"Nitrous acid causes all possible transitions (A<-- -->G, C<-- -->T). If a protein has an alanine at position 65, what amino acid(s) would be likely to replace alanine as a result of nitrous acid mutagenesis?a. isoleucine

b. "valine, threonine"

c. threonine

d. "isoleucine, threonine, valine"

e. leucine

Answer:

Smooth cells are involuntary non-striated muscle cells.

Explanation:

They line the hollow organs like stomach,intestine,urinary bladder, uterus, arteries. They work automatically. They are of two types:

1. Visceral smooth muscle

2. Multi unit smooth muscle

The matching can be done as;

Matching is the assigning the right terms to the right definitions.

For instance, Scales serves as a internal support system that allowed for large, heavy bodies.

Amniotic egg serves as structure that develops within the mesoderm and allowed for more complex internal organ formation.

CHECK THE COMPLETE QUESTION BELOW;

Match each feature to its description.

1.) Scales

2.) pharyngeal gill slits

3.) notochord

4.) endoskeleton

5.) coelom

6.) Amniotic egg

A.) Internal support system that allows fir heavy bodies.

B.) feature that arose in reptiles and helped vertebrates to colonize land

C.) flexible structure that supports the nerve cord and led to the evolution of the backbone.

D.) structure that develops within the mesoderm and allowed for more complex internal organ formation.

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THE TWO MOST COMMON REASON FOR MODIFYING  THE  LAND IN SOUTH AND SOUTH ASIA IS

(i) first is significantly, the area of lands categorized as under forest/woodland and wetlands declined .At the same time, cultivated area increased .

(ii) another is requirement of additional land due to increase in population.

Explanation:

Between 1880 and 1980, the South and Southeast Asian landscape underwent dramatic modification.The 81% of the lost forest and wetland vegetation appears to have been converted into expanded agricultural land.

Human population in this region more than tripled between 1880 and 1980, producing an enormous demand for additional land for cultivation.

Answer:

C-  

agriculture and forestry

Explanation:

In a marine ecosystem, the leopard seal's most efficient prey from an energy transfer viewpoint would be the organism closest to it in the trophic level - likely fish or penguins. This is because less energy is lost to the environment each time it moves up a trophic level.

In a marine ecosystem, it's important to remember that energy is transferred from one trophic level to the next with a general efficiency of about 10%. This means that for a leopard seal, the prey that would provide the most efficient energy transfer would likely be the one closest to it in the food chain.

Let's say the leopard seal's potential prey includes krill, fish, and penguins. The krill, being primary consumers that eat producers (like phytoplankton), would have the most energy available. Next would be the fish that eat the krill, and then the penguins - secondary and tertiary consumers respectively.

Therefore, the most efficient prey for a leopard seal, from an energy transfer perspective, would be the one closest to it in trophic level - likely the fish or the penguins. This is because less energy is lost to the environment as it moves up each trophic level.

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Answer:

H. pylori uses the enzyme urease to breakdown urea into ammonia (NH3) & carbon dioxide (CO2), where NH3 can act as a buffer to the acidic solution in the stomach.

Explanation:

H. pylori is a bacteria that has the enzyme urease to breakdown urea into ammonia (NH3) & carbon dioxide (CO2). The compound of interest here would be ammonia, or NH3. NH3 is a base, although relatively weak to other stronger bases, which means it has a pH above 7. In the stomach, the pH is acidic, or below 7. By synthesizing ammonia, H. pylori is able to buffer the stomach solution in a manner so that it isn't entirely acidic, but more toward the basic side, thereby allowing for its survival.

Answer:Intermediates of a multistep reaction sequence do not dissociate from the enzyme complex.

Explanation:

Substrate channeling is the moving of intermediate metabolic product of one enzyme to another enzyme without it be lost to another enzyme.

Channeling makes metabolic pathway more fast and efficient that it will be when the enzyme are at random.

It avoid the use up of intermediate formed by other reaction catalyze by another enzyme.

For pyruvate dehydrogenase channeling prevent the intermediate from dissolving and been used up by other reaction.

Substrate channeling in the PDH complex improves reaction speed, minimizes side reactions, and sequesters reactive intermediates. It raises efficiency by avoiding the limitations of diffusion constants and reducing the dissociation of intermediates.

