Answer:
A) A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation.
D) The gene encoding miR-5 in a region of heterochromatin.
B) D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
Explanation:
MicroRNAs (miRNAs) are small regulatory RNA sequences (approximately 20-24 nucleotides in length) that regulate gene expression by RNA interference (RNAi) mechanisms. These sequences (miRNAs) bind by base complementarity to messenger RNAs and thus inhibit protein translation and/or trigger mRNA degradation. In this case, miR-5 binds to the 3’UTR of the mRNA transcript of the UCD gene, thereby inhibiting/slowing protein UCD synthesis. The UCD protein is an allosteric regulator that binds and activates the expression of the p101 protein, thereby the miR-5 RNAi pathway also indirectly decreases the expression of the p101 gene. Moreover, hypoacetylation is an epigenetic mark generally associated with gene silencing (heterochromatin is a transcriptionally inactive state of chromatin).
Digestion is primarily controlled by the _____.
What is a community?
1 all the animals that live in a habitat
2 a single species that lives in a habitat
3 all the species that live in a habitat
4 a population that lives in a single habitat
Answer:
3. All the species that live in a habitat.
A community is where all the species live in a habitat. Hence the correct option is 1.
A community is an ecological term that encompasses all the different species of organisms that coexist and interact within a specific habitat or geographic area. It includes plants, animals, fungi, and microorganisms that share the same environment and form intricate ecological relationships with each other.
These relationships can be competitive, predatory, symbiotic, or other forms of interactions that influence the dynamics and structure of the community. Understanding the composition and interactions within a community is vital in studying the biodiversity, ecosystem functioning, and overall health of a given habitat.
Hence the correct option is 1.
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Organisrns that transfer diseases to hurnans are
O hosts
O pathogens
O parasites
O vectors
The nitrogen cycle is the using and reusing of nitrogen in an ecosystem. True or false?
Answer:
True
Explanation:
Nitrogen is a fundamental component of both inorganic and organic compounds, where it is the main constituent of biomolecules such as nucleic acids (DNA, RNA) and proteins. The nitrogen cycle refers to the biogeochemical processes by which nitrogen circulates between the components of an ecosystem, i.e., between organisms (like plants and decomposers), and non-living things (i.e., soil, water, air). This cycle consists of several processes which include, among others, nitrogen fixation (i.e., the process by which nitrogen in the atmosphere is converted into ammonia), nitrification (i.e., the oxidation of ammonia is oxidized into nitrite and subsequent transformation of nitrites into nitrates), denitrification (where nitrate is reduced), anaerobic ammonia oxidation and putrefaction.
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I will mark the Brillianest
Answer:
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a special effects artist mixes two different liquids in a bowl. both liquids are white. when she heats the bowl, a new compound forms. will the new compound be a white liquid?
Answer:
if a special effects artist mixes two different liquids in a bowl and they both are too white !
after heats those liquids
its forms curd like substance
its either will be white, its can increase its colour as darkish white or lightly yellowishblack
when she heats the bowl, a new compound forms and it will be white before heating and after heating it may be white or other dark color.
What are the properties of a compound ?A compound composed up of two or more elements which are chemically combined with a fixed proportion by their mass.
For example, Water, Sodium Chloride, Ammonium Chloride etc.
They have fixed melting and boiling points.
Formation of compound is a change in chemical reaction and the components of compound are mixed in a fixed proportion.
Two major types of compounds are covalent and ionic compounds.
Covalent compounds refers to formation of covalent bond among two nonmetals, like water or methane. These molecules are neutral and weak.
When metal react with non metal it form Ionic compounds, are held together by opposite charges, so the bond is stronger than covalent compound.
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DNA is a nucleic acid involved in heredity, or the passing down of genetic traits from one generation to the next. DNA consists of four different types of nucleotide monomers.Which part of the nucleotides' structure is responsible for the incredible variation that exists amongst all types of organisms
Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).
