b) globulin fraction in embryo and scutellary proteins
c) Prolamins
d) Glutelin
What must happen before meiosis can begin ?
(URGENT) Which of these statements is correct about the ocean waves? (100 points)
A) Wave 1 and Wave 2 have the same wavelength.
B) Wave 2 and Wave 3 have the same wavelength.
C) Wave 1 and Wave 2 have the same amplitude.
D) Wave 2 and Wave 3 have the same amplitude
Answer:
D
Explanation:
Wave 2 and Wave 3 have the same amplitude.
