Answer:
Eye changes or adjusts itself according to the focus of our eye.Explanation:
Pupil changes its shape when the eye focuses on near or far object. When we observe something far away, the pupil gets widen. But when we look at something near to us, pupil get smaller. So, pupil adjust itself according to the focus of our eye. If the partner switches from far to near, the pupil changes from wide to small. Pupil changes its shape according to bright as well as dim light also.
50.Botox, which is used to reduce wrinkles for a limited time interval, inhibits membrane fusion by targeting a protein that allows overcoming the repulsive force of the membranes. Based on this information, what is the protein that is inhibited by Botox?
Answer: Botox functions by inhibiting the function of SNARE protein.
Explanation:
The primary function of the SNARE protein is to mediate the vesicle fusion, which shows the fusion of the target membrane bound compartment with the vesicle.
The botox functions by inhibiting the function of this protein. This is a drug which weakens or paralyzes the muscles.
in small doses it is used to reduce the wrinkles on the face. This drug is made of bolulinum toxin which acts on the SNARE protein.
One of Mendel’s Laws says that the probability of inheriting a maternal allele for a gene does not affect the probability of inheriting the maternal allele for another gene. If each gene has two alleles, what is the probability that one individual inherited the paternal allele at three different genes?
Answer:
There's 2 alleles for every quality and there are 3 qualities. The likelihood of acquiring the fatherly allele of those 3 qualities would be 1/2 x 1/2 x 1/2. This is on the grounds that you have a 50/50 possibility of getting the allele from your mom or your dad for three unique alleles. Since you are making sense of the likelihood of acquiring 3 qualities from your dad you duplicate at that point to get 1/8.
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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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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In avocets, curved beaks result from a Z-linked recessives gene. A curve-beak male is mated with a flat-beaked (normal) female. The F1 are interbred to produce the F2. What is the probability of producing a curve-beak female in the F2
Answer:
50%
Explanation:
Z O
z Zz zo
z Zz zo
The males will have two copies of the alleles of the genes present on the Z chromosomes. As females have only one Z chromosome, they will have only one allele for the genes present on the Z chromosome.
The results from the punnet square show that there will be chance that the females of the F2 progeny will have curved-beaks.
Final answer:
The probability of producing a curve-beak female in the F2 generation when a curve-beak male is mated with a flat-beaked female, and their F1 offspring are interbred, is 25%. This calculation assumes the trait for a curved beak is Z-linked recessive, and it uses the basic principles of genetics and the specific sex chromosome composition of birds.
Explanation:
The question asks about the inheritance of a Z-linked recessive trait in avocets, specifically the probability of producing a curve-beak female in the F2 generation when a curve-beak male is mated with a flat-beaked female. In birds, males have two Z chromosomes (ZZ) and females have one Z and one W chromosome (ZW). A Z-linked recessive trait, like the curved beak in avocets, will manifest in females (ZW) if the Z chromosome carries the recessive allele since there is no corresponding allele on the W chromosome to mask it.
First, consider the parental generation (P): The male is curve-beaked and therefore has the genotype ZZc where c indicates the recessive allele for the curved beak. The flat-beaked female, assuming she does not carry the recessive allele (as it's not expressed and she's normal), would have the genotype ZW.
The F1 generation, resulting from their mating, would consist of males (ZcZ) and females (ZcW), all of which would have flat beaks because the curved beak is recessive.
When these F1 individuals interbreed, the F2 generation could have the following genotypes:
ZZ (flat-beaked males)
ZcZ (flat-beaked males carrying the recessive gene)
ZW (flat-beaked females)
ZcW (curve-beaked females)
The only way to produce a curve-beaked female is from an F1 male (ZcZ) mating with an F1 female (ZcW), resulting in the ZcW genotype. Therefore, the probability of this outcome is 1/4 or 25% since there are four possible combinations (ZZ, ZcZ, ZW, ZcW) and only one produces a curve-beaked female.
A young couple went to see a genetic counselor because each had a sibling affected with cystic fibrosis. (recessive disease, neither member of the couple is affected. Their four parents are also not affected))(a) What is the probability that the female of this couple is a carrier?(b) What are the chances that their child will be affected with cystic fibrosis?(c) What is the probability that their child will be a carrier of the cystic fibrosis mutation?
