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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what is the definition of wildlife?
Answer:
wild animals collectively; the native fauna (and sometimes flora) of a region.
Explanation:
I hope this helped :)
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.
what is haemopoiesis??
Haemopoiesis is from greek meaning “ to make. new blood” •
Explanation:
It refers to the formation of blood cellular. components.
Organisrns that transfer diseases to hurnans are
O hosts
O pathogens
O parasites
O vectors
Human being get energy from
i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
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.
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:
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
how much water was retained by soil C
Answer:
we dont know sorry but i dont know
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
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?
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:
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
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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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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....where in an embryo are the instructions located for how to build organs
Answer:In the nucleus
Explanation:
The information for all bodily functions resides in DNA in every cell.
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.
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
The mass of the Sun is 1.99 × 1030 kg. Jupiter is 7.79 × 108 km away from the Sun and has a mass of 1.90 × 1027 kg. The gravitational force between the Sun and Jupiter to three significant figures is × 1023 N.
Answer:
4.16
Explanation:
edge2021
Answer:
4.16
Explanation:
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.
The most basic organization level of life is a ____________. A. membrane B. tissue C. cell D. organ
[tex]Hello[/tex] [tex]There![/tex]
The answer is...
C. Cell.
Hopefully, this helps you!!
[tex]AnimeVines[/tex]
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.
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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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¿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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Question 5
Not yet answered
Marked out of 1.00
P Flag question
Complete the following sentence: "The interior of living cells is more
than the exterior because more
ions are expelled than ions are taken in by the sodium-potassium pump."
Select one:
O a. electropositive. Nak
O b. electronegative, Na.
O C. electronegative, Na, K
O d. electropositive, Na+, K+
Question 6
Not yet answered
Marked out of 1.00
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The high concentration of protons in the inner mitochondrial space relative to the mitochondrial matrix represent
O
В.
77°F
AGD (0)
10-25 PM
7/28/2021
Answer:
i really really need the brainly points
Explanation:
sry for this answer, i need the answers for myself
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