Answer:
Green Fluorescent Protein Labelled Bacteria
Explanation:
Green Fluorescent Protein labeling is very useful in studying prokaryotes. It is highly likely that the bacteria was also labelled. That's why after lysis and density level differentiation in a centrifuge the pellet start glowing green.
At the neuromuscular junction, an electrical signal from the motor neuron is translated to a chemical signal and then back to an electrical signal in the muscle cell. What proteins interact with the released chemical to cause the electrical signal in the muscle cell?
a.Sodium voltage-gated channels
b.Potassium ligand-gated channels
c.Cation ligand-gated channels
d.Calcium mechanical-gated channels
Answer:
Cation ligand-gated channels
Explanation:
At the neuromuscular junction, the arrival of nerve impulse triggers the release of chemical signals called neurotransmitters in the cleft. Binding of these chemical signals to the receptor present on the motor endplate results in the opening of a ligand-gated cation ion channel. These channels are present in the receptor. Opening of ion channels allow cations, specifically Na+, to flow across the membrane. The entry of cations makes the inside of the muscle fiber more positively charged and triggers a muscle action potential.
Cation ligand-gated channels on the muscle cells interact with the neurotransmitter (chemical signal) released from the motor neuron to generate an electrical signal in the muscle cell at the neuromuscular junction.
Explanation:The proteins that interact with the released chemical to cause the electrical signal in the muscle cell at the neuromuscular junction are the cation ligand-gated channels. Here's a brief step-by-step of what happens: An electrical signal or action potential travels down the motor neuron. When this signal reaches the end of the neuron, it causes the release of a neurotransmitter called acetylcholine. The acetylcholine diffuses across the neuromuscular junction and binds to the acetylcholine receptors - which are a type of cation ligand-gated channels - on the muscle cell. This binding causes these channels to open and allow positively charged ions to flow into the muscle cell, creating an electrical impulse in the muscle cell that ultimately leads to muscle contraction.
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in describing sound, what is a more common name for propagation velocity?
Answer: It is called Velocity factor.
Explanation:
Velocity factor or propagation velocity of a transmission medium is the ratio of speed of electromagnetic signal or radio signal move through the transmission medium to the speed of light in a vaccum. It means a ration of speeds is confined in sound cables or computer network
Suppose a species of tulip has three alleles for the gene that codes for flower color. The C R allele produces red tulips, the C p allele produces purple tulips, and the C w allele produces white tulips. C R is dominant over C p and C w , and C p is dominant over C w . For each cross, determine the expected phenotype ratio of offspring flower color.
Answer and Explanation:
Available data:
The Cr allele produces red tulipsThe Cp allele produces purple tulipsThe Cw allele produces white tulips. Cr is dominant over Cp and CwCp is dominant over CwCross 1
Parental) CrCp x CpCw
Gametes) Cr Cp Cp Cw
Punnet Square) Cr Cp
Cp CrCp CpCp
Cw CrCw CpCw
F1 phenotype ratio)
2/4=1/2 Red flowers, CrCp and CrCw
2/4=1/2 Purple flowers, CpCp and CpCw
Cross 2
Parental) CrCw x CpCw
Gametes) Cr Cw Cp Cw
Punnet Square) Cr Cw
Cp CrCp CpCw
Cw CrCw CwCw
F1 phenotype ratio)
2/4=1/2 Red flowers, CrCp and CrCw
1/4 Purple flowers, CpCw
1/4 White flowers, CwCw
Answer: The expected phenotype ratio of cross between CR and Cp is 1:1:2.
1 for CC,
1 for Cp, purple
2 for CR and Rp(red) all which are offsprings from the cross.
For cross between Cp and Cw, phenotype ratio is 1:1:2.
1 for CC,
1 for Cw, white,
2 for Cp and pw,purple.
Explanation:
Phenotype gives the observable characters of an individual.
Here, the character observed is flower color of tulip species.