Answer: A: I would say a 50% chance that she's the carrier SF
B: 25%
C. The child will most likely be a carrier
Explanation: I'm not really sure if i'm correct but how I figured it out was. Mapping out the family tree and shading the circles if the had SF and if they did not have SF and was a carrier I only shaded half of it. After I finished shading everything out I made a punnet square to predict the probability of the child.
Make a hypothesis about how you think the two species of Paramecium sp. will grow alone and how they will grow when they are grown together.
I hypothesize that Paramecium species will grow faster alone due to reduced competition. When together, resource competition might slow growth.
I expect that when each species of Paramecium is cultured alone, their growth rates will be relatively higher due to the absence of interspecific competition for resources. In a mixed culture, where both species coexist, I anticipate a potential decrease in growth rates compared to individual cultures.
This could be attributed to the sharing of limited resources, leading to heightened competition between the two species. Consequently, the presence of interspecific interactions might result in a more constrained growth environment, affecting the growth trajectories of both Paramecium species when compared to their growth in isolation.
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When grown separately, both Paramecium aurelia and Paramecium caudatum will probably proliferate successfully. However, when they share the same environment and compete for resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis relies on previously observed behaviors, but must still be validated by experimental testing.
Explanation:A hypothesis about the growth of two species of Paramecium, specifically Paramecium aurelia and Paramecium caudatum, could be: When each species is grown separately, they will both proliferate successfully. However, if they are grown together and forced to compete for the same resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis is based on the observations shared in figures 45.24, 36.25, and 19.19 that reveal this competitive dynamic.
Beyond competition, these species might engage in symbiosis, which represents different types of close, long-term interactions between species. Depending on the conditions and specific interactions, these could be commensal (one species benefits, the other is unaffected), mutualistic (both species benefit), or parasitic (one species benefits at the expense of the other).
The interaction between these Paramecium species could potentially mirror other observed species interactions, such as between different groups of minnows in a shared environment or different mussel cultures in a lab. It's important to note however that any hypothesis must be tested experimentally to verify its accuracy and be subjected to further refinement based on findings.
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Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the mean realtive fitness within this population? Please give your answer to two decimal places.
Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the expected allele frequency change for A after one generation with selection? Please give your answer to two decimal places.
Final answer:
The mean relative fitness of the population is 0.85. The expected allele frequency change for allele A after one generation with selection is a decrease to approximately 0.62 due to the relative fitness differences among the genotypes.
Explanation:
The question at hand requires an understanding of genetic structure, allele frequency, and the Hardy-Weinberg principle to calculate both the mean relative fitness within a population and the expected allele frequency change for allele A after one generation with selection.
Mean Relative Fitness Calculation:
To calculate the mean relative fitness (w), we multiply the relative fitness of each genotype by its proportional representation in the population and sum the results:
w(AA) = 1.00 × (500/1000) = 0.50,w(Aa) = 0.80 × (250/1000) = 0.20,w(aa) = 0.60 × (250/1000) = 0.15.The sum is the mean relative fitness: wμ = 0.50 + 0.20 + 0.15 = 0.85.
Expected Allele Frequency Change:
The next step is to calculate the expected change in allele frequency using the Hardy-Weinberg equation. First, determine the frequency of the alleles (p for A and q for a). The current frequency of A (p) = (2×500 + 250) / (2×1000) = 0.625. The frequency of a (q) = 1 - p = 0.375. To calculate the expected frequency of allele A after one generation of selection, we must account for the relative fitness of each genotype:
p' = (p² × w(AA) + pq × w(Aa)) / wμ,p' = (0.625² × 1.00 + 0.625 × 0.375 × 0.80) / 0.85 ≈ 0.62.Hence, the expected frequency of A after one generation is approximately 0.62, indicating a slight decrease due to selection.
The mean relative fitness within the population is 0.85, rounded to two decimal places.
To find the mean relative fitness within the population, we need to calculate the weighted average of the relative fitnesses of each genotype, considering their frequencies.
Given:
- AA genotype count [tex](n_AA)[/tex] = 500
- Aa genotype count [tex](n_Aa)[/tex] = 250
- aa genotype count [tex](n_aa)[/tex] = 250
- Relative fitness of AA genotype [tex](w_AA)[/tex] = 1.00
- Relative fitness of Aa genotype [tex](w_Aa)[/tex] = 0.80
- Relative fitness of aa genotype [tex](w_aa)[/tex] = 0.60
First, calculate the total number of individuals in the population (N):
[tex]\[ N = n_{AA} + n_{Aa} + n_{aa} \][/tex]
[tex]\[ N = 500 + 250 + 250 \][/tex]
[tex]\[ N = 1000 \][/tex]
Next, calculate the contribution of each genotype to the total fitness:
[tex]\[ Contribution_{AA} = n_{AA} \times w_{AA} = 500 \times 1.00 = 500 \][/tex]
[tex]\[ Contribution_{Aa} = n_{Aa} \times w_{Aa} = 250 \times 0.80 = 200 \][/tex]
[tex]\[ Contribution_{aa} = n_{aa} \times w_{aa} = 250 \times 0.60 = 150 \][/tex]
Now, sum the contributions of all genotypes:
[tex]\[ Total \, Contribution = Contribution_{AA} + Contribution_{Aa} + Contribution_{aa} \][/tex]
[tex]\[ Total \, Contribution = 500 + 200 + 150 = 850 \][/tex]
Finally, calculate the mean relative fitness (w_mean):
[tex]\[ w_{mean} = \frac{Total \, Contribution}{N} \][/tex]
[tex]\[ w_{mean} = \frac{850}{1000} \][/tex]
[tex]\[ w_{mean} = 0.85 \][/tex]
So, the mean relative fitness within this population is 0.85, rounded to two decimal places.
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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The cornea is the transparent outer layer of the human eye. Because it must be transparent to light, it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion. Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.
During operation, the cornea produces waste in the form of CO2 gas that must be expelled to keep the eye healthy and functioning. This is accomplished by the simultaneous diffusion of CO2 from the cornea to the surrounding atmosphere, which generally features a low CO2 concentration.
It is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become susceptible to scarring. The body may also react by growing additional blood vessels into the eye, which can damage the cornea.
If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, which of the following contact lens / ambient condition modifications is not likely to be useful?
Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.
(a) Increase the contact lens thickness
(b) Increase the diffusivity of oxygen gas by increasing the contact lens porosity
(c) Increase the ambient temperature
(d) Increase the ambient partial pressure of oxygen gas
(e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen
the answer is not c
Increasing the thickness of a contact lens would decrease the flow rate of oxygen to the cornea, not increase it. Thus, this modification would not be useful if the goal is to increase the steady-state flow rate of oxygen to the cornea.
Explanation:The best answer is (a) Increase the contact lens thickness. In the context of diffusion, the flow rate of a molecule is inversely proportional to the thickness of the layer it has to go through. Increasing the contact lens thickness would effectively reduce the rate at which oxygen reaches the cornea, thereby reducing the flow rate, not increasing it as desired. On the other hand, solutions (b), (c), and (d) could potentially increase the flow rate of oxygen to the cornea. Increasing the diffusivity, ambient temperature, or ambient partial pressure of oxygen gas would all potentially increase the rate of oxygen diffusion to the cornea.
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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.
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:
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.
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.
How does hemoglobin function as a pH buffer?a.Hemoglobin binds hydrogen ions when carbon dioxide exits the red blood cell. b.Hemoglobin releases hydrogen ions after carbon dioxide enters the red blood cell. c.Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell. d.Hemoglobin releases hydrogen ions when oxygen exits the red blood cell.
Answer: Option C.
Haemoglobin binds Hydrogen ion after carbondioxide enters red blood cells.
Explanation:
Haemoglobin is the protein in the red blood cells that help to transport oxygen in the blood. It is an iron compound. Haemoglobin acct as buffer by binding to acid or hydrogen ion in the blood when carbondioxide enters the blood, to remove the acid in the blood before it changes the blood pH.
Final answer:
Hemoglobin acts as a pH buffer by binding free hydrogen ions when carbon dioxide is converted into bicarbonate ions within red blood cells, preventing significant pH changes. So, the correct option is c : Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell.