Substrate channeling in the Pyruvate Dehydrogenase (PDH) complex provides several advantages in biochemical reactions. The benefit of substrate channeling lies mainly in three aspects: improvement of reaction speed, minimization of side reactions, and sequestration of reactive intermediates.

Since the active sites of different enzymes are closely aligned in substrate channeling, the intermediates do not need to diffuse out into the aqueous solution before they are used by the next enzyme. This prevents reaction progress from being limited by the diffusion constant and increases the overall reaction speed. Furthermore, it prevents the intermediates of a multi-step reaction sequence from dissociating from the enzyme complex, thus minimizing the chance of side reactions.

Another advantage is that reactive or unstable intermediates are sequestered within the enzyme complex, preventing unwanted reactions with other cellular components. Therefore, substrate channeling enhances the efficiency and control of metabolic pathways.

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Answer:

the answer is the stage happens in the cytoplasm and glucose is broken down. IG....

Explanation:

Answer:

Evolution

Explanation:

Evolution consists of changes in the heritable traits of a population of organisms as successive generations replace one another

Answer:

The cells in a population die at a constant rate

Explanation:

Microbial death is the loss of the ability of microbes to reproduce and survive in an environment. When a given microbial population is given a treatment, the microbial cells die at a constant rate. Microbial death rate  is not dependent on the specie and nor on the antimicrobial agent.

Therefore, the microbial cells in a population does not die at once but die at a constant logarithmic rate; the cells decreases exponentially as nutrients decreases and waste product increases.

For example, if 500,000 microbes are treated or in a nutrient depleted environment and 50,000 microbes is left after 1 minute, by the next minute under the same condition 5,000 microbial cells will be left and this pattern will continue, this explains exponential decrease

Microbial death typically follows a pattern of exponential death, where the cells in a population die at a constant rate.

The pattern of microbial death can vary depending on the species and the antimicrobial agent used. However, in general, the cells in a population die at a constant rate, following a pattern known as exponential death. This means that the population shrinks by a constant fraction over a given time period, resulting in a gradual decrease in the number of viable cells.

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Answer:

Choose different ways to get around your city. Walk, bike, skateboard, scooter, take public transit, as many times a week as you can. Focus especially on those short trips–for example, buy a shopping trolley and walk to the grocery store if possible.

Answer: BY FOLLOWING THE BELOW GIVEN IDEAS WE CAN IMPROVE OUR CITY STREETS.

Explanation:

The quality of life in many European cities is affected by the

negative impacts of increasing traffic

levels. This chapter looks at ways in

which a dominance of car traffic

affects our lives in urban areas, and

suggests that there is a growing

consensus, from the global to the

local level, that the situation is

unsustainable.

technically challenging and politically

sensitive planning option in urban

areas where road congestion is

already a problem.This chapter

brings together best practice from a

wide range of expertise and

experience in dealing with these

issues, in particular that drawn from

the schemes described in this

document. The objective is to assist

politicians and planners working to

develop more sustainable transport

strategies for Europe’s towns and

cities.

The traditional response to the problem of traffic congestion has

been to increase the road space  available for cars. In this chapter, the

theory of ‘traffic evaporation’ is

explored as a concept which

challenges the logic of this approach.

This theory supports the proposition

that reducing road capacity for cars

in congested city centres can

represent a sustainable, efficient

planning solution. In addition, once

freed from domination by car traffic,

reclaimed urban spaces can become

accessible, vibrant ‘living’ places.

• Kajaani, Finland

• Wolverhampton, England

• Vauxhall Cross, London, England,

• Nuremberg, Germany

• Strasbourg, France

• Gent, Belgium

• Cambridge, England

• Oxford, England

This chapter presents the experiences of a small selection of

European cities where urban

planners, with the political support of

local leaders, have had the vision and

the courage (often in the face of

considerable opposition) to take

away congested road space from

private cars. In each case study, after

an initial settling-in period, the

predicted traffic chaos did not

materialise and some of the traffic

‘evaporated’.

Answer:

the correct option is : "translation sometimes begins before transcription is complete because prokaryotes have no nucleus."

Explanation:

Prokaryotic cells do not have a membrane around the nucleus, or around their organelles, and store their genetic material in a single large DNA molecule. All prokaryotic organisms are single-celled, while eukaryotes can have one or more cells

Answer:

1. commensalism ( A )

2. mutualism ( C )

3. Parasitism ( B )

4. amensalism ( D )

Explanation:

Mutualism ; in this kind of relationship both parties benefit.

commensalism ; in this kind of relationship on party benefits without causing harm or damage to the other.

Parasitism ; Only one party benefits here by inflicting damages on the other.

amensalism ; a relationship between two different organisms in which one inhibits or destroy the other without been affected.

Answer:

1. commensalism "Aerobic bacteria in the human colon consume oxygen, making it possible for anaerobic species to survive"

2. mutualism "Flagellates live in the gut of termites, feeding on the wood consumed by the termite and allowing the termite access to nutrition and energy in wood that they could not otherwise digest"

3. parasitism "A helminth takes up residence in a human digestive tract, consuming nutrients"

4. amensalism "A mold produces a chemical that kills bacteria without apparently benefiting"

Explanation:

1. commensalism: one species benefits while the other species neither benefit nor prejudiced  

2. mutualism: two species benefit

3. parasitism: one species benefits while the other species is prejudiced  

4. amensalism: one species is unaffected while the other species is prejudiced

Answer:

A. The tails do not interact with the DNA

Explanation:

The acetylation refers to the transfer of the acetyl group from Acetyl-CoA to the N-terminal of the histone protein.

Lysine residues (positively charged amino acid) are present at the end of the N-terminal of the histone protein which is neutralized by the acetyl group.

This loses the compaction between the positively charged histone and the negatively charged DNA and the DNA becomes more relaxed. This relaxed state allows the transcription factors to easily bind the DNA and therefore the DNA becomes transcriptionally active.

Thus, Option-A is correct

Answer:

Explanation:

Radiata

In the early part of the 19th century, Echinodermata was recognized as a distinct group of animals and was occasionally associated with the cnidarians and selected other phyla in a division of the animal kingdom known as the Radiata; the concept of a superphylum called Radiata is no longer valid.

Answer:

Plants, animals, or pathogens that are non-native (or alien) to the ecosystem under consideration and whose introduction causes or is likely to cause harm.”

Explanation:

so, yes

Answer:

No Humans are not ivasive species

Notable examples of invasive plant species include the kudzu vine, Andean pampas grass, and yellow starthistle. Animal examples include the New Zealand mud snail, feral pigs, European rabbits, grey squirrels, domestic cats, carp and ferrets.

The answer is [ B. The Calvin cycle will produce less glucose. ]

A plant needs to take in carbon dioxide to produce glucose. So, if a plant were to not receive enough CO2, then the plant will produce less amounts of glucose. Glucose is created through the photosynthesis process. Plants use the sun to create glucose and oxygen from water and CO2.

Best of Luck!

Answer:

B

Explanation:

Answer:

photosynthesis

Explanation:

please rate and mark as brainliest

Answer:

RR - homozygous red

Rr- heterozygous red

rr - homozygous white

Explanation:

Given -

Allele for red color of flower is "R" and allele for white color flower is "r"

Also, R is dominant over r

In the F1 cross a homozygous red plant is crossed with a homozygous white plant.

Genotype of homozygous red plant - RR

Genotype of homozygous white plant - rr

RR * rr

Rr, Rr, Rr, Rr

In the F2 cross, offsprings of F1 generation are crossed.

Rr* Rr

RR, Rr, Rr, rr

Genotype & Phenotype

RR - homozygous red

Rr- heterozygous red

rr - homozygous white

Answer:

Explanation:

Bacteria can be classified traditionally into two broad categories according to their cell wall, which are Gram-positive bacteria and Gram-negitive bacteria.

Gram-positive bacteria are bacteria that produce a positive result when Gram stain test is performed. It takes up the crystal violet stain of the test, and then show a purple-coloured appearance when seen through an optical microscope. This is as a result of the thick "peptidoglycan" layer in the bacterial cell wall which retains the stain used in the cell after washing it away from the rest of the sample.

