Answer:
50%
Explanation:
When you use a monohybrid cross you will get Bb, Bb,bb,bb
Since the brown eyed are heterozygous.
Environmental scientists cite several challenges to protecting endangered species. Which of the following is NOT a difficulty faced by those working to protect an endangered species? Scientists who are focused on a single species may not recognize another species in peril. Extinction is natural in nature and its rate of occurrence has remained steady. Protection efforts may be slowed because of difficulty enforcing new regulations. Fundraising can be more difficult if the species is not perceived as valuable.
Answer:
Extinction is natural in nature and its rate of occurrence has remained steady.
Explanation:
As of right now we are going through what is known as the sixt extinction, and human acticity is one of the Main causes. (which has never happened before our time) Species are going extinct at a rapid rate, and considering how all life is connected it makes it easy to understand why the extinction rates can increase due to a biotic factor (niche) being "removed". (:
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
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:
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.
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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Digestion is primarily controlled by the _____.
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.
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
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:
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
Which of the following is true about oncogenes
Answer:
genes involved in the cell cycle following a mutation become oncogenes.
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 test is used to ensure that the results you expect are the same as the results you are observing?
A. Epistatic test
B. Chi-squared test
C. Observational confirmation test
D. Punnett square test
Reset Selection
Answer:
b. Chi Squared
Explanation:
The chi squared test is best to find signifigance between differences in observed and expected values.
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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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.
Cystathioninuria can be caused by two different mutations in the enzyme cystathionase. Cystathioninuria caused by mutation 1 can be overcome by providing cells with increasing concentrations of the coenzyme pyridoxal phosphate. Which of the following statements describe the most likely changes in the binding affinities of the two mutant enzymes?
A. Both mutant enzymes have lost the ability to bind the substrates.
B. The enzyme with mutation 1 has lost the ability to bind to the substrates, whereas the enzyme with mutation 2 has lost the ability to bind to pyridoxal phosphate.
C. The enzyme with mutation 1 has decreased affinity for pyridoxal phosphate, whereas the enzyme with mutation 2 has lost the ability to bind to the substrates.
D. Both mutant enzymes have lost the ability to bind pyridoxal phosphate.
Answer:
C. The enzyme with mutation 1 has decreased affinity for pyridoxal phosphate, whereas the enzyme with mutation 2 has lost the ability to bind to the substrates.
Explanation:
A coenzyme is an organic cofactor that binds with an enzyme in order to initiate or aid the function of the enzyme. A coenzyme binds to the active site of the enzyme (where the reaction occurs), thereby triggering its activation by modifying protein structure during the reaction. Some examples of coenzymes include Coenzyme A and Adenosine triphosphate (ATP). Pyridoxal phosphate is a coenzyme (it is the active form of vitamin B6) that is required for the function of cystathionase. Moreover, cystathionase is an enzyme that enables cells the synthesis of cysteine from methionine (transsulfuration pathway). The binding of pyridoxal phosphate to the enzyme increases the binding affinity of the enzyme for the substrate, thereby influencing its activity. In this case, it is expected that mutation 1 reduces the binding affinity of the enzyme to the cofactor, and thereby the cofactor is required at a higher concentration to restore normal enzyme activity.
State whether the following statements are true or False. Hormones in plants travel by the vascular bundle.
Answer:
True
Explanation:
In plants, hormones travel large throughout the body via the vascular tissue (xylem and phloem) and cell-to-cell via plasmodesmata. In contrast, many animal hormones are produced only in specific glands. Plants do not have specialized hormone-producing glands.
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.
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?
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.
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
One Reason Why The Temperature needs To Be Kept Constant
Answer:
the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance
Explanation:
It is due to the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance hence the temperature stays constant. Hence the temperature remains constant as all the heat is used up and no external heat is released or absorbed
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.
In a hydrogen ion pump, the energy is used to join small molecules together
to
make larger ones. Which factor most likely has the greatest effect on the
number of molecules mitochondria can produce?
Answer: The number of H+ ions moving down the channel
Explanation:
Human being get energy from
Which of the following is an example of an enzymatic cycle?
Answer:
Catabolism
Explanation:
The process of catabolism degrades the bacterial and fungal enzymes into simple inorganic molecules.
A substance, without being a reactant, which speeds up a chemical process is referred to as a catalyst. Enzymes are known as catalysts for biological reactions in living organisms. Although ribonucleic acid (RNA) molecules behave as enzymes, they are usual proteins. Enzymes.
Enzymes carry out the essential role of reducing the activated energy of a reaction — that is, the amount of energy needed to start the process. Enzymes work by attaching and retaining reactant molecules so that the chemical bonding and bonding activities are carried out more easily.
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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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
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.
-----------------------------------------------------------------------------------------------------------
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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i need help in biology questions please G10?
Answer:
ok where is it
we can help only if there is something attached
