To determine the distance between the eyes based on the width of one eye, we need to use a formula known as the "interpupillary distance" or IPD. The IPD is the measurement of the distance between the centers of the pupils in each eye, and it is typically expressed in millimeters.
To calculate the IPD based on the width of one eye, we can use the following equation: IPD = 2 x (Width of one eye)
Using the information given in the question, we can convert the width of one eye from inches to millimeters by multiplying by 25.4 (the number of millimeters in one inch):
Width of one eye = 1.75 inches x 25.4 mm/inch = 44.45 mm
Plug this value into the IPD equation to find the distance between the eyes: IPD = 2 x (44.45 mm) = 88.9 mm
The distance between the eyes is approximately 88.9 millimeters. It's important to note that this is an estimate, as everyone's IPD can vary slightly.
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We have learned that the response to selection is equal to the selection differential times the narrow-sense heritability and that the narrow-sense heritability includes only the additive genetic variance. Why arenât the dominance genetic variance and the gene interaction variance included? Why donât they contribute to the genetic variation that is acted on by selection?
We have learned that the response to selection is equal to the selection differential times the narrow-sense heritability (R = h^2 * S), where narrow-sense heritability includes only the additive genetic variance. The reason dominance genetic variance and gene interaction variance are not included is that they do not contribute to the genetic variation that is acted on by selection in the same way as additive genetic variance. So, only additive genetic variance is included in narrow-sense heritability, as it provides a reliable measure of the genetic variation that is consistently acted on by selection across generations.
Additive genetic variance directly contributes to the resemblance between parents and offspring, and therefore, it is an essential component for predicting the response to selection. In contrast, dominance genetic variance and gene interaction variance involve interactions between alleles at the same locus or between different loci, which can cause non-additive effects. These non-additive effects do not consistently transfer from one generation to the next, making them less predictable in terms of how they will affect future generations.
As a result, only additive genetic variance is included in narrow-sense heritability, as it provides a reliable measure of the genetic variation that is consistently acted on by selection across generations.
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Please help and hurry!!
Describe one of the four inventions for deep-sea exploration covered in the lesson, and explain the specific uses for which it is designed.
One invention for deep-sea exploration is the remotely operated vehicle (ROV), designed for scientific research, sample collection, and underwater surveying.
One of the inventions for deep-sea exploration covered in the lesson is the remotely operated vehicle (ROV). ROVs are robotic underwater vehicles designed to explore the depths of the ocean where humans cannot easily reach. They are equipped with cameras, lights, and a variety of scientific instruments. The specific uses of ROVs include conducting research, collecting samples, and surveying underwater ecosystems. ROVs can be used to study marine life, geology, hydrothermal vents, and archaeological sites. They play a crucial role in gathering data and imagery for scientific research, oil and gas exploration, and underwater construction and maintenance. With their maneuverability and real-time video feeds, ROVs allow scientists and engineers to observe and document the underwater environment, enabling a better understanding of the deep-sea world and its resources, as well as supporting various industries that operate in the oceanic realm.For more such questions on Deep-sea exploration:
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Label the diagram below with the following terms :photosynthesis, cellular respiration, h2o,o2,co2,atp,c6,h12,o6
The diagram shown in the image shows the process of photosynthesis with its various components involved.
What is photosynthesis?Photosynthesis is a biochemical process in which inorganic energy such as sunlight is converted into organic energy such as glucose molecules. The energy from this process will be stored in NADPH and ATP. This energy will also be used for the manufacture of sugar molecules together with the reduction of carbon dioxide molecules.
In our ecosystem they will be involved in these photoautotrophic processes such as plants, algae, cyanobacteria and red bacteria. For this reason, life on earth depends on solar energy.
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what specific differences do you think you will see between the animal and plant cells using the compound microscope?
Answer: Plant cells have a cell wall, as well as a cell membrane. In plants, the cell wall surrounds the cell membrane. This gives the plant cell its unique rectangular shape. Animal cells simply have a cell membrane, but no cell wall.
Explanation:
Plant cells:
1. Have a cellulose cell wall outside the cell
membrane.
2. Have a cell membrane.
3. Have cytoplasm.
4. Have a nucleus.
5. Often have chloroplasts containing
chlorophyll.
6. Have prominent one or more vacuoles.
7. Often regular in shape.
Animal cells:
1. Have no cell wall.
2. Have no chloroplasts.
3. Have only small vacuoles.
4. Often irregular in shape.
5. Do not contain plastids.
6. Have complex and prominent Golgi
apparatus.
7. Cytoplasm fills almost the entire cell.
Explain the effect of temperature and pH on enzyme action.
Needs to be done ASAP.
Answer:
enzymes loose their form in both unsuitable ph and temprature.
Explanation:
at very high tempratures enzyme molecules denatures due to high speed movements, at low tempratures enzymes works too slow to be useful enough.
low ph solotions contain lot of H+ and high ph solutions have few H+ , and the number of these H+ ions have polar effects on enzymes. some Enzymes like pepsin works perfect in acidic environment ( ph=2) while others might work at high ph or average ph( like Trypsin works well at ph=7).
The nucleotide sequences found in two different species are made of the same base pairs. This is evidence of
Answer:
The correct answer is - closely related species.
Explanation:
If two species have similar gene sequences or nucleotide sequences with the same base pairs then it means they both are closely related and have diverged from a similar and latest ancestor.
The same gene sequence results in the same proteins that could result in the same phenotypic behavior. Such cases establish that both species are closely related and have recently diverged. These species must have a common ancestor.
In the average person, what percentage of alcohol is absorbed through the small intestine?.
In the average person, 70-80 percentage of alcohol is absorbed through the small intestine.
Alcohol is not treated the same way as other food nutrients. In fact, the digestive system exerts additional effort to remove it from the body, giving alcohol removal priority over other nutrients including lipids (fats), carbs, and proteins. All parts of the digestive tract take in alcohol. In contrast to other nutrients, alcohol is quickly absorbed in the small intestine and enters the bloodstream immediately via the stomach lining. Although other bodily cells can also metabolize alcohol, the liver is where alcohol metabolism mostly takes place. Acetaldehyde, a poisonous chemical, is created when alcohol is broken down.learn more about alcohol here: https://brainly.com/question/27427140
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visible characteristics of snake
Answer:
The jaws of snakes are loosely jointed and extremely flexible. There are no ears or movable eyelids; the eyes are covered by transparent spectacles, or ocular scales.
Explanation:
Snakes have good vision.There are no ears or movable eyelids; the eyes are covered by transparent spectacles, or ocular scales.
2
Which of the following is an example of stewardship towards the environment?
OA. giving money to charities
OB. helping find a lost dog
OC. pouring unused medicine down the drain
OD. volunteering to help plant trees
Answer: installation of water saving fixtures, grey water recycling systems, reuse of production waste water, rain water harvesting systems
Explanation: installation of water saving fixtures, grey water recycling systems, reuse of production waste water, rain water harvesting systems, etc. Material Conservation Conserve materials used at your site by changing process, product or formulations, participate in waste exchanges, reuse operational waste, etc.
Answer:
volunteering to help plant trees
Explanation:
OD
Find the slope between (2,4) and (4,6). Reduce it.
show your work
Answer:
\(\displaystyle m=1\)
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightAlgebra I
Coordinates (x, y)Slope Formula: \(\displaystyle m=\frac{y_2-y_1}{x_2-x_1}\)Explanation:
Step 1: Define
Point (2, 4)
Point (4, 6)
Step 2: Find slope m
Simply plug in the 2 coordinates into the slope formula to find slope m
Substitute in points [Slope Formula]: \(\displaystyle m=\frac{6-4}{4-2}\)[Fraction] Subtract: \(\displaystyle m=\frac{2}{2}\)[Fraction] Divide: \(\displaystyle m=1\)nvestigations were carried out in a science lab to explore the topic of chemical and physical changes.
Investigation A
Step 1. Add 5 tsp. salt to 100 ml warm water and stir until most or all of the salt is no longer visible.
Step 2. Heat the salt solution on a burner until only a white solid remains.
Investigation B
Step 1. Mix 10 tsp. white sugar into 100 ml water and stir until most or all of the sugar is no longer visible.
Step 2. Heat the sugar solution on a burner until the solution thickens and turns brown.
In which step(s) did a chemical change most likely occur?
Investigation A, Step 1 and Investigation B, Step 1
Investigation A, Step 2 and Investigation B, Step 2
Investigation A, Step 2 only
Investigation B, Step 2 only
In Investigation B, Step 2 only a chemical change has occurred. Caramelization, which is a chemical change, takes place when the sugar solution is heated on a burner until the solution thickens and turns brown. All the other steps are physical changes.
How to decide whether an alteration is chemical or physical?Physical changes involve a change in the substance's shape or appearance but not its composition. In contrast, a chemical shift alters the nature of the matter and results in the formation of at least one new substance with novel features.