Explanation:
Hemoglobin functions as a pH buffer in the bicarbonate buffer system, which plays a crucial role in the body's regulation of blood pH. Carbon dioxide (CO2) enters red blood cells and is converted into carbonic acid (H2CO3) by the enzyme carbonic anhydrase (CA). This unstable intermediate quickly dissociates into bicarbonate ions (HCO3-) and hydrogen ions (H+). The key to buffering is that hemoglobin within the red blood cells binds to these free H+ ions, thereby preventing a drastic shift in pH levels. This buffering capability keeps the blood's pH within a narrow and healthy range. When carbon dioxide enters the red blood cell and is converted into bicarbonate ions, hemoglobin binds hydrogen ions. Therefore, the correct answer to the original question is that hemoglobin functions as a pH buffer by binding hydrogen ions after carbon dioxide enters the red blood cell (option c).
The genetic condition Xeroderma pigmentosum, which can lead to skin cancer, results from A. inability to correct UV induced lesions B. inability to process phenylalanine. C. inability to produce functional hemoglobin. D. inability to correct transitions E. breaks in the X chromosome
Answer:
Option A
Explanation:
Xeroderma pigmentosum arises as a result of the cell being unable to correct lesions induced by UV. This can be as a result of mutations in the enzymes which include XP A-E needed for correction of the lesions. Failure to correct these lesions leads to their accumulation and then damage to the cell.
Which term is not correctly matched with its proper definition? Prototroph: possesses the same metabolic/growth capabilities as the parent, wild type strain. The prefixes chemo- and photo-: define the electron source Macronutrient: phosphorous and sulfur are in this category Chemoorganotroph: a heterotroph could be considered this Growth factor: components, either defined or undefined, that promote/support growth
Answer:
The prefixes chemo- and photo-: define the electron source
Explanation:
The prefix 'chemo' indicates something that is a “chemical” or “chemically induced” while the prefix "photo" relates to light. An electron source is that which produces electrons. The prefixx chemo and photo do not relate to sources of electron production.
The term 'Prototroph' is incorrectly matched with its description. A Prototroph is a microorganism that can synthesize all compounds needed for its growth and does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain. Other terms are correctly matched with their definitions.
Explanation:In the list of terms and their corresponding definitions, the term that isn't aptly matched is Prototroph. The definition of Prototroph usually refers to a microorganism that, unlike an auxotroph, does not require any additional nutrient other than a source of organic carbon for growth. It does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain.
Specifically, a Prototroph is a strain of microorganism that possesses a complete set of genes, enabling it to synthesize all compounds needed for its growth, thereby not requiring any growth factors. Similarly, a wild strain usually refers to the 'normal' or 'natural' condition of an organism, but it may or may not always have the same growth capabilities as its parent strain.
On the other hand, the terms and definitions of Macronutrient, Chemoorganotroph, Growth factor, and terms with the prefixes chemo- and photo- are correctly matched. These terms are standard in the field of Microbiology.
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Catecholamines mediate their effects throughI. α-adrenoreceptors.II. β-adrenoreceptors. III. receptor tyrosine kinases.IV. monomeric G protein receptors
Explanation:
Catecholamines are hormones made by your adrenal organs, which are situated on your kidneys. Models incorporate dopamine; norepinephrine; and epinephrine (this used to be called adrenalin or adrenaline). Our adrenal organs send catecholamines into our blood when you're genuinely or sincerely focused. They cause you to inhale quicker, raise your pulse, and send more blood to significant organs, similar to your cerebrum, heart, and kidneys.What is the expected Hardy-Weinberg frequency of heterozygotes in a population of green anoles, where the locus under investigation has two alleles and the frequency of one allele is 0.005 (give your answer to five decimal places)?
Answer:
0.00995
Explanation:
If the population is in Hardy Weinberg Equilibrium, the sum total of frequencies of the alleles of a locus would be one. The frequency of one allele for a locus is 0.005. So, the frequency of the other allele for the same locus would be=1-0.005= 0.995
According to the given information, there are only two alleles for the locus in the population. So, the frequency of heterozygous genotype in a population= 2 x frequency of one allele x frequency of other allele= 2 x 0.005 x 0.995= 0.00995
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.
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.
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.
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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Saturated fatty acids are degraded by the stepwise reactions of beta-oxidation, producing acetyl-CoA. Under aerobic conditions, how many ATP molecules would be produced as a consequence of removal of each acetyl-CoA?