The specific arrangement of these four bases within the DNA of each organism gives that organism its unique traits; here are the arrangements:
-Adenine is paired with Thymine (think of A for apple and T for tree)
-Cytosine is paired with Guanine (think of C for car and G for garage)
search "DNA base pairs" and go to images for better understanding
The part of the nucleotides' structure is responsible for the incredible variation that exists among all types of organisms are Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).
Who is responsible for passing down of genetic traits from one generation to the next?DNA is a nucleic acid involved in heredity, or the passing down of genetic traits from one generation to the next. DNA consists of four different types of nucleotide monomers.
The specific arrangement of these four bases within the DNA of each organism gives that organism its unique traits and here are the arrangements are mentioned below:
Adenine is paired with Thymine (think of A for apple and T for tree)Cytosine is paired with Guanine (think of C for car and G for garage)Therefore, The part of the nucleotides' structure is responsible for the incredible variation that exists among all types of organisms are Nitrogenous base DNA consists of four unique nucleotides that each contain one unique nitrogenous base—adenine (A), thymine (T), cytosine (C), or guanine (G).
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give a reason why lack of roughage in diet often leads to constipation
Answer:
The main cause of constipation is intake of a low fiber diet.the bulk and soft texture of fibre helps prevent hard dry stools that are difficult to pass thereby reducing constipation.
I hope this helps
how much water was retained by soil C
Answer:
we dont know sorry but i dont know
What is the chain of custody?
A) the order of who gets custody of the crime scene
B) the paper trail of every person that has been in contact with the evidence throughout its seizure, collection, preservation, analysis and
disposition
C) the order of suspects
D) the people who have custody of children until the case is closed
You are studying an enzyme that is inactivated by phosphorylation and create a mutant in which the threonine that is normally phosphorylated is replaced with glutamate. Predict the impact of this change on the activity of this enzyme. Group of answer choices
Answer:
always active
Explanation:
Phosphorylation is a posttranslational modification that consists of the addition of phosphate groups to specific amino acids on the protein. Phosphorylation acts as a molecular switch for proteins that are phosphorylated (i.e., in some situations phosphorylation acts to activate protein function, whereas in other situations phosphorylation can inactivate protein function). Phosphorylation modifies the three-dimensional structure of the protein, thereby affecting, for example, the accessibility of the active site of a phosphorylated enzyme to its substrate. Phosphorylation can occur only at the side chains of three amino acids: Serine, Threonine and Tyrosine. In this case, the enzyme is inactivated by phosphorylation on the Threonine residue, so it is expected that the mutant enzyme cannot be phosphorylated, remaining in an active state.
H. pylori cannot grow in other microenvironments of the human body because the conditions are unsuitable for its growth, but other species require different conditions.
a. True
b. False
Answer:
it should be an true statement
Our body needs both vitamin and mineral in a small quantity ,still they are important why?
Answer:
Vitamins and minerals are considered essential nutrients—because acting in concert, they perform hundreds of roles in the body. They help shore up bones, heal wounds, and bolster your immune system. They also convert food into energy, and repair cellular damage.
The body regulates the amount of hormones are released by using feedback loops. A __ feedback loop increases the response whereas a __ feedback loop decreases the response.
Positive feedback loop increases the response whereas a negative feedback loop decreases the response.
What is positive feedback?Positive feedback is the amplification of a body's response to a stimulus. For example, in childbirth, when the head of the fetus pushes up against the cervix (1) it stimulates a nerve impulse from the cervix to the brain (2).
A feedback mechanism resulting in the inhibition or the slowing down of a process.
Examples of processes that utilise negative feedback loops include homeostatic systems, such as thermoregulation (if body temperature changes, mechanisms are induced to restore normal levels), blood sugar regulation (insulin lowers blood glucose when levels are high ,glucagon raises blood glucose when levels are low).
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How does pollution affect biodiversity
Answer:
All forms of pollution pose a serious threat to biodiversity, but in particular nutrient loading, primarily of nitrogen and phosphorus, which is a major and increasing cause of biodiversity loss and ecosystem dysfunction. ... In addition, nitrogen compounds can lead to eutrophication of ecosystems.
