which of the following is an example of horizontal transmission? choose one: a. pcr b. mitosis c. transformation d. cell division

Answer:

Saturated

Explanation:

Plants are organisms made up of cells that have both a cell membrane and a cell wall.

PLANTS:

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Most people who are lactose intolerant are not able to produce lactose as an adult, reducing their ability to digest the sugar lactose, the enzyme model that best represents lactose intolerance is the lock and key model.

This lock and key model illustrates how enzymes work by showing that enzymes have a specific shape that fits with a specific substrate (or molecule) like a key fits into a lock. In the case of lactose intolerance, the enzyme lactase, which breaks down lactose into simpler sugars, is not produced or is produced in insufficient amounts. This means that lactose, the sugar found in milk and dairy products, cannot be properly digested and can lead to symptoms such as bloating, gas, and diarrhea.

The lock and key model shows how lactase, like all enzymes, has a specific shape that allows it to fit with lactose and break it down into smaller molecules. Without this enzyme, lactose cannot be broken down and the body cannot properly digest it, this leads to the symptoms of lactose intolerance. Most people who are lactose intolerant are not able to produce lactose as an adult, reducing their ability to digest the sugar lactose, the enzyme model that best represents lactose intolerance is the lock and key model.

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Answer:

where is the 20 points!!!!!!!

These pigments ( chlorophyll b and carotenoids ) can absorb more light wavelengths and, consequently, more energy than chlorophyll an alone can. They increase the range of light that can be employed in the light reactions since they are a component of photosystems' light-harvesting complexes.

More wavelengths of light (and hence more energy) can be absorbed by these pigments than by chlorophyll an alone. They increase the range of light that can be employed in the light reactions as a component of the light-harvesting complexes in photosystems.

Instead, they deliver the excitation energy straight to the chlorophyll molecules, which subsequently deliver it to the reaction centers and photosynthetic pathway. Chlorophyll and carotenoids combined make form the light-harvesting antenna within cells, making carotenoids known as accessory pