Answer:
I would assume you are talking about just basic general biology, in which case it's probably just looking for cellular respiration.
Explanation:
Please help me with these questions
I will mark the Brillianest
Answer:
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Eutrophication occurs when excess nutrients are supplied to a region, leading to an algae bloom and ultimately ______
A. Coral bleaching
B. Ocean deoxygenation
C. Ocean acidification
D. Overfishing
Answer:
The correct answer is - B. Ocean deoxygenation.
Explanation:
Eutrophication is the process in which a water body gets excessively rich in nutrients that leads to the algal growth or plankton growth in this region and covers the complete surface or most of the water body.
Due to this algal and plankton growth, there is a significant decrease in the concentration of the dissolved oxygen in water bodies that result in the incapability of supporting the lives found in it. The primary and main reason for this deoxygenation is eutrophication. Ocean deoxygenation is the reduction of the oxygen concentration of the oceans.
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
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.
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....¿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:
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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
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 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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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?
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
Some of the largest mountains in the world, including the Himalayas, occur where
Select one:
a.
two oceanic plates diverge.
b.
two continental plates converge.
c.
an oceanic and a continental plate diverge.
d.
an oceanic and a continental plate converge.
Answer:
b. two continental plates converse
Organisrns that transfer diseases to hurnans are
O hosts
O pathogens
O parasites
O vectors
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.
In a certain breed of dog, the alleles B and b determine black and brown coats respectively. However, the allele Q of a gene on a separate chromosome is epistatic to the B and b color alleles resulting in a gray coat (q has no effect on color). If animals of genotype B/b ; Q/q are intercrossed, what phenotypic ratio is expected in the progeny
Answer:
12 gray , 3 black, 1 brown
Explanation:
If Q allele of a gene on a separate chromosomes is epistatic to the B (black) and b (brown) color alleles, in cross between two animals with genotypes BbQq produces 12 gray coat color, 3 black coat color and 1 brown coat color animals.
BbQq x BbQq
Gray coat Gray coat
BQ Bq bQ bq
BQ BBQQ(gray) BBQq(gray) BbQQ(gray) BbQq(gray)
Bq BBQq(gray) BBqq(Black) BbQq(gray) Bbqq(Black)
bQ BbQQ(gray) BbQq(gray) bbQQ(gray) bbQq(gray)
bq BbQq(Gray) Bbqq(Black) bbQq(gray) bbqq(brown)
So the phenotypic ratio is Gray : Black : Brown
= 12 : 3 : 1
i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
what is haemopoiesis??
Haemopoiesis is from greek meaning “ to make. new blood” •
Explanation:
It refers to the formation of blood cellular. components.
how much water was retained by soil C
Answer:
we dont know sorry but i dont know
Human being get energy from
what is the definition of wildlife?
Answer:
wild animals collectively; the native fauna (and sometimes flora) of a region.
Explanation:
I hope this helped :)
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:
Amoeba sisters video recap Biomagnification
Biomagnification refers to the presence of higher concentration of chemical toxins as a result of the accumulation of toxins in organisms.
What is biomagnigication?Biomagnification is best explained as a condition in which the chemical concentration of a toxin is amplified in an organism compared to the environment in which the organism is found.
Biomagnification usually is observed as one goes higher in the trophic levels of organisms.
For example, chemical pollutant found in water may be present at tolerable levels. However, in organisms, living in the water, the concentration of the pollutant is higher as these organisms accumulate these toxins in their tissues.
Therefore, biomagnification refers to the accumulation of toxins in organisms higher than found in their environment.
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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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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:
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.
Hopefully, this helps you!!
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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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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.
The meaning of ALARA in radiation?
Answer:
The guiding principle of radiation safety is “ALARA”. ALARA stands for “as low as reasonably achievable”.
Occurs in mitochondria.
Select one:
a. Both Photosynthesis and Cellular Respiration
b. Neither Photosynthesis nor Cellular Respiration
c. Photosynthesis
d. Cellular Respiration
Answer:
d. cellular respiration
Explanation:
mitochondria is where cellular respiration occurs, and photosynthesis occurs in the chloroplast, so d would be the most reasonable answer.
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.
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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.
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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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