The expected phenotype ratio of cross between CR and Cp is 1:1:2. The offsprings from the cross are CC, Cp,CR and Rp
1 for CC,
1 for Cp, purple
2 for CR and Rp red, since red is dominant over purple.
For cross between Cp and Cw, phenotype ratio is 1:1:2. The offsprings from the cross give CC, Cw, Cp and pw
1 for CC,
1 for Cw, white,
2 for Cp and pw,purple since purple is dominant over white.
The _______________ community has low diversity with microorganisms primarily in the Leptospirillum groups II and III and from the Ferroplasma types I and II.
Answer:
The Acidophilic Microbial Community has low diversity with microorganisms primarily in Leptospirillum groups II and III and from Ferroplasma types I and II.
Explanation:
An acidophilic microorganism or plant is one which grows best in acidic conditions.
They are also referred to as microorganisms which occur in acidic natural (solfataric fields, sulphuric pools) and man-made (eg. Acid mine drainage) environments.
Acidophilic Microbes otherwise known as Acidophiles are an ecologically and economically important group.
They possess networked cellular adaptations for regulating intracellular pH. Several extracellular enzymes from acidophilic microbes are known to be functional at much lower pH than that inside the cells.
Acid stable enzymes have applications in several industries such as starch, baking, fruit juice processing, animal feed and pharmaceuticals, and some of them have already been commercialized. Acidophiles are widely used in bio-leaching of metals from low grade ores
Recent studies show that acidophiles are currently being considered to be utilized in bio-conversion and bio-remediation, as well as in microbial fuel cells to generate electricity.
Acidophilic microbes of similar characteristics are classifed in groups for ease of study and identification.
Leptospirillum Group II and II as well as Ferroplasma types I and II are groups of acidophilic microorganisms within the Acidophillic Microbial community.
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A researcher decides to investigate whether there are any significant changes in personality from early adulthood through late adulthood. The researcher begins with a sample of 21-year-olds and plans to follow this group until they are 80 years old. The type of research design is:
Answer:
Longitudinal research design
Explanation:
Longitudinal research design is an observational design where data is collected for the same subjects over a long period of time which can range from months to years.
Therefore, in our case the researcher begins with a sample of 21 year olds and is planning to follow this group until they are 80 years in order to investigate whether there are any significant changes in personality from early adulthood through late adulthood.
This makes his research design to be Longitudinal research.
A particular tissue was discovered to have cells that lie in small chambers, separated by a matrix that is solid yet flexible, and fibers that could be stretched while still being able to return to their original length. Name this tissue.
Answer: Fibrous Tissue
Explanation:
A simple strong or repair tissue which consists of the twisted strands made of collagen and laid down by cells called as fibroblast.
Fibrous tissues are abundant in the body and are found almost everywhere. This tissue has the ability to repair after any scar or wound.
These tissue also gets repaired once it is damaged even in the case of some disease. This is a type of connective tissue.
Mechanical energy is the energy of _____.
Answer:
Mechanical energy is the energy of potential and kinetic energy i.e sum of potential and kinetic energy.
Explanation:
Mechanical energy is the energy posses by an object which enable it to work due to it's potion or movement.
Mechanical energy can be inform of potential energy( energy stored or due to object position) or kinetic energy( object in motion).
Mechanical energy is the sum of potential energy and kinetic energy. This energy is associated with object's position or motion.
What is one advantage of sexual reproduction over asexual reproduction?
a decrease in genetic variation
an increase in genetic mutation
a continuation of genetic weaknesses
an increase in genetic variation
Answer:
an increase in genetic variation
Explanation:
1. _______________________ is a process of gaining and testing knowledge & must be proven through experimentation. 2.___________________or non-science claims to be scientific but does not use experimentation. Cryptozoology is an example of this.
Answer:
SciencepseudoscienceExplanation:
Science is the study of nature and behavior of everything around us. It is the knowledge on the physical or material world obtained through observation and experimentation.
Pseudoscience incorporates convictions, speculations, or practices that have been or are viewed as logical, however have no premise in scientific fact. This could mean they were invalidated experimentally, can't be tried logically, or need proof to help them. The term commonly has a negative implication. At the point when utilized, somebody's intimating the theme did not depend on logical discoveries and is, in this manner, ailing in truth.
Which of the following is a true statement about antibiotic resistant bacte
they attack the immune system
they can spread through the air
they are not affected by antibiotics
they only attack human, not animals
Answer: They are not affected by antibiotics.
Explanation:
Antibiotics resistant bacteria are bacteria that are resistant or not affected by antibiotics rather they develop adaptations to antibiotics by living and surviving in the presence of antibiotics.
They resist to antibiotics by changing in a way that reduces the effectiveness of antibiotics. Examples include penicillin resistant enterococcus, methicillin resistant staphylococcus aureus e.t.c.
Answer:
they are not affected by antibiotics
Explanation:
Resistance Bacteria prevent antibiotics from affecting them. They build immune against the activity of the antibiotics in there system by preventing its function.
This can be done by pushing the antibiotic out of the system reducing its concentration. They also change membrane system and preventing permeability of the antibiotic in the membrane. Some bacterial also destroy the active ingredient in the antibiotics thereby preventing the activities.
Hence antibiotic resistance bacteria are not affected by the antibiotic by inhibiting there activities.
Meselson and Stahl designed an experiment that would allow them to discern whether DNA replication occurs in a dispersive, semiconservative, or conservative manner. They started with E. coli that had been growing for many generations in medium containing 15N. They then transferred the bacteria into medium containing only 14N, and allowed the bacteria to undergo two rounds of DNA replication. After each round of replication, the scientists performed density-gradient centrifugation of the DNA. The scientists reasoned that each of the three models would predict different DNA banding patterns after the two rounds of replication. Can you identify the banding patterns predicted by each model after the first round of replication
After the first round of replication in the Meselson-Stahl experiment, conservative replication predicts two bands -- one heavy and one light. Semiconservative replication predicts one intermediate band, as does dispersive replication.
Explanation:The Meselson-Stahl experiment was designed to test the three theories of DNA replication: dispersive, conservative, and semiconservative. After the first round of replication in 14N medium:
Conservative replication would predict two bands -- one at the heavier 15N density and one at the lighter 14N density. This represents the original parent 15N DNA and the new 14N DNA separately. Semiconservative replication would predict one band at an intermediate density. This represents the hybrids of one strand of 15N-parent DNA and one new 14N strand. Dispersive replication would also predict one band at an intermediate density, though the model suggests that each individual DNA strand would be a mixture of old and new DNA, unlike the semiconservative model. Learn more about Meselson-Stahl experiment here:
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You are an epidemiologist in charge of county health services for low income, HIV positive individuals in NJ. Which data would be most important for planning the long-term health care needs of this community, incidence or prevalence?
Answer:
The correct answer is - prevalence.
Explanation:
Prevalence is the number of individidual cases that are alive with the disorder at a particular time frame or particular time. It provides the measure more precise.
The given question is comes under Prevalence as the factor of prevalence of HIV cases was based, on average, on number of infected people with HIV in New Jersey by ethnicity, race and gender for the most recent 1 year time frame.
Thus, the correct answer is - prevalence.
Prevalence is the most important data for planning the long-term health care needs of a low-income, HIV positive community.
Explanation:Prevalence and incidence are two important measures in epidemiology. Prevalence refers to the total number of both new and existing cases in a population over time, while incidence measures the number of new cases of a disease during a specific time period. When planning the long-term health care needs of a low-income, HIV positive community, prevalence would be the most important data to consider. This is because prevalence gives an indication of the overall burden of the disease in the community and helps in estimating the resources and services needed to address the health care needs of the population over time.
which of the following is not associated with translation
ribosome
dna polymerase
anti codons
mrna
Answer:
The DNA polymerase
Explanation:
In translation, the mRNA produced after transcription, tRNA that bring the brings the amino acids, ribosome; the organize involved in translation and all other factors eg initial factors...are involved in the process of translation. DNA polymerase is involved in the process of replication of the DNA.
