Answer:
h² = 0.6
Explanation:
Before answering the question, we need to know a few concepts.
Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.
Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.
The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,
SD = X₀ - Xs
Heritability in the strict sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.
h² = R/SD
The response to selection (R) refers to the metric value gained from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.
R = h²SD
R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),
R = X₀ - X₁
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Now that we know these concepts and how to calculate them, we can solve the proposed problem.
Available data:
You are selecting rice´s decreased period of maturation. The population of rice has a mean maturation time of 30 days → X₀ Parental selected average period to maturation is 25 days → Xs F1 plants mature on average in 27 days → X₁ N arrow sense Heritability → h²According to what we sow previously, we need to find out the value of h².
We know that h² = R/SD, so we need to get R and SD first.
R = X₁ - X₀
R = 27 - 30
R = -3
SD = Xs - X₀
SD = 25 - 30
SD = -5
Knowing this, we can calculate h²
h² = R/SD
h² = -3/-5
h² = 0.6
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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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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Human being get energy from
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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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:
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.
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.
yinto uxinzelelo lwengqondo?
Answer:
ni se la respuesta estoy respondiendo sólo para ganar puntos
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.
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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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.
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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.
-------------------------------------------------------------------------------------------------------------
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 haemopoiesis??
Haemopoiesis is from greek meaning “ to make. new blood” •
Explanation:
It refers to the formation of blood cellular. components.
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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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
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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 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.
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
Cellular respiration produces
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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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?
Organisrns that transfer diseases to hurnans are
O hosts
O pathogens
O parasites
O vectors
Lectins often bind their ligands via multiple weak interactions. bind their ligands with relatively low specificity. prevent viruses from binding to their target cells. are carbohydrates that bind to receptor proteins.
Answer:
The correct answer is - B.often bind their ligands via multiple weak interactions.
Explanation:
Lectins are specific types of proteins that identify and bind to specific carbohydrates present on the cell surfaces. They have an essential role in interactions and communication between various cells for identification and recognition.
Binding sites of lectins on the surface of one cell bind to the Carbohydrates on the surface of another cell. A lectin usually has two or more binding sites for carbohydrate units.
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.
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i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
how much water was retained by soil C
Answer:
we dont know sorry but i dont know
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
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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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.