Growers often wrap potted plants in plastic sleeves prior to shipping. If plants remain in these sleeves too long, their leaves curl and fall off. Imagine you are working at a garden center and have been asked to make a presentation to a local garden club. Unit 5 DQ Question: Explain the processes that may cause the leaves to drop and fall.
Answer:
Shock, age and climate.
Explanation:
The processes that may cause the leaves to drop and fall are shock, age and climate. The plants drop their leaves due to shock when they experience by transplanting from place to another. When the leaves get older having yellow colour so due to no function it falls. The climatic condition is also responsible for the dropping of leaves. Those plants which can't grow in cool season so when the winter comes they shed their leaves and wait for the summer season.
State whether the following statements are true or False. Hormones in plants travel by the vascular bundle.
Answer:
True
Explanation:
In plants, hormones travel large throughout the body via the vascular tissue (xylem and phloem) and cell-to-cell via plasmodesmata. In contrast, many animal hormones are produced only in specific glands. Plants do not have specialized hormone-producing glands.
Which of the following traits are characteristic of all mammals?
Mammals are air-breathing, warm-blooded, and have a backbone, yet these characteristics do not distinguish them from the rest of the animal kingdom. Mammals have the ability to regulate their body temperature using their metabolism and sweat glands in a way that no other animal can.
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QUESTION 11 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the allele's frequencies experimentally derived, calculate the frequency of the H4/H5 genotype that would be expected if the class were a population in Hardy-Weinberg equilibrium. 1. 0.28 2. 0.51 3. 0.19 4. 0.72 5. 0.14 6. 0.24 7. 0.41 QUESTION 12 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the genotype frequencies derived assuming that the class were a population in Hardy-Weinberg equilibrium, calculate the number of H4/H4 individuals that would be expected in the class (rounded numbers). 1. 19 2. 57 3. 72 4. 147 5. 171 6. 120 7. 96 QUESTION 13 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and comparing the observed and the expected number of individuals for the three genotypes, calculate the value of the Chi-square statistic 1. 2.69 2. 0.05 3. 28.67 4. 14.59 5. 0.50 6. 22.31 7. 3.84 QUESTION 14 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium, which is the (correct) null hypothesis tested by Chi-square? 1. The whole class represents a population that is in Hardy-Weinberg equilibrium 2. The whole class represents a population that may not be in Hardy-Weinberg equilibrium 3. The whole class represents a population that is not in Hardy-Weinberg equilibrium 4. The whole class represents a population that may be in Hardy-Weinberg equilibrium QUESTION 15 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis? 1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom, I conclude that P>0.05. Hence, I fail to reject the null hypothesis. 3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis.
According to Hardy-Weinberg, when a population is in equilibrium, it will have the same allelic frequencies generation after generation, meaning that they are stable, they are not evolving.
When any evolutive force is acting on the population, this equilibrium breacks, and allelic and genotypic frequencies change through generations, differing from the expected ones.
A) Option 7 is the correct answer ⇒ 0.41
B) Option 6 is the correct answer ⇒ 120
C) Option 7 is the correct answer ⇒ 3.84
D) Option 1 is the correct answer ⇒ The class represents a population that is in H-W equilibrium
E) Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.
-------------------------------------------
Allelic frequencies in a locus are represented as p and q, referring to the
allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (H0m0zyg0us dominant), 2pq (H3ter0zygous), q² (H0m0zyg0us recessive). Populations in H-W equilibrium will get the same
allelic frequencies generation after generation.
The sum of the allelic frequencies equals 1, this is p + q = 1.
In the same way, the sum of genotypic frequencies equals 1, this is
p² + 2pq + q² = 1
Being
p the dominant allelic frequency,
q the recessive allelic frequency,
p² the h0m0zyg0us dominant genotypic frequency
q² the h0m0zyg0us recessive genotypic frequency
2pq the h3ter0zyg0us genotypic frequency
Situation: Through PCR, we have determined the PER3 genotypes for a class of students as follows:
H4/H4 = 125 individuals;
H4/H5 = 85 individuals;
H5/H5=24 individuals.