Gram-negative bacteria after the decolorization don't retain the violet stain,the outer membrane of gram-negative cells is degraded by the alcohol.The peptidoglycan layer is positioned and between the membrane and a bacterial outer membrane, causing them to take up the counterstain and made appear red or pink.

Therefore, Deinococcus is one genus of the bacterial phylum that have resistant to environmental hazards. They posses thick cell walls that give them Gram-positive stains, but they also posess second membrane that made them them closer in structure to Gram-negative bacteria, which present an interesting conundrum to scientists who try to classify them as either gram-positive or gram-negative.

Which disease can destroy red blood cells?

I) Hemolytic anemia is a decease which destroys red blood cells .

Which disease infects T cells?

ii) chromosomal breakage syndromes (CBSs) is one one of those decease that infect T cells.

Which disease produces a scaly, stinging, and itchy rash on the feet?

iii) Eczema , it's a skin disorder .

thanks

Answer:

b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.

Explanation:

The expected outcome if branch migration occurs is that mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.

Final answer:

The correct option is b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele. In homologous recombination involving a heterozygote with alleles differing by a single base pair, mismatch repair following branch migration may convert the D allele to a d allele or vice versa, leading to gene conversion.

Explanation:

When homologous recombination occurs in a heterozygote where the alleles D and d differ by a single base pair, the Holliday junction and branch migration can lead to heteroduplex formation across the differing positions. If the D allele has a G (a GC base pair) and the d allele has a C (a CG base pair) at a particular position, branch migration across this position may result in mismatches of G-G and C-C in the heteroduplex region. The process of mismatch repair will recognize these mismatches and can potentially convert a D allele to a d allele or vice versa. This could result in gene conversion, where one allele is changed to match the sequence of the other without affecting the flanking sequences.

Therefore, the correct option is b. Mismatch repair will identify abnormal G-G and C-C base pairs and may convert the D allele to a d allele or the d allele to a D allele.

Answer:

b. Prions respond to environmental changes by bonding with other prion variants in the population and potentially forming new variants able to handle the change.

d.Environmental changes cause some prions to break into small segments that may be better able than larger prions to survive and reproduce in the new environment.

Explanation:

The first hypothesis that tried to explain how prions replicate in a protein-only manner was the heterodimer model. This model assumed that a single PrPSc molecule binds to a single PrPC molecule and catalyzes its conversion into PrPSc. The two PrPSc molecules then come apart and can go on to convert more PrPC.

prion hypothesis states that the misfolded prion protein, rather than virus or bacteria, is the infectious agent that results in the transmission of prion disease.

Answer:

These are lands which are set aside for the purpose of increasing wildlife numbers by protecting wildlife and key habitat. One of the major goals of a wildlife management areas is to protect at least a minimal number of animals so the population can increase.

Explanation:

Hope this helps

Answer:

It entirely depends on location and what wildlife you are trying to protect. The obvious response would be the land that best suits the wildlife you are trying to protect. For example, the most important areas to southeastern beach mice are the dune areas at beaches they are found.

Explanation:

Answer:

a)  Cell concentration when the dilution rate is one-half of the maximum is  0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex]   is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d)  the cell productivity at 0.8  [tex]D_{max}[/tex]   is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield  on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

[tex]D = \frac {D_{max}[S] }{ks+[S]}[/tex]

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

[tex]0.5 = \frac{S}{1.3+[S]}[/tex]

[tex]\\\0.65=0.5[S][/tex]

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

b)

Substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex]    is calculated as:

[tex]D = \frac {D_{max}[S] }{ks+[S]}[/tex]

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore ,  the substrate concentration when the dilution rate is 0.8 [tex]D_{max}[/tex]   is 5.2g/l

c)

Maximum dilution rate is calculated using the expression [tex]D_{max} = \frac{1}{time}[/tex]

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

d)

The cell productivity at 0.8 [tex]D_{max}[/tex] can be calculated by multiplying the amount  of the cell yield with the amount of the substrate consumed at 0.8[tex]D_{max}[/tex]  

Cell yield = [tex]\frac {cell \ productivity \ at \ 0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}[/tex]

Cell productivity at 0.8 [tex]D_{max}[/tex]    = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8  [tex]D_{max}[/tex]   is 2.40g cell/L

To find the cell concentration at half the maximum dilution rate, calculate the biomass concentration using the Monod equation. Use the same equation to find the substrate concentration when the dilution rate is 0.8 Dmax. The maximum dilution rate can be determined using the saturation constant and cell yield on acetate. Cell productivity at 0.8 Dmax is found by multiplying biomass concentration by the dilution rate.