Answer:anti codons
Explanation:
Chromatin structure can increase or decrease transcription of a gene according to the:
A. combination of amino acid modifications in the histone tails.
B. coding sequences in the messenger RNAs for histone proteins.
C. combination of histone proteins found within the nucleosome.
D. proportion of arginine and lysine amino acids in the histone proteins.
Answer:
D. proportion of arginine and lysine amino acids in the histone proteins.
Explanation:
Epigenetic modifications are one of the main reasons of methylation and de-methylation, acylation and de-acylation of histone proteins that trigger the transcriptional process. All of these mentioned modification causes the change in the ration of lysine and arginine residues and these are the main reasons that affect the structure of chromatin as required.
Plant cells form a cleavage furrow or indentation of membrane between new daughter cells.
Animal cells lack centrioles and no spindle forms during cell division.
The cell plate is the final partitioning of plant cells.
Plant cells resort to binary fission.
There is no difference.
Plant cells and animal cells undergo the same cellular processes during mitosis.
Explanation:
Plant cells and animal cells undergo the same cellular processes during mitosis.The centrioles absent in the plant.Animals cells have centrioles, to what delivers the shaft strands is one of the differences among plant and creature mitosis. The periods of mitosis and goes on in each stage is the equivalent aside from when it comes to how cytokinesis is practiced. Plant cells separate the cytoplasm between the two new cores by framing a cell plate (new cell wall)between the cores. Animals cells achieve this by shaping a cleavage wrinkle which is a squeezing in at the center of the cell until two new cells have been framed with cytoplasm and organelles for each new nuclei.Imagine a cell whose membrane is permeable to water, but impermeable to solute X. If solute X is more concentrated outside the cell than inside, which way will water move by osmosis?
A. Water will move equally in both directions, in and out of the cell.
B. Water will enter the cell.
C. Water will not move in either direction, in or out of the cell.
D. Water will leave the cell.
Answer: D
Explanation:osmosis is the movement of molecules from a region of lower concentration to a region of higher concentration through a semi permeable membrane.
Final answer:
Water will move from inside the cell to the outside by osmosis because the concentration of solute X is higher outside the cell, making the outside a region of lower water concentration.
Explanation:
If solute X is more concentrated outside the cell than inside, and the cell's membrane is permeable to water but impermeable to solute X, water will move by osmosis from inside the cell to outside. This is because osmosis is the diffusion of water across a membrane from an area of higher water concentration (or lower solute concentration) to an area of lower water concentration (or higher solute concentration). Therefore, the correct answer is D. Water will leave the cell.
Suppose that the resistance between the walls of a biological cell is 7.0 × 109 Ω. (a) What is the current when the potential difference between the walls is 80 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.85 s?
Answer:
Explanation:
Using Ohm's law
V ( voltage) = I (current A) × Resistance R in ohms
R = 7.0 × 10⁹Ω
V = 80 mV = 80 / 1000 = 0.08 V
0.08 V = I × 7.0 × 10⁹Ω
a) I = 0.08 V / 7.0 × 10⁹Ω = 1.142857 × 10 ⁻¹¹ A
b) quantity of charge = I × t = 1.142857 × 10 ⁻¹¹ A × 0.85 s = 9.7142857 × 10⁻¹² C
number of Na⁺ ions ( q = +e) = 9.7142857 × 10⁻¹² C / 1.6 × 10⁻¹⁹ C = 60714285.714 Na⁺ ions
"A parakeet with green plumage is crossed to a white parakeet. The progeny are all green. Crossing these progeny together gave the following offspring: 3 white, 8 blue, 29 green, and 9 yellow. Propose a genetic hypothesis to explain the results. -g
Answer: This is due to the theory of independent assortment of gene.
This means that gene separate and assort with each other, independently. There a gene that is recessive in the parents may not be recessive in the offspring.
A recessive gene is the gene that makes up the physical appearance of the offspring. While a dominant gene is the gene that exist in the offspring, but is not among the gene that makes up the physical appearance of the offspring.
For the two pure stock parent plants, their have yellow and blue as the dominant gene, while green and white are the recessive gene.
That's is why the yellow and blue gene has a tendency of occurrence in future generations.
For their first progeny which appears to be all green, is because the green gene has sorted themselves independently to be recessive, while others become dominant.
For the second progeny, the recessive gene from the parent plants which was carried by the offspring as dominant gene, now became recessive in some plants.
The genetic pattern observed in parakeets suggests the presence of two gene pairs determining plumage color, influenced by dominance and epistasis. A Punnett square could be used to predict offspring ratios, reflecting complex inheritance patterns.
The question describes a genetic phenomenon observed in parakeets when crossed, which showcases inheritance patterns akin to those discovered by Gregor Mendel. A green parakeet was crossed with a white parakeet, and all progeny were green. When these progeny were crossed, various colors were observed, suggesting a relationship between different gene pairs influencing plumage color. Using Mendelian genetics principles, we can hypothesize that two gene pairs are interacting, where one gene pair determines the presence of color (G for green dominant over white g), while the other pair determines the color type (Y for yellow and B for blue, both dominant over white).
To align with the observed progeny, we might consider that the green color is dominant and that the white is recessive, masked by the presence of the green trait. We possibly analyze this using a Punnett square to predict the offspring ratios. Additionally, the appearance of blue and yellow parakeets indicates a pattern where other genetic factors or allele interactions impact the final phenotype. If we consider epistasis, where one gene affects the expression of another, we may infer that the yellow gene could be epistatic to the white color, and the green could be showing a typical Mendelian 3:1 ratio when the y allele is not present. This hypothesis aligns with the observation that crossing the progeny results in multiple plumage colors, adhering to complex inheritance patterns beyond simple dominance.
10
Select the correct text.
Which of the following is the correct definition of biomass?
the total amount of organic matter from plants and animals in a habitat
all materials that form from the remains of living organisms
hydrocarbons and nitrogen oxides that are released when materials are burned
Answer: the total amount of
Organic matter from plants and animals in a habitat.
Explanation:
Answer:
the total amount of organic matter from plants and animals in a habitat
Explanation:
One advantage to the Palo Verde Nuclear Power station to the ASU West solar array is that the nuclear power station can generate electricity at night, but the solar array does not generate electricity when the Sun is down.
a. True
b. False
The statement 'nuclear power station can generate electricity at night, but the solar array does not generate electricity when the Sun is down' is TRUE.
A nuclear power plant is a station in which the energy from nuclear fission is used to create heat in a nuclear reactor, which is then used to generate electricity.Conversely, a solar station is a large-scale photovoltaic system capable of generating electricity by using solar panels.Solar panels cannot produce energy at night because photovoltaic cells need the light energy from the Sun to generate electricity.In conclusion, the statement 'nuclear power station can generate electricity at night, but the solar array does not generate electricity when the Sun is down' is TRUE.
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Do we expect a son/daughter to have a trait if the trait is inherited in the following way (for each, please answer Y/N/P/W for yes, no, possibly, or ‘what!? – that doesn’t make sense!!’). Please fill out the grid for each type of inheritance. For example, a X-linked dominant female-limited trait would look like this:
Answer: incomplete
Explanation:
In the "arms race" between plants and herbivores, herbivores (insects) can overcome the toxic effects of plant specialized metabolites through natural selection. Which of the following would be an adaptation that would promote insect herbivory?
Plants express a novel gene that encodes for an inhibitor of larval development.
Insects express a gene coding for a new enzyme that degrades the plant toxin.
Plants develop a mutation that promotes the development of hairs and spines.
Insects evolve new taste receptors that trigger an avoidance reaction to the toxin.
The following that be an adaptation that would promote insect herbivory is :
D) Insects evolve new taste receptors that trigger an avoidance reaction to the toxin.
"Adaptation of Insects"The following that be an adaptation that would promote insect herbivory is the Insects evolve new taste receptors that trigger an avoidance reaction to the toxin.
Plants are regularly fed upon herbivorous creepy crawlies. to get security against such assaults plant has created a few instruments like generation of poisons and unstable substance that either slaughter the bother or pull in the normal adversaries of the creepy crawlies respectively.
Thus, the correct answer is D.
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Final answer:
An adaptation promoting insect herbivory in the plant-herbivore arms race is insects evolving an enzyme that degrades plant toxins, allowing them to counteract plant defenses and feed on previously toxic plants.
Explanation:
In the "arms race" between plants and herbivores, an adaptation that would promote insect herbivory is insects expressing a gene coding for a new enzyme that degrades the plant toxin. This represents a direct adaptation by the herbivores to overcome the plant's defenses, enabling them to feed on plants that would otherwise be toxic to them. The coevolution observed in nature often results in such adaptations, where insects evolve mechanisms to detoxify or evade plant defenses, such as specialized enzymes that can neutralize toxic compounds. This form of adaptation allows insects to exploit a wider range of host plants, giving them a selective advantage in habitats where their preferred food sources might be scarce or heavily defended.
The question below refers to the following information: 1. reception 2. transmission 3. transduction 4. perception 5. amplification For the events above, which of the following is the correct sequence as it relates to being able to hear sound
Answer:
Reception, Amplification, transmission, transduction, perception.
Explanation:
In the ear, the external ear collects sound pressure waves (reception) and funnels them toward the tympanic membrane, The vibrations (amplification) from the eardrum/tympanic membrane set the ossicles into motion. The middle ear transmits sound from the outer ear to the inner ear (transmission). The middle ear consists of three bones: the hammer (malleus), the anvil (incus) and the stirrup (stapes), the oval window, the round window and the Eustrachian tube. Sound vibrations are then transduced into electrical energy/impulses by hair cells in the inner ear. These electrical impulses are finally perceived by the brain as sound.
If the map distance between genes J and K is 20 map units and the map distance between genes K and L is 35 map units, what is the map distance between genes J and L?
The map distance between genes J and L can be calculated by adding the map distance between genes J and K (20 map units) and the map distance between genes K and L (35 map units), resulting in a total map distance of 55 map units.
Explanation:The map distance between genes is determined by their recombination frequency, which corresponds to the likelihood of their alleles being inherited separately. Given the map distance between genes J and K is 20 map units and the map distance between genes K and L is 35 map units, we add these two distances together to calculate the total map distance between genes J and L. Hence, the map distance between genes J and L is 20 + 35, which equals 55 map units.
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Given that the mean cuteness of the current herd is 39.7 OMGs, the BAF workers picked the 16 cutest adults and allow them to reproduce with each other. The mean cuteness of these chosen individuals is 41.5 OMGs. What is the predicted mean cuteness of the next generation of alpacas in the BAF environment?
a. 39.9
b. 40.4
c. 41.5
d. 42.5
Answer: 41.5 OMGs is the predicted mean cuteness in the next generation.
Explanation:
Cuteness is a way to measure relative ability of individuals with a certain genotype to reproduce successfully.
Mean cuteness is given as the summation of individual cuteness. Mean cuteness also changes in the next generation.
The mean cuteness of the current herd was given as 39.7 OMGs and after 16 individuals were chosen, the mean was given as 41.5.
