Answer:
Interphase - There are 3 subphases called - G1 , S and G2 . In these subphases, DNA material is duplicated, and essential proteins for cell division are formed.
Prophase - the nuclear envelop and nucleolus disappears and condensed DNA forms chromosomes that are short in size.
Metaphase - the form of two sister chromatids are arranged in the middle of the metaphase/equatorial line. Two centrioles are found at two poles of the cell which initiate the formation of spindle fibers.
Anaphase - spindle fibers are formed and can be observed. These fibres attach with the kinetochore present in the centromere of chromosomes and these chromatids now start separated by pulling and pushing force of spindle fibers in opposite poles
Telophase - chromatids reach at destination pole and then nuclear envelop starts creating again in both poles. Nucleolus reappears and now in a cell two nucleus are found.
A small group of mice are released on an island without mice but with abundant food for mice and no predators. After the population size stabilizes for several years, a hurricane drastically reduces it. We can now say that:________.
A) the biotic potential of the population has been reduced.
B) its new population size is a result of density-dependent regulation.
C) its new population size is a result of density-independent regulation.
D) it can now act as a sink metapopulation.
The correct answer is option C) The new mice population size is a result of density-independent regulation.
The carrying capacity might be affected by different factors, known as limiting factors, which might be a result of the population density (for example, competition) or might be density-independent. This last case refers to dense-independent factors, and among these, we can mention human impact or natural disasters (fires, volcanic eruption, flooding). Natural disaster causes damages in an ecosystem, reducing the available resources such as food or shelter, and consequently decreases the number of individuals. Natural disasters reduce the carrying capacity of the environment
In the exposed example, mice got to stabilize on the island. The population had enough food and no predators. But the occurrense of the huricane reduced drastically the population size. This is an example of a natural dissaster acting as a limiting dense-independent factor affecting the population size.
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which specialized senses, do not rely on hair-like appearing cell structures or cilia to transduce stimuli?
Answer: Touch.
Explanation:
Cilia are short, mobile prolongations which contain a central structure made up of microtubules and proteins, enveloped by the cytosol and the plasma membrane. They are involved in cell movement, transport of materials, displacement of fluids, among others. Cilia are important for many biological processes, such as the senses of taste, hearing, smell, sight and balance.
For example, each taste bud of the tongue is made up of a set of cells, among which are the taste cells that have cilia that come into contact with the substances dissolved in the mouth by saliva. As for the sense of hearing, when there is a sound, the endolymph in the cochlea moves and this stimulates the cilia of the internal sensitive cells, which communicate with the acoustic nerve that informs the brain of what the sound is like. In the sense of smell, the receptors are the olfactory cilia of the olfactory neurons, which are located in the mucosa of the upper portion of the nostril, above the level of the superior concha. In the sense of vision, cones and rods are the two types of photoreceptor cells that capture light energy and convert it into electrical signals. They are highly specialized cells and can be differentiated into several regions: an outer segment, an inner segment containing the nucleus and a synaptic terminal. The outer segments are modified cilia and consist of flattened membranous sacs or disks. As for the balance system, the vestibular system consists of the utricle and the saccule, which are chamber-shaped organs filled with endolymph. The maculae of the saccule are located in a vertical plane and effectively capture the accelerations of the upward and downward movements of the head, and therefore of the gravitational forces. The hair cells of the maculae are responsible for transforming the mechanical energy produced by movement into nerve signals. The activity of these cells is determined by their morphofunctional polarization or ciliary organization, which is different in the utricle and in the saccule.
Thus, the only sense that does not depend on cilia to transmit stimuli is touch. The skin contains nerve endings, as well as glands, blood vessels and hair follicles. These nerve endings detect pain, touch, pressure and temperature.
Compared with a eukaryotic cell, a prokaryotic cell Select one:______
a. lacks organelles beyond ribosomes.
b. is larger.
c. does not require energy.
d. is not living.
e. has no method of movement.
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.
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.
When the concentration of ADH increases, __________.
Explanation:
Less urine is produced this will help u
Cells control or regulate the flux through metabolic pathways by means of I. allosteric control of enzymes. II. covalent modification of enzymes. III. genetic control of the concentrations of enzymes. IV. genetic expression of allosteric regulators.
Answer:
I, II, III
Explanation:
Allosteric regulation is a common mechanism of regulation of enzyme activity, which generally involves key enzymes in metabolic pathways. Allosteric modulation occurs when a substance/molecule called 'allosteric regulator' binds to an enzyme at a site other than the enzyme's active site, thereby triggering a conformational change in the enzyme and thus affecting its activity. Moreover, covalent modifications such as phosphorylation, acetylation, methylation, etc, are also involved in the regulation of enzymatic activity in metabolic pathways. For example, phosphorylation is a type of reversible covalent modification of proteins consisting of the addition of phosphate groups at specific amino acid residues (i.e., serine, threonine, and tyrosine) by specific enzymes known as protein kinases. Phosphorylation alters the three-dimensional (3D) structure of proteins, thereby turning the substrate or cellular pathway to active and/or inactive. Finally, both transcriptional and post-transcriptional mechanisms are also able to control the expression of enzymes involved in signaling pathways. Transcriptional mechanisms are capable of limiting the amount of mRNA that is produced from a given gene (in this case, an enzyme encoding gene), whereas post-transcriptional mechanisms such as, for example, RNA interference pathways, control the translation of messenger RNA (mRNA) molecules into proteins.
A man bought a goldfish in a pet shop. Upon returning home, he put the goldfish in a bowl of recently boiled water that had been cooled quickly. A few minutes later the fish was found dead. Explain what happened to the fish
Answer:
lack of oxygen in the water
Explanation:
The fish most likely died from lack of oxygen in the water. This is because fishes actually use their gills to extract and breathe in the oxygen from the water while also expelling carbon dioxide from their lungs. Similar to how humans breathe. When the water was boiled it caused the dissolved gases to be expelled, which includes oxygen. Therefore, without the necessary oxygen in the water, the fish ultimately suffocated.
Boiling the water reduce its amount of dissolved oxygen which is needed by the fish to breathe, that's why the fish died after few minutes.
What is dissolved oxygen?Dissolved oxygen is the amount of oxygen present in the water.
The organisms live to consume dissolved oxygen to breathe.
The amount of dissolved oxygen is high in the current water like rivers than in the still water like pond.
If the amount of DO is high in the water, it causes bubble gas disease in the aquatic organisms.
If the amount of DO is low in the water than, fishes and other aquatic organism cant survive due to low oxygen level.
Thus, boiling the water reduce its amount of dissolved oxygen which is needed by the fish to breathe, that's why the fish died after few minutes.
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Classify the descriptions as pertaining to nucleosides, nucleotides, or both nucleosides and nucleotides.
a. Do not contain a phosphate group
b. The product when a base bonds to carbon 1 of ribose or deoxyribose
c. Contain a base and a monosaccharide monosaccharide, and a phosphate group
The correct description is as follows:
Do not contain a phosphate group - nucleosidesThe product when a base bonds to carbon 1 of ribose or deoxyribose - nucleosideContain a base and a monosaccharide and a phosphate group - nucleotidesWhat is a nucleoside and nucleotide?Nucleoside is a a compound consisting of a nitrogenous base (purine or pyrimidine) linked to a sugar e.g. adenosine.
Nucleotide is a compound consisting of a nucleoside linked to a phosphate group i.e. nucleoside + phosphate group = nucleotide
Therefore,
A substance that do not contain a phosphate group - nucleosidesThe product when a base bonds to carbon 1 of ribose or deoxyribose - nucleosideA substance that contain a base and a monosaccharide and a phosphate group - nucleotidesLearn more about nucleotides at: https://brainly.com/question/13185536
What else is produced during the replacement reaction of silver nitrate and potassium sulfate?
2AgNO3 + K2SO4 Ag2SO4 + ________
KNO3
2KNO3
K2
2AgNO3
Answer:
2KNO3
Explanation:
Pls mark it brainliest
hope it helps u
Answer:
The answer is B.) 2KNO3
Explanation:
Fertilization, Fruit and Seed Formation x3 3. a. name the two processes that lead to seed formation in flowering
Answer:
Pollination, the transfer of pollen from flower-to-flower in angiosperms or cone ... In angiosperms, the process of seed development begins with double ...
Missing: x3 | Must include: x3
Ecosystems rely on interdependence between species to keep balance. Which of the following is a threat to a stable
ecosystem?
A. Loss of biodiversity
B. High biodiversity
C. Low biodiversity
D. Increase in biodiversity
Answer:
loss of biodiversity
Explanation:
Biodiversity- refers to the variety of life on Earth at all its levels, from genes to ecosystems, and can encompass the evolutionary, ecological, and cultural processes that sustain life.
loss in biodiversity affect food chains greatly
thanks
hope it helps
i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
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
can anyone explain what is stroke volume?
