Answer:
C. 25%
Explanation:
A bacterial cell gives rise to two daughter cells by one division. If a single bacterial cell enters into cell division, it will form a total of 2^3= 8 cells after three rounds of cell division. The two DNA strands of the parent DNA duplex having N15 will be present in two bacterial cells out of the total 8 cells. This would occur since the process of DNA replication forms a new DNA duplex with one parental strand and one new strand. So, after three round of cell division 25% cells will have DNA with N15 (2/8 x 100= 25%)
Final answer:
After three divisions in 14N medium starting with 15N-labeled DNA, due to semi-conservative replication, 25% of bacterial cells will contain DNA with 15N label per Meselson and Stahl's experiment.
Explanation:
The experiment by Meselson and Stahl with E. coli growing in heavy nitrogen (15N) and then switching to light nitrogen (14N) demonstrates the semi-conservative model of DNA replication. After the first division in 14N medium, each DNA molecule contains one strand with 15N and one with 14N, making it fifty percent 14N. By the principle of semi-conservative replication, after three divisions, there will be one more division than there are strains that have any 15N isotopes. Thus, one quarter of the bacterial cells should contain 15N since those are the ones that did not undergo the last division and still contain one original heavy strand. This makes the answer (C) 25%.
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.
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
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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Case Study: Parents of a 3-year-old boy noticed that their son had a waddling gait, fell frequently and had difficulty getting up again, and was not able to run. By age five, there was progressive muscular weakness and muscle wasting, and weakness of the trunk muscles led to abnormal posture. By age 9 he was confined to a wheelchair. Contractures appeared, first in the feet, as the gastrocnemius muscles tightened.1. This hereditary X-linked recessive disease characterized by progressive muscular weakness is ______.2. What does dystrophy mean? Why is this term used to describe this case?
Answer:
Muscular dystrophy
Explanation:
Muscular dystrophy is an inheritable genetic condition that involves mutations in genes that encodes the production of muscle proteins to build and to preserve healthy proteins. The disease is characterised by progressive weakness and loss of muscle mass.
Dystrophy is referred to as a disorder in which the a tissue or an orgasm progressively wastes away. Thé dystrophy term is used because the disease involves the gradual and progressive wasting away of the muscle which is a type of tissue causing muscular weakness allowing the child to experience these symptoms.
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Linda wrote a few steps to describe how carbon circulates between the atmosphere and living organisms. Step 3 is missing.
Step 1: Carbon enters the atmosphere as carbon dioxide from respiration and combustion
Step 2: Plants use carbon dioxide to make glucose
Step 3:
Step 4: Animals exhale carbon dioxide
Which of the following best describes step 3?
a. Dead animals are buried underground and form fossil fuels.
b. Carbon present in glucose is converted into oxygen by animals.
c. Carbon present in glucose is transferred into plant-eating animals.
d. Dead animals decompose and recycle the carbon in the form of carbon dioxide.
Answer:D
Explanation:decomposition of dead plants and animals.
Answer:
c
Explanation:
edit: yup its c i just did the test and got it right
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.
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.
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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After you performed the freeze-thaw cycle to lyse the bacteria, you pelleted the debris using a centrifuge. What two reasons did this pellet fluoresce green? (Yes, I know it contained GFP…. but why?)
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.
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:
What do you think might be the evolutionary benefit of the milk production regulation mechanism in breastfeeding?
Answer: The benefit of evolutionary benefit of milk production regulation mechanism is that the mother will not waste her energy or nutrients required for her growth and development to produce milk at that time rather it will be during her late pregnancy stage where the mammary gland will be activated by oxytocin hormone to produce milk automatically to breast feed infants at birth.
Explanation:
Evolution of lactation or mammary glands in mammals has been very important to feed the infants.
Lactation refers to production of milk from the mammary glands. Lactation normally occur at post pregnancy stage. At this stage the infants are born and milk is been secreted by the mammary glands to breast feed the young ones. Lactation doesn't occur all the time but it occur during late to post pregnancy stage. This is activated by oxytocin which help to implant embryo in the uterus and help in lactation.
The milk production regulation mechanism in breastfeeding is a complex process that is controlled by a variety of hormones, including prolactin, oxytocin, and estrogen. This mechanism ensures that the mother produces enough milk to meet the needs of her baby, but not so much that it becomes a burden.
There are several possible evolutionary benefits of this mechanism.
First, it helps to ensure that the baby is always fed, even if the mother is not constantly producing milk. This is important for the survival of the baby, as it needs to be fed frequently in order to grow and develop.
Second, the mechanism helps to prevent the mother from producing too much milk, which can lead to mastitis, an infection of the breast tissue. Mastitis can be painful and can interfere with breastfeeding.
Third, the mechanism helps to ensure that the milk is of the right composition for the baby. The composition of milk changes over time to meet the changing needs of the baby. For example, the milk produced in the early days after birth is high in colostrum, which is a nutrient-rich liquid that helps to protect the baby from infection.
Hence, the milk production regulation mechanism in breastfeeding is a complex and efficient process that has evolved to ensure the survival and health of the baby.
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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 .
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.
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.
A 35-year-old man presents with anemia, neutropenia, thrombocytopenia, myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes. Cytochemistry examination demonstrates peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. This hematologic picture is consistent with: A) Acute lymphoblastic leukemia (ALL) B) Acute myeloblastic leukemia (AML) C) Chronic myelocytic leukemia (CML) D) Chronic lymphocytic leukemia (CLL) E) None of the above
Answer:
B) Acute myeloblastic leukemia (AML
Explanation:
It is a type of cancer that affects the blood as well as the bone marrow. It causes a situation whereby there is excessive immature white blood cells thus affecting the production of matured or normal white and red blood cells, blood platelets through the interference of the myeloid cells.