This implies that there was a change already as mean cuteness will either increase or decrease in the next generation after selection. Here, 39.7 increased to 41.5 .
The predicted mean cuteness is therefore 41.5 OMGs.
An inbred strain of plants has a mean height of 74cm. A second strain of the same species from a different geographical region also has a mean height of 66cm. When plants from the two strain are crossed together, the F1 plants are a uniform height of 70cm. However, the F2 generation shows a wide range of heights: the majority are like the P1 and F1 plants, but approximately 1/1024 are only 50cm tall, and about 1/1024 that are 90cm tall. A.) What mode of inheritance is occuring here? B.) How many gene pairs are involved? C.) How much does each gene contribute to plant height? D.) Indicate one possible set of genotypes for the original P1 parents and the F1 plants that could account for these results? PLEASE EXPLAIN your reasoning. E.) Indicate three possible genotypes that could account for F2 plants that are 58cm high and 3 that account for F2 plants that are 74cm high? PLEASE EXPLAIN you reasoning.
Answer:
Explanation: see attachment
The plant height variation in the F2 generation indicates polygenic inheritance, with four gene pairs involved. Each gene pair contributes 2.5cm to the plant's height. The P1 parents could have been AABBCCDD and aabbccdd, and F2 genotypes causing specific heights would have different combinations of alleles.
Explanation:The scenario described suggests a polygenic inheritance pattern, where multiple genes contribute to the phenotypic expression of height in these plants. Since the F2 generation exhibits a wide range of heights and specifically shows extreme phenotypes (50cm and 90cm tall plants) at a frequency of 1/1024, we can infer that several gene pairs are involved.
To determine the number of gene pairs involved, we must consider that each gene pair segregates independently and follows Mendel's laws of inheritance. With a 1/1024 occurrence for the extreme phenotypes, this points towards a 4th power of 1/4 (since 1/4 is the probability of a homozygous recessive phenotype from a heterozygote), indicating that four gene pairs are involved. Each gene pair contributes equally to the overall height phenotype, so to find the contribution of each gene, we examine the difference in height between the extreme phenotypes (90cm - 50cm = 40cm) and divide it by the number of gene pair combinations that result in the extreme phenotypes (2^4, where 2 is for dominant or recessive allele, and 4 is the number of gene pairs), therefore each gene contributes 2.5cm to the plant height.
The P1 parental genotypes could have been AABBCCDD (for the 74cm strain) and aabbccdd (for the 66cm strain), resulting in all F1 offspring having the genotype AaBbCcDd (70cm tall, intermediate height). The F2 generation would exhibit a wide range of phenotypic variation due to different combinations of alleles.
For F2 plants that are 58cm in height, with each gene contributing 2.5cm, these individuals are 16cm shorter than the F1 intermediate height, suggesting they have 6 recessive alleles. Therefore, three possible genotypes could be Aabbccdd, aaBbccdd, or aabbCcdd. For F2 plants that are 74cm high, the genotype must be the same as the original 74cm P1 plant, or include combinations that also result in the same height due to the same number of dominant alleles; possible genotypes include AABBccdd, AABbCcDd, or AaBBCcDd.
Which of the following forms of cellular transport is being illustrated below?
Wy-laigh molecule
outside
CELL MEMBRANE
Bit formas
Inside
membrane
encloses
yeslcſó forma
D
A,
facilitated diffusion
(В.
exocytosis
C. pinocytosis
OD. endocytasis
It sounds like the described process is facilitated diffusion, which is a type of cellular transport that helps molecules cross the cell membrane from areas of high concentration to lower concentration using transport proteins. This process does not require energy as it's a passive form of transport.
Explanation:Without a visual, it's difficult to definitively identify which form of cellular transport is being described here. However, if we have a scenario where molecules are moving from the outside to the inside of the cell through the membrane, it sounds like facilitated diffusion. This process involves specific transport proteins that act as channels or carriers, helping molecules that cannot cross the cell membrane alone to do so. Facilitated diffusion does not require energy as it's a type of passive transport, and the movement is down the concentration gradient, from a high concentration outside the cell to a lower concentration inside.