Answer:
4 ATP molecules
Explanation:
Normally, about 11 ATP molecules are generated as the result of Beta-Oxidation of saturated fatty acid in Kreb's Cycle. But the total removal of acetyl-CoA under certain aerobic condition decreases the overall yield and approximately 4 ATP molecules comes out from each removal of acetyl-CoA.
Final answer:
Each acetyl-CoA molecule produced from the stepwise degradation of saturated fatty acids through beta-oxidation can yield around 10 ATP molecules.
Explanation:
Saturated fatty acids are degraded through beta-oxidation, which produces acetyl-CoA. Each acetyl-CoA molecule can yield a certain number of ATP molecules when it enters the citric acid cycle and oxidative phosphorylation. One mole of acetyl-CoA metabolized by the citric acid cycle yields about 10 ATP molecules. Therefore, the energy produced by the removal of each acetyl-CoA molecule from the beta-oxidation of saturated fatty acids can be estimated to be around 10 ATP molecules.
Information transfer is fundamental to all living organisms. For the following examples, explain in detail how the transfer of information is accomplished.
1. The genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation?
2. The genetic material in one eukaryotic cell is copied and distributed to two identical daughter cells.
3. A gene in a eukaryotic cell is transcribed and translated to produce a protein.
Answer:
Explanation:
We are to explain in details how
genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation.
For transformation, here, there is genetic alteration of a cell. It is a process of horizontal gene transfer where bacteria take up foreign genetic material from the environment.
Transduction is genetic recombination in bacteria where genes from a host cell are carried into the genome of a bacterial virus and then carried to another host during infection cycle.
Conjugation. In this, DNA plasmid is transferred from one bacterium to another through pilus.
2. We explain also in details how genetic material in one eukaryotic is copied and distributed.
Genetic material is copied during DNA replication and genetic material is distributed during mitosis.
The followings take place during DNA replication.
• DNA is copied during interphase, S phase of cell cycle.
• Site origin of DNA is known.
• Unwinding of DNA at the origin
• Synthesis of new strands with enzyme called helicase.
• Formation of replication forks.
• Proteins associated with replication forks help in the initiation and continuation.
• DNA polymerase synthesizes new strands by adding nucleotides that complement each strand.
Mitosis.
• Chromosomes are doubled
• Prophase: phase of condensation of chromosomes, formation of spindle.
• Metaphase: Alignment of chromosomes
• Anaphase: Chromatids are separated
• Telophase: reformation of nuclear membrane, division of cells, cell cycle control.
3. We also explain how transcription and translation of gene in eukaryotic.
Transcription
• Here we have DNA sequence to RNA sequence.
• Occurs in the nucleus
• DNA is carried into nucleosomes.
• RNA polymerase: add complementary RNA nucleotides to DNA template.
• Growth of new strands
• Leads to mRNA processing.
Translation
• Here we have mRNA base sequences to amino acids initiation
• Sequence of events such as complexes, small unit of ribosomes.
• Also we have the first tRNA
• Structure of ribosomes formed.
• Complete description of two subunits, 2 action sites, rRNA and proteins.
Mutations in which of the following genes lead to transformations in the identity of entire body parts?A) morphogensB) segmentation genesC) egg-polarity genesD) homeotic genesE) inducers
Answer: Option A
Explanation:
A morphogen gradient is an important concept in case of the developmental biology. It produces the signals from the part of embryo which decides the fate of development of the parts.
The differentiation of the cell takes place by the help of the signals produced by the embryo in order to decide where will which part of the embryo grow.
The surrounding cells receive the signals and the development of entire body parts takes place.
So, the mutation in this gene will transform the entire body part.
Answer: Option A.
Explanation:
Morphogens are signaling molecules that originate from a specific location that spread from their origin via diffussion to form a concentration gradients acting at long distances to induce responses in cells base on the gradients the cells interact with. It determines the pattern of tissue development and morphogens gradients cause the differentiation of unspecialized stem cells into different cell types.
An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene
Answer: 5
Explanation:
Since 1 out of 200 people carry the defective gene that causes inherited colon cancer.
In a sample of 1000 individuals, we say
1 = 200
y = 1000
cross multiply
y x 200 = 1000 x 1
200y = 1000
y = 1000/200
y = 5
Thus, in a sample of 1000 individuals, approximately 5 will be carriers
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.
Learn more about Cellular Transport here:https://brainly.com/question/32778892
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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.
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: