For every 100ml of deoxygenated blood delivers approximately _____ml of CO2 to the alveoli.
Answer:
For every 100ml of deoxygenated blood delivers approximately __4___ml of CO2 to the alveoli.
Answer:
4ml
Explanation:
For every 100ml of deoxygenated blood delivers approximately 4 ml of CO2 to the alveoli.
Hope it is helpful....The meaning of ALARA in radiation?
Answer:
The guiding principle of radiation safety is “ALARA”. ALARA stands for “as low as reasonably achievable”.
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.
What component of Earth's atmosphere exists entirely as a result of photosynthesis?
oxygen pas
n mas
O water vapor
O nitrogen gas
O carbon dioxide gas
Answer:
Explanation:
Carbon dioxide
¿Qué afirmación(es) es(son) correcta(s)?:
a) El dióxido de azufre de la quema de carbón en centrales
eléctricas protege contra la lluvia ácida.
b) Los óxidos de nitrógeno de las emisiones de combustión
provocan el aumento del pH en el agua de lluvia.
c) Una solución ácida de dióxido de carbono en agua de lluvia se
llama lluvia ácida.
d) El dióxido de azufre junto con los óxidos de nitrógeno son las
principales causas de la lluvia ácida.
Explanation:
jwjejeueuru4ibwsbbefbrvbrbebdbfhhdhdhdbdbdfbd3 eebdddd
Digestion is primarily controlled by the _____.
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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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
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.
Learn more about goldfish
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what are the advantage of the presence of hydrogen in large scale in the sun ?
hydrogen is essential to nuclear fusion. hydrogen is the fuel the sun burns. we don't have to burn solar energy, so there's no smoke and pollution
Answer:
Hydrogen fusion reaction is the one which provides so much light and heat (in short, energy) which escapes into space and spread. If Hydrogen on sun ends, It will lose its warmth (and may be soon become a planet, but will surely be a dead star).
Explanation:
Answer from Gauth math
identify the following organism and state to which phylum it belongs
Answer:
it belongs to Coelenterata phylum.
Which of the following is true about oncogenes
Answer:
genes involved in the cell cycle following a mutation become oncogenes.
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
What variable should Anurag change in his experiment
Answer:
For geological carbon sequestration, the reaction of aqueous CO2 with silicate rock permits carbonate formation, achieving permanent carbon sequestration.A
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?
List 3 variable that Anurag should keep the same
Answer:
seawater ,upside-down funnel and seaweed
Explanation:
Why should body temperature not be allowed to fluctuate too much?
Answer:
Because that can destroy the helpful enzymes in the body, and therefore cause a lot of problems or maybe cause death
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:
Which BEST
describes the
amount of air in
a typical soil?
A. about 90%
B. hardly any
C. about 25%
Answer:
C
Explanation:
it's about 20%-30%, so C would appropriately about that range since it the best answer :)
hope it helps
In New York, the worst-hit state, a total of 350,121 cases have been confirmed and 28,232 deaths have been reported. There are now 29 states with tens of thousands of cases, and more than 50,000 in New York, New Jersey, Massachusetts, Illinois, California, Pennsylvania and Michigan
Answer:
Thats true
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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A large section of tropical rainforest is cleared to build roads for mining. Select the least likely outcome of this habitat fragmentation from the options below. a.) Species that require open grazing areas may not survive. b.) Species will quickly adapt and re-populate the area. c.) The environment on the habitat edges will support different plants and animals. d.) Subpopulations of species may emerge.
Answer:
B
Explanation:
B - Species, especially ones with small niches, that looses their niches will likely die out or move to another location. Roads Will, in general, also disrupt the ecosystem (or the remains of it) and its inhabitants, which might lead to migration once again. That would be my reasoning at least. (:
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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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.
some strains of __________________ produce proteins that kills certain insects such as lepidopterans , coleopterans and dipterans
Answer:
Bacillus thuringiensis
Explanation:
some strains of Bacillus thuringiensis produce proteins that kills certain insects such as lepidopterans , coleopterans and dipterans
Answer:
Bacillus thuringiensis
i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
Phân tích các quy luật hoạt động thần kinh cấp cao ở trẻ và vận dụng trong thiết lập thói quen học tập và kỉ luật ở học sinh tiểu học.
Answer:
very different than ducks do u want it is not the
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
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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Human being get energy from