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You fill a shallow pan with water and place a drop of red ink in one end of the pan and a drop of green ink in the other end. Which of the following is true at equilibrium? a. The red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan. b. The red and green inks are uniformly distributed throughout the pan. c. Each ink is moving down its concentration gradient. d. The concentration of each ink is higher at one end of the pan than at the other end.
Explanation:
When we fill a shallow pan with water and place a drop of red ink in one end of the pan and a drop of green ink in the other end we see that the red nd green inks are uniformly distributed throughout the pan. This is due to the Diffusion of red ink and green ink in the water. Diffusion is the process where the movement of particles takes place from the region of higher concentrations to the region of lower concentartions. Until the concentration of substances is uniform throughout, this phenomena continues.
At equilibrium, the red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan.
Explanation:At equilibrium, the correct statement is:
a. The red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan.
Equilibrium is reached when the concentration of each ink becomes uniform throughout the pan. This happens because the molecules of the red and green ink are constantly moving due to Brownian motion. Over time, they will mix and spread evenly, resulting in a uniform distribution of the inks in the pan.
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Archeological evidence suggests that Neanderthals might have been aware of the medicinal properties of plants over 60,000 years ago. Imagine you are a modern-day ethnobotanist and have identified a compound from fossilized tree pollen that binds to the alpha subunit of the most common G-proteins. To test how whether the compound affects the activity of Gα, you treat liver cells with the compound alongside epinephrine. You observe that the cells fail to produce glucose. Which could be the activity of the compound?
Answer:
It could be prevent Ga from binding to the epinephrine receptor.
Explanation:
Hindering GTPase action will make G-protein to tie to adenyl cyclase for all time so Glucose is delivered persistently.
Diminishing the proclivity for GDP will build the opportunity of authoritative of GTP to G protein which thus will initiate the pathway.
On the off chance that G protein ties with adenyl cyclase it will invigorate it.
Yet, in the event that the G-protein is kept from official with epinephrine receptor, at that point the receptor can't enact trade the guanine nucleotide to G-protein for initiation. Thus the G-protein stays in latent state.
The activity of the compound should be that It could prevent Ga from binding to the epinephrine receptor.
How to make G-protein:In the case when GTPase activity should be inhibiting for binding the adenyl cyclase in a permanent way due to this the glucose should be generated in a continuous way. In the case when it reduced the affinity for GDP it raised the chance of binding for GTP to G protein that turned the activation of the pathway.
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Pigment in mouse fur is only produced when the C allele is present. Individuals of the cc genotype are white. If color is present, it may be determined by the A, a alleles. AA or Aa results in agouti color, while aa results in black coats. What are the phenotypic ratios in the F2 generation of mice if an AACC parent and an aacc parent are bred to yield F1 mice which are then crossed to generate an F2 generation? Enter your answer as a fraction in the blank next to the appropriate coat color.
agouti ______
colorless_______
black________
Answer:
Agouti: colourless:black=9:4:3 so the agouti coat color would be 9/16 , Colourless coat color would be 1/4 , and Black would be 3/16 .
Explanation:
The type of genetic pattern that is followed in this cross is epistasis, in this pattern the phenotypic expression of a particular gene in one locus changes the effect of the gene on other locus.
In this case allele C present will leads to the color but homozygous C denotes colorless and A denotes agouti where aa denotes black. With the first cross all F1 individual will be AaCc . Hence, in the second cross for F2 would be AaCc × AaCc
Thus, the phenotypic ratio of all three type of coat color woyld be agouti: colourless:black=9:4:3 so the agouti coat color would be 9/16 , Colourless coat color would be 1/4 , and Black would be 3/16 .