Why caramelization is a chemical change?A non-enzymatic browning reaction is known as caramelization. The process results in the release of volatile molecules that give caramel its distinctive flavor. Both the breakdown of the sugar and the elimination of water (as steam) are components of the reaction. The type of sugar affects the caramelization reaction.
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Give examples of the type of plants that would grow using these vocabulary words vegetation,colonize,pioneer species, sere,hardwood,softwood,substrate.
Vegetation refers to all plant life that exists in a particular area. Plants that are able to colonize a new area are referred to as pioneer species. In the process of plant succession, sere refers to the sequence of plant communities that follow one another in the course of time on a given area.
The following types of plants would be able to grow using the vocabulary words mentioned above:
Softwood: Softwood refers to the type of trees that retain their leaves or needles throughout the year. Conifers, such as pine, spruce, and fir, are examples of softwood trees. These trees have a greater tolerance for nutrient-poor soils and are often pioneer species.
Hardwood: Hardwood refers to trees that shed their leaves or needles annually. Oak, maple, and birch trees are examples of hardwood trees. They grow in nutrient-rich soils and are generally not pioneer species, but are usually found in more mature plant communities.
Vegetation: Vegetation refers to all plant life that exists in a particular area. Coastal vegetation, such as beach grass and salt cedar, is an example of vegetation that is adapted to salt spray and sandy soils.
Colonize: Plants that are able to colonize a new area are referred to as pioneer species. Lupines, fireweed, and dandelions are examples of plants that can colonize an area quickly.
Sere: In the process of plant succession, sere refers to the sequence of plant communities that follow one another in the course of time on a given area. For example, a sere that develops after a forest fire could include pioneer species such as grasses and shrubs, followed by conifers, and finally hardwood trees.
Substrate: The term substrate refers to the underlying soil or rock material on which plants grow. Lichens are an example of organisms that can colonize a bare substrate and help create conditions for other plants to grow.
Softwood trees like pine, spruce and fir can grow in nutrient-poor soil and are often pioneer species while hardwood trees like oak, maple, and birch prefer nutrient-rich soils and are not pioneer species. Beach grass and salt cedar are examples of coastal vegetation that is adapted to salt spray and sandy soils while lupines, fireweed, and dandelions can colonize an area quickly. After a forest fire, the sere could include pioneer species such as grasses and shrubs, followed by conifers, and finally hardwood trees. Lichens can colonize a bare substrate and help create conditions for other plants to grow.
To summarize, vegetation refers to all plant life in an area, pioneer species are plants that colonize a new area, a sere is the sequence of plant communities that follow each other in the course of time, hardwood and softwood trees grow in different soils and substrates refers to the underlying soil or rock material on which plants grow.
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Name and describe the physical and compositional layers into which scientist divide the earth
Answer:
Core. Compositional layer of Earth just below the mantle to the center. ...
Crust. Compositional outermost solid layer of Earth. ...
Mantle. The compositional layer of Earth between the core & the crust. ...
Lithosphere. ...
Asthenosphere. ...
Mesosphere. ...
Outer Core. ...
Inner Core.
Explanation:
now help me with my math please
The earth is divided into three main layers crust, mantle and core.
The earth is a rocky planet that is made up of three layers that are divided into the upper and lower parts. In order to understand the geologic composition of the earth.
We must know that earth's lithosphere is made up of slabs or plates. Which is known as the upper crust. It has a thickness of 0 - 30 meters. The lower layer is divided by the oceanic or aquatic lithosphere. Below the crust there is a gap and that gap us represented by the mohorovicic discontinuity or moho that exists between the upper mantle and the lower crust. The upper mantle is known as asthenosphere and is sold as semi-solid and has a thickness of 2850 meters. Between the lower mantle and upper core or outer core there lies the gap of Gutenberg discontinuity. The third layer that is core is divided into the inner and outer core the inner core 1270 km thick and outer core is about 2200 km thick.Hence the earth is divided into the physical characteristics.
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Which stage of the cell cycle can lead to uncontrollable reproduction in mutated cells?
0 G2
OM
OS
OG1
Answer:M
Explanation:
Saw some guy say it
Answer:
M is correct
Explanation:
Did the test.
5. This mallard duck and this moose both have bones in their feet called metatarsals. However, they have different types
of feet. The duck has webbed feet that help it swim in water. The moose has feet with hooves that help it walk on hard
ground. What most likely explains why both the duck and moose have feet with metatarsals and why the feet are
different?
Answer:The duck has webbed feet that help it swim in water. The moose has feet with hooves that help it walk on hard ground.
Explanation:
Which of the following structures in an ascomycete is haploid? A. basidium B. ascus C. ascospore D. ascocarp
The structure in an ascomycete that is haploid is the ascus.
In ascomycetes, the ascus is the structure where sexual reproduction takes place. It contains the ascospores, which are the result of meiotic division. The ascus is formed from the fusion of two haploid nuclei, one from each parent, to form a diploid nucleus. This diploid nucleus then undergoes meiosis, resulting in the formation of haploid ascospores. These ascospores are the reproductive units that can be dispersed and give rise to new haploid individuals. Therefore, among the given options, the ascus is the structure that is haploid, as it contains the haploid ascospores. The basidium is a structure found in basidiomycetes, and the ascocarp is the fruiting body of an ascomycete, both of which are not haploid structures.
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convert coenzyme a (given) to a high energy form by combining it with acetic acid.
Coenzyme A (CoA) is a molecule that plays a vital role in metabolism, especially in the breakdown of fats and carbohydrates. To convert CoA to a high-energy form, it needs to be combined with acetic acid to form acetyl-CoA.
This reaction is catalyzed by the enzyme pyruvate dehydrogenase complex (PDC) and occurs during the oxidative decarboxylation of pyruvate in the mitochondria.
Acetyl-CoA is considered a high-energy molecule because it contains a thioester bond that has a high potential energy.
This bond can be hydrolyzed to release energy, which is used to power many metabolic processes in the cell, including the citric acid cycle and oxidative phosphorylation.
In summary, combining CoA with acetic acid to form acetyl-CoA is an important step in energy metabolism, and the resulting molecule is considered high energy due to its thioester bond.
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middle which diagram shows the changes in appearance of a plant cell when a remains in a concentrated sugar solution for thirty minutes? what is th answer
The osmosis diagram is the one that most accurately represents how a plant cell might change in appearance after spending thirty minutes in a concentrated sugar solution.
Osmosis is the process by which water molecules move from an area with a lower concentration of solutes to an area with a higher concentration of solutes across a semi-permeable membrane.
If a concentrated sugar solution were present, the solute concentration outside the plant cell would be higher than that inside.
Water from inside the plant cell would naturally try to exit the cell as it is submerged in the concentrated sugar solution in an effort to balance the solute concentration on either side of the cell membrane.
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why is a heat-stable dna polymerase from a thermophilic bacterium (the taq polymerase) used in the polymerase chain reaction rather than a dna polymerase from e. coli or humans?
The heat-stable DNA polymerase from a thermophilic bacterium, such as Taq polymerase, is used in the polymerase chain reaction (PCR) for several reasons:
1. Thermostability: Thermophilic bacteria, which thrive in high-temperature environments, have evolved enzymes that can withstand and function optimally at elevated temperatures. Taq polymerase, isolated from the bacterium Thermus aquaticus, is highly thermostable. It remains active even at the high temperatures (usually around 94-98°C) required for DNA denaturation in PCR. In contrast, DNA polymerases from E. coli or humans would denature at these temperatures, rendering them ineffective for PCR.
2. Temperature cycling: PCR involves repeated cycles of heating and cooling to denature the DNA strands, anneal primers, and extend them by polymerization. Taq polymerase can withstand these temperature fluctuations without losing its activity, making it suitable for the PCR process. Other DNA polymerases that are not heat-stable would be inactivated or degraded during the high-temperature denaturation steps of PCR.
3. Efficient DNA synthesis: Taq polymerase exhibits robust DNA synthesis activity and high processivity, meaning it can efficiently synthesize long DNA strands during PCR. This feature is crucial for the amplification of target DNA sequences. While DNA polymerases from E. coli or humans are capable of DNA synthesis, they may not perform as well as Taq polymerase in terms of efficiency and processivity.
4. Minimal contamination risk: Taq polymerase is typically purified from Thermus aquaticus, a bacterium that does not naturally inhabit the human body. Therefore, using Taq polymerase in PCR reduces the risk of contaminating the reaction with other DNA sequences, which could lead to false-positive results. In contrast, DNA polymerases from E. coli or humans may carry the risk of introducing additional DNA contaminants into the PCR reaction.