To find the cell concentration when the dilution rate is one-half of the maximum, you need to calculate the biomass concentration in the chemostat. This can be done using the Monod equation, which relates the specific growth rate of the bacteria to the substrate concentration. When the dilution rate is one-half of the maximum, the specific growth rate is also half of the maximum growth rate. Using the Monod equation, you can then find the corresponding biomass concentration.

To find the substrate concentration when the dilution rate is 0.8 Dmax, you can use the Monod equation again. This time, you need to rearrange the equation to solve for the substrate concentration, given a specific growth rate. The specific growth rate at 0.8 Dmax can be determined by multiplying the maximum growth rate by 0.8.

The maximum dilution rate is the dilution rate at which the specific growth rate is at its maximum. To find it, you need to determine the maximum specific growth rate. This can be done using the saturation constant and cell yield on acetate. Once you have the maximum growth rate, you can determine the dilution rate at which it occurs.

The cell productivity at 0.8 Dmax can be calculated by multiplying the biomass concentration at 0.8 Dmax by the dilution rate at that point.

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Answer:

it stands for deoxyribonucleic acid,

Explanation:

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Answer:

The answer is D) A cell is the smallest unit of an organism that is considered alive.

Explanation:

The statement that a cell is the smallest unit of an organism that is considered alive is true about living things.

Hence, the correct option is d. A cell is the smallest unit of an organism that is considered alive.

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Answer:

The mentioned symptoms exhibited by the woman in the given question show that damage has occurred in the occipital and the frontal lobe, which is, primarily in the cerebral cortex. The cerebral cortex is the place where memory processing takes place. Thus, any kind of damage to the mentioned parts of the brain will eventually result in personality changes, loss in memory, and mental confusion.  

In the given case, the cranial nerves getting affected will be the one taking part in the visual activity, that is, optic nerves and oculomotor nerves. Damage to these cranial nerves would have resulted in the visual conditions as mentioned in the given case.  

Answer:

1) check the attached file below

2)  recognition spacer sequences are separated by 12 or 23 base pairs

Explanation:

Answer:

coevolution

abiogenesis

Endosymbiont Theory

It's estimated that over 99 percent of the species that have existed on Earth at some point in time are extinct today.

Coevolution implies that the evolution of one species is dependent on and works in relation to the evolution of another species. This may cause positive or negative impacts and could be beneficial to both organisms or only to one.

This theory states that the first building block of life that allowed for reproduction of organisms was the development of self-replicating RNA. This hasn't been able to be fully demonstrated in any science experiment but is based on the idea that there are RNAs that can catalyze biochemical reactions on their own without proteins.

This process points to the development of structures through the envelopment of smaller cells that perform specific functions. This is how eukaryotic cells evolved from prokaryotic cells.

This experiment was important because it showed that, in the right primordial soup, organic compounds could develop from inorganic compounds.

Explanation:

penn foster

1. coevolution

2. abiogenesis

3. Endosymbiont Theory

Coevolution occurs when two or more species reciprocally affect each other's evolution through the process of natural selection. The term sometimes is used for two traits in the same species affecting each other's evolution, as well as gene-culture coevolution.

Abiogenesis or the origin of life is the natural process by which life has arisen from non-living matter, such as simple organic compounds.

The endosymbiotic theory states that some of the organelles in eukaryotic cells were once prokaryotic microbes. Mitochondria and chloroplasts are the same size as prokaryotic cells and divide by binary fission. Mitochondria and chloroplasts have their own DNA which is circular, not linear.

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Answer:

B. The liquid is less dense than water .

Explanation:

If something is more dense it will sink and if something is less dense it will rise, therefore the ice is more dense than the liquid. If the ice is more dense than the liquid is less dense.