No spam
Answer:
The definition of stroke volume is the volume of blood pumped out of the left ventricle of the heart during each systolic cardiac contraction.
Explanation:
The stroke volume is the volume of blood pumped from the left ventricle per beat. It is calculated by using the measurements of ventricle volume from echocardiogram and subtracting the end systolic volume from end diastolic volume.
Answer:
Stroke volume is the volume of blood pumped out of the left ventricle of the heart during each systolic cardiac contraction
Explanation:
The left ventricle is one of the four chambers in the heart that pumps blood full of oxygen to the body. Stroke volume is basically how much blood pumps out of it per beat.
Hope this helped
Which of the following is true about oncogenes
Answer:
genes involved in the cell cycle following a mutation become oncogenes.
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.
What do MSDS sheets do, in your own words
Answer: The MSDS lists the hazardous ingredients of a product, its physical and chemical characteristics (e.g. flammability, explosive properties), its effect on human health, the chemicals with which it can adversely react, handling precautions, the types of measures that can be used to control exposure
Explanation:
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?
Human being get energy from
During the performance of the simple staining procedure, you failed to heat fix your E-coli smear preparation. Upon microscopic examination , how would you expect this slide to differ from the correctly prepared slides ?
OMG I'm a business study Student so I really don't know about this stuff
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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The image below shows a club fungus. A club fungus has an erect, above-ground fruiting body and long, branching, underground hyphae.
How do the structures of the fruiting body and hyphae relate to their functions?
A.
The fruiting body is tall to allow for spore dispersal, while the hyphae are branched to increase the absorption of nutrients.
B.
The fruiting body is curved to help the uptake of decaying plant material, while the hyphae are long to increase gas exchange.
C.
The fruiting body is tall to maximize photosynthesis, while the hyphae are branched to support the release of waste products.
D.
The fruiting body is curved to improve absorption of water, while the hyphae are long to allow for reproduction.
Answer:
The fruiting body is curved to improve absorption of water, while the hyphae are long to allow for reproduction.May 12, 2020
Explanation:
If it helps you mark me as a brainleast
Answer:
Explanation:
d
which life cycle stage is found in plants but not animals
Answer:
Multicellular haploidOAmalOHopeO
Plants have multicellular haploid and multicellular diploid stages in their life cycle.
Gametes develop in the multicellular haploid gametophyte . Fertilization gives rise to a multicellular diploid sporophyte, which produces haploid spores via meiosis.What is multicellular haploid stage?The haploid multicellular stage produces specialized haploid cells by mitosis that fuse to form a diploid zygote.The zygote undergoes meiosis to produce haploid spores. Each spore gives rise to a multicellular haploid organism by mitosis.To know more about diploid stage here
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What will the percentage of cytosine be in this segment of DNA?
Answer:
15%
Together, adenine and thymine compose 70% of the segment. This means that 30% of the section is composed of guanine-cytosine pairs. Since these two bases will be equal in quantity, 15% of the DNA section will be cytosine bases.
What is the function of the mitochondria?
A. Stores the cell's DNA
B. Builds proteins
C. Produces energy for the cell by respiration
OD. Stores the cell's glucose
Reset Selection
Answer:
Produces energy for the cell by respiration
Explanation:
The glucose obtained from food is broken down to pyruvic acid in the cytoplasm. This pyruvic acid is broken down into oxygen, water and energy rich ATP molecules in the Mitochondria.
Which of these is a covalent bond in which the electrons are not evenly shared?
Answer:
polar covalent bond
Explanation:
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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What is a function of the integumentary system?
1 forms blood clots
2 grows new tissue
3 absorbs water
4 break down chemical bonds in food
Answer:
absorbs water is a function of the integumentary system. hope it help you.
A function of the integumentary system is to grow new tissues. Thus, the correct option B.
What is the Integumentary system?The integumentary system may be defined as a unique type of system that primarily consists of the skin, hair, nails, and exocrine glands. It is the set of organs that construct an outermost layer of an animal or human body.
The integumentary system plays a significant role in the growth of new tissues within the body. The cells of the skin generally hold the mechanism of renewal or replacement of injured or damaged cells which are present on the epidermis part of a specific organ.
The integumentary system also protects the body from infections, and injuries provide defense against pathogens, etc.
Therefore, a function of the integumentary system is to grow new tissues. Thus, the correct option B.
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Digestion is primarily controlled by the _____.