Overall, the choice of Taq polymerase in PCR is driven by its thermostability, resistance to high-temperature cycling, efficient DNA synthesis, and low contamination risk, all of which make it a reliable and widely used enzyme for DNA amplification in the PCR process.
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The heat-stable DNA polymerase from a thermophilic bacterium, such as Taq polymerase, is used in the polymerase chain reaction (PCR) instead of DNA polymerase from: coli or humans because of its ability to withstand the high temperatures required during the PCR process. The correct option is (e).
PCR involves cycles of high-temperature denaturation, annealing, and extension. The denaturation step requires heating the DNA sample to separate the double-stranded DNA into single strands.
Then, during the annealing and extension steps, the DNA primers bind to the target sequences and the DNA polymerase synthesizes new DNA strands.
Taq polymerase, derived from the thermophilic bacterium Thermus aquaticus, is ideal for PCR due to its unique property of being highly heat-stable.
It can withstand the high temperatures (typically around 95°C) used for DNA denaturation without losing its enzymatic activity. In contrast, DNA polymerases from E. coli or humans are not heat-stable and would denature at these high temperatures, rendering them ineffective for PCR.
The use of Taq polymerase ensures that the DNA polymerase remains active throughout the PCR cycles, enabling efficient amplification of the target DNA.
This makes Taq polymerase a crucial component in PCR and has revolutionized various fields of research, diagnostics, and biotechnology.
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Someone please help me!
Answer:
It's A
Explanation:
Trust.
AWNSER FAST PLEASE BRAINLIEST 100 POINT AWNSERS Both plant and animal cells contain mitochondrion however they are not in equal amounts. Between the cheek and onion cells, which one do you think would contains the greater amount of mitochondrion and why?
Answer:
well I think the cheek cell contains the greater amount of mitochondrion
Explanation:
when you look at both the onion and cheek cells under the microscope you will notice that
It has been demonstrated that ectotherms (such as insects and reptiles) in warm climates tolerate higher temperatures than ectotherms in cooler climates. Angilletta hypothesized that urban heat islands could lead to similar differences in heat tolerance between urban and rural populations of leaf-cutter ants. He predicted that ants in urban areas would tolerate heat better than ants in nearby rural areas.
Select the two best experimental designs for comparing the thermal tolerance of ants in urban areas to ants in nearby rural areas.
a. Measure the heat tolerance of many ants of the same species as a function of the distance they live from a city center.
b. Measure the heat tolerance of many ants of the same species from a city center and from a rural area in the same region.
Due to the presence of densely located buildings, the wind speed is reduced and the convective cooling effect is diminished.
These heat islands form because urban surfaces, such as roads and rooftops, absorb and emit heat to a greater extent than most natural surfaces. Heat reflected from a building is trapped by nearby taller buildings, known as the urban canopy. Cities with temperatures warmer than their surrounding natural environments are called urban heat islands. Man made materials, such as pavement, heat up quickly during the day. These surfaces radiate heat, which causes the air temperature around the city to rise. Trees, green roofs, and vegetation can help reduce urban heat island effects by shading building surfaces, deflecting radiation from the sun, and releasing moisture into the atmosphere.
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Denise just received the results of a complete physical that found her body is not producing enough insulin. which part of the endocrine system is affecting her body's ability to produce insulin?
The part of the endocrine system which affects her body's ability to produce insulin is referred to as the pancreas.
What is Insulin?This is defined as substance which is secreted into the bloodstream by the organ known as the pancreas and is leaf shaped. Its mode of action is to allow glucose to enter the body's cells to provide energy.
Insulin helps to regulate the blood sugar or glucose level in the body and a deficiency will lead to diabetes which comes with a lot of complications and risks.
This is therefore why pancreas was chosen as the most appropriate choice.
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An example of a compulsion is
A) Anxiety about contracting HIV infection from a sneeze
B) Uncontrollable worry about an accident
C) The impulse to hurt a family member
D) Constant and repetitive checking that the alarm is set
An example of a compulsion is D) Constant and repetitive checking that the alarm is set.
Compulsions are repetitive behaviors or mental acts that individuals with certain mental health conditions, such as obsessive-compulsive disorder (OCD), feel driven to perform. These compulsions are often performed in response to obsessive thoughts or to alleviate anxiety or distress. Compulsions are typically not connected to the actual event or situation they are intended to address, but rather serve as a way to temporarily relieve anxiety or prevent perceived harm.
In the given example, constant and repetitive checking that the alarm is set is a classic example of a compulsion. The individual feels compelled to repeatedly check the alarm to ensure it is set, even though they may have already checked it multiple times before. This behavior is driven by an irrational fear or anxiety that something negative will happen if the alarm is not set correctly. The act of checking the alarm becomes a repetitive behavior intended to alleviate the anxiety associated with the fear.
It is important to note that anxiety about contracting HIV infection from a sneeze (A) and uncontrollable worry about an accident (B) are examples of obsessive thoughts or worries, which are often experienced in conjunction with compulsions in OCD. However, in the given options, they do not specifically represent compulsions.
The impulse to hurt a family member (C) does not represent a compulsion. It is important to differentiate between typical impulses or aggressive thoughts and the repetitive, ritualistic behaviors associated with compulsions in conditions like OCD.
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Sara's mother gets a flat tire on her car while driving Sara to school. They use a jack to
change the flat tire. It exerts a force of 5,000 newtons to lift the car 0.25 meters. How much
work is done by the jack? Use commas as appropriate.
Answer:
1250Joules
Explanation:
The parameters given are:
Force=5,000N
Distance= 0.25m
Work done by the Jack=
force×distance
= 5,000×0.25
=1250joules
Therefore, the Workdone by the Jack is 1250Joules
Some anaerobic prokaryotes use other terminal electron acceptors other than O2. Using standard reduction potentials listed in Table 14-4 and assuming 100% efficiency, how much ATP could be synthesized by the oxidation of 1 molecule of NADH by the following? Show calculations.
Nitrate (NO3-)
Elemental Sulfur (S)
How does this compare to the oxidation of NADH by ½ O2?
The oxidation of 1 molecule of NADH by nitrate (NO3-) would yield approximately 4.3 ATP molecules per NADH molecule due to its reduction potential at 0.54 V.
The standard reduction potential of elemental sulfur (S) is at 0.04 V and therefore in the oxidation of 1 NADH molecule, it would synthesize around 1.2 ATP molecules per NADH molecule. The oxidation of NADH by ½ O2 molecule yields approximately 2.5 ATP molecules per NADH molecule due to the reduction potential of oxygen at 0.82 V.
The amount of ATP molecules generated from each of the terminal electron acceptors in NADH oxidation show the significant influence that the reduction potentials of the different terminal electron acceptors have on the amount of ATP the bacteria can synthesize.
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suggest which factor carbon dioxide oxygen or water might limit the rate of photosynthesis
Answer:
Explanation:
Oxygen is not required for photosynthesis so we should just cross it out of the list. Secondly, the factors upon which photosynthesis depend are sunlight, carbon dioxide and temperature. So this also puts an end to water. As for carbon dioxide, if its amount is increased, the stomata of leaves will close and will not allow more of it to enter and thus process of photosynthesis will eventually decrease. And if we decrease the amount of carbon dioxide, plant will not be able to make its food properly.
A characteristic that diffrrnattites cells from prokaryotic cells is that eukaryotic cells have
Answer:
Prokaryotic Cells DO NOT have a nucleus and membrane-bound organelles, Eukaryotic Cells do.
Explanation:
An easy way to remember this is Pro sounds like NO, and EU sounds like DO.
A vaccine against HIV proteins made by a genetically-engineered vaccinia virus that has infected a eukaryotic cell line is a(n)
The current population of a threatened animal species is 1.6 million, but it is declining with a half-life of 20 years. How many animals will be left in 40 years? in 65 years? The population after 40 years will be animals. (Round to the nearest whole number as needed.) The population after 65 years will be animals. (Round to the nearest whole number as needed.)
The population after 40 years will be 400,000 animals, and the population after 65 years will be approximately 336,000 animals.
To determine the population after a certain time period, we can use the formula for exponential decay;
N(t) = N0 × \((1/2)^{(t/T)}\)
Where;
N(t) is the population at time t
N0 is the initial population
t is the time elapsed
T is the half-life of the population
Given;
N0 = 1.6 million
T = 20 years
Let's calculate the population after 40 years;
N(40) = 1.6 million × \(1/2^{(65/20)}\)
= 1.6 million × (1/2)²
= 1.6 million × (1/4)
= 400,000
Therefore, the population after 40 years will be 400,000 animals.
Now, let's calculate the population after 65 years;
N(65) = 1.6 million × \((1/2)^{(65/20)}\)
= 1.6 million × \((1/2)^{(13/4)}\)
≈ 1.6 million × 0.210
≈ 336,000
Therefore, the population after 65 years will be approximately 336,000 animals.
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