⇒ Total number of individuals= 125 + 85 + 24 = 234
⇒ Genotypic frequencies, F(xx):
F(H4/H4) = 125/234 =0.534
F(H4/H5) = 85/234 = 0.363
F(H5/H5) = 24/234 = 0.102
⇒ Allelic frequencies, f(x):
f(H4) = p = F(H4/H4) + 1/2 F(H4/H5) = 0.534 + 0.363/2 = 0.534 + 0.182 = 0.716
f(H5) = q = F(H5/H5) + 1/2 F(H4/H5) = 0.102 + 0.363/2 = 0.102 + 0.182 = 0.284
Questions:
A) According to the theoreticall frame, we know that 2pq is the h3ter0zygous genotypic frequency. So,
F(H4/H5) = 2pq = 2 x 0.716 x 0.284 = 0.408 ≅ 0.41 ⇒ Option 7 is the correct answer.
-------------------------------------------------------------------------------------------------------------
B) According to the theoreticall frame, we know that p² is the h0m0zyg0us genotypic frequency. So,
p = 0.716
p² = 0.5126 ≅ 0.513 ⇒ This is the genotypic frequency.
To calculate the number of individuals carrying this genotype, we need to multiply it by the total number of
individuals.
H4/H4 individuals = p² x total number of individuals = 0.513 x 234 = 120
Option 6 is the correct answer.
-----------------------------------------------------------------------------------------------------------
C) Up to here we know that 2pq = 0.41 and p² = 0.513
Now we need to calculate q ²
q = 0.284, then q² = 0.284² = 0.08
These are the expected frequencies if the population was in H-W equilibrium.
The expected number of individuals with each genotype are:
H4/H4 = 0.513 x 234 = 120 individuals
H4/H5 = 0.41 x 234 = 96 individuals
H5/H5= 0.08 x 234 = 18 individuals
The observed number of individuals with each genotype are:
H4/H4 = 125 individuals
H4/H5 = 85 individuals
H5/H5=24 individuals
X² = ∑ (Observed - Expected)²/Expected)
X² = ((125-120)²/120) + ((85 - 96)²/96) + ((24-18)²/18)
X² = 0.21 + 1.26 + 2 =
X² = 3.47
The clossest option is option 7 = 3.84. The difference might be related to decimals and rounding.
-------------------------------------------------------------------------------------------------------------
D) The correct answer is 1 ⇒ The whole class represents a population that is in Hardy-Weinberg equilibrium
The null hypothesis always predict that populations are in H-W equilibrium.
-----------------------------------------------------------------------------------------------------------
E)
X² = 3.47
Freedom degrees = n - 1 = 3 - 1 = 2
Table p value: 7.82
Significance level, 5% = 0.05
Table value/Critical value = 5.991
5.991 > 0.347
Meaning that the difference between the observed individuals and the expected individuals is statistically significant. Not probably to have differe by random chances. There is enough evidence to reject the null
hypothesis.
Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.
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The most basic organization level of life is a ____________. A. membrane B. tissue C. cell D. organ
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The answer is...
C. Cell.
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Environmental scientists cite several challenges to protecting endangered species. Which of the following is NOT a difficulty faced by those working to protect an endangered species? Scientists who are focused on a single species may not recognize another species in peril. Extinction is natural in nature and its rate of occurrence has remained steady. Protection efforts may be slowed because of difficulty enforcing new regulations. Fundraising can be more difficult if the species is not perceived as valuable.
Answer:
Extinction is natural in nature and its rate of occurrence has remained steady.
Explanation:
As of right now we are going through what is known as the sixt extinction, and human acticity is one of the Main causes. (which has never happened before our time) Species are going extinct at a rapid rate, and considering how all life is connected it makes it easy to understand why the extinction rates can increase due to a biotic factor (niche) being "removed". (:
what is haemopoiesis??
Haemopoiesis is from greek meaning “ to make. new blood” •
Explanation:
It refers to the formation of blood cellular. components.
i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
The development of DNA technology is bringing profound changes to science, agriculture and healthcare. Provide one example of a DNA technology and provide at least one advantage and one example of a concern or problem associated with its use.
Answer:
The CRISPR-Cas9 genome editing system can be used to edit genes and correct mutations associated with inherited diseases. However, this technology also has the potential to edit genes in germline cells in order to irreversibly modify the human species and the natural evolution of life
Explanation:
The CRISPR-Cas9 (Clustered regularly interspaced short palindromic repeats and CRISPR-associated protein 9) system is a natural prokaryotic defense system used by bacteria to defend against invading DNA. In the laboratory, the CRISPR-Cas9 system has been repurposed to create a versatile genome-editing tool that allows us to modify the genome of mammalian cells in a targeted fashion. The CRISPR-Cas9 is a simple gene-editing tool that consists of a single guide RNA (sgRNA) that guides the Cas9 enzyme to the exact genomic location where Cas9 needs to make a cut, which is then repaired by different DNA repair mechanisms. During DNA repair, nucleotides can be replaced and/or deleted, thereby producing desired genomic modifications. The CRISPR-Cas9 has an enormous potential to repair mutations in genes associated with inherited genetic disorders and cancer (i.e., oncogenes might be reversed in vivo by using this technology). However, the CRISPR-Cas9 genome editing system is also a subject of concern due to its dual use. For example, this technology can be used to modify the genome of germline cells by inducing mutations that can be passed across generations, thereby irreversibly modifying human DNA and altering the normal course of evolution.
Carnivore that feeds on primary consumers
Question 3 Multiple Choice Worth 3 points
(01.01 LC)
Which of the following is an example of a decomposer?
What about the structure of DNA allows it to copy itself?
A. It is held together mostly by hydrogen bonds, which are easy to open for replication
B. It has enzymes built into the helix that make a copy when needed.
C. It is double stranded, so each side serves as a template for a new strand.
D. DNA is not copied, it is only passed down through each generation.
Answer:
C. It is double stranded, so each side serves as a template for a new strand
Explanation:
The structure of DNA is double stranded, so each side serves as a template for a new strand. The correct option is C.
What is DNA replication?Replication is the procedure by which the DNA of the genetic code is copied in cells.
Before dividing, a cell must duplicate (or replicate) its a whole genome so that each eventually results daughter cell has its own entire genome.
The DNA replicates itself multiple times during the process of replication. It is a biological polymerization that goes through the steps of commencement, elongation, and termination.
The reaction is catalyzed by enzymes. The main enzyme in the process of replication is DNA Polymerase.
DNA replication occurs in eukaryotic cells' interphase nuclei. At the S-stage (synthesizing) of the cell cycle, DNA replication occurs prior to mitosis.
Thus, the correct option is C.
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Which of the following is an example of an enzymatic cycle?
Answer:
Catabolism
Explanation:
The process of catabolism degrades the bacterial and fungal enzymes into simple inorganic molecules.
A substance, without being a reactant, which speeds up a chemical process is referred to as a catalyst. Enzymes are known as catalysts for biological reactions in living organisms. Although ribonucleic acid (RNA) molecules behave as enzymes, they are usual proteins. Enzymes.
Enzymes carry out the essential role of reducing the activated energy of a reaction — that is, the amount of energy needed to start the process. Enzymes work by attaching and retaining reactant molecules so that the chemical bonding and bonding activities are carried out more easily.
3. When recording medication administered to a client on the medical administration record, always add
next to each medication you deliver.
A. the client's weight
B. the client's initials
C. your initials
D. the client's resting heart rate
Answer:
option B
Explanation:
Answer:
your initials
are the ones that should be in the MAR. since your the one who gave the medication.
functions of insulin
Answer:
Insulin helps control blood glucose levels by signaling the liver and muscle and fat cells to take in glucose from the blood. Insulin therefore helps cells to take in glucose to be used for energy. If the body has sufficient energy, insulin signals the liver to take up glucose and store it as glycogen.
Explanation:
Human being get energy from