It is treatable medical condition that requires medical diagnosis. It requires laboratory tests with imaging always needed. It is also known as Acute myelogenous leukemia, acute granulocytic leukemia and acute nonlymphocytic leukemia.
Answer: The hematologic picture is consistent with Acute myeloblastic leukemia (AML). Therefore the correct option is B.
Explanation:
Leukaemia is a type of cancer that affects the blood forming tissues such as the bone marrow, leading to excessive or over production of abnormal blood cells ( usually the white blood cells). They occur in two categories; it can be Acute leukaemia or Chronic leukaemia.
There are different forms of acute leukaemia which is classified according to the lineage ( myeloid or lymphoid). They include:
- Acute Myeloblastic Leukaemia (AML)and
- Acute Lymphoblastic Leukaemia ( ALL).
In AML, the following Laboratory diagnosis differentiates it from other leukaemia, these includes:
- myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes, and
-peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. Therefore the correct option is B
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 organisms with large genomes, inversions are more likely to be tolerated if the breakpoints occur in: reciprocal translocations. noncoding DNA. open reading frames. coding DNA. closed reading frames.
Answer:
Non-coding DNA.
Explanation:
Inversion is a type of chromosomal abnormality in which the sequence of a segment of a chromosome is inverted or rotated at an angle of 180 degrees. This type of abnormality can change the reading frame of the gene and can cause mutations.
But if the genome sequence is non-coding that is not involved in the formation of protein synthesis than even if the reading frame is inverted will not affect the phenotype of the cell. Also, the non-coding sequences are removed by the splicing mechanisms.
Thus, Non-coding DNA is correct.
Inosine monophosphate is a branch point for the synthesis of a variety of purine nucleotides. Match the appropriate reaction for the synthesis of either AMP or GMP:
1) AMP A) condensation with Asp
2) GMP B) oxidative hydration
C) transamination with Gln
D) release of fumarate
1. --> 1:A, C; 2: B, D
2. --> 1:C, D; 2: A, B
3. --> 1:A, B; 2: C, D
4. --> 1:A, D; 2: B, C
Answer:
1:A, D; 2: B, CExplanation:
1.Purine Bases Can Be Synthesized by two pathways: i) de Novo and ii) Salvage Pathways.
2.Purine nucleotides synthesis starts with Phosphoribosyl pyrophosphate (PRPP), and it leads to the synthesis of nucleotide, inosine 5'-monophosphate (IMP). Inosine monophosphate is a branch point for the synthesis of many purine nucleotides amd leads to the synthesis of a variety of purine nucleotides.
3.The first reaction of purine synthesis is catalyzed by the enzyme glutamine phosphoribosylpyrophosphate amidotransferase.
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 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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"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.
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
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.
Cheers!
a) There are abiotic and biotic factors that keep population size of species under control Provide two examples that affect (or have affected) population size of native species in WA (one example for density dependent and one for density independent). You need to pick two real examples, provide specific species names and habitats/ecosystems where those species occur. You need to use different examples from the ones I used in lectures
b) Explain why species that overlap a great deal in their fundamental niches have a high probability of competing. Now explain why species that overlap a great deal in their realized niches and live in the same area, probably do not compete significantly.
c) What are two important pieces of ecological knowledge that you could apply to the decisions you make about your own life? Mainly from an ecological point of view, but it is good that you make a connection with other aspects of your life (personal or professional).
Answer:
Answer A: A key example of biotic and abiotic protagonists is the presence of rocky structures that exist in the mountains of southern Argentina (Patagonia) that thanks to them exist the ideal temperatures for the procreation of animals such as the Patagonian condor and the possibility of form nests in these rocky structures that secure their young.
Answer B:
The explanation for this is that they have populations that increase in number and have their niches realized and constant resources necessary to develop the species in this way, considering the dominant species in the area and the one that predominates, in the case of fundamental niches. , they are necessary and even obligatory niches for many populations, which if populations do not go to them, they die, extinguish, or decrease their number, that is why in the face of this reality made up of a fundamental niche since it is fundamental to persist life of the population.
Answer c:
Two key principles are: one, maintain the equilibrium, that is why one seeks to have an environmentalist position preserving nature, and two considering that we are part of that balance but that does not give us the right to break it with environmental contamination, intervention of natural food systems and many other factors.
Explanation:
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.g Imagine two different sigma factors with different promoter recognition sequences. What would happen to the overall gene expression profile in the cell if one sigma factor were artificially overexpressed?
Answer:
Transcription is affected via the modulation of the concentrations of the different types of holoenzymes, so saturated promoters are only weakly affected by sigma factor competition. However, in case of overlapping promoters or promoters recognized by two types of sigma factors, we find that even saturated promoters are strongly affected. Active transcription effectively lowers the affinity between the sigma factor driving it and the core RNAP, resulting in complex cross-talk effects. Sigma factor competition is not strongly affected by non-specific binding of core RNAPs, sigma factors and holoenzymes to DNA. Finally, we analyze the role of increased core RNAP availability upon the shut-down of ribosomal RNA transcription during the stringent response. We find that passive up-regulation of alternative sigma-dependent transcription is not only possible, but also displays hypersensitivity based on the sigma factor competition.
Many chemotherapy drugs target rapidly dividing cells, such as cancer cells. Why might cancer cells divide, and therefore evolve, more quickly than other cells
Answer:
The faster that cancer cells divide, the more likely it is that chemotherapy will kill the cells, causing the tumor to shrink. They also induce cell self-death or apoptosis. Chemotherapy drugs that kill cancer cells only when they are dividing are called cell-cycle specific.
Explanation:
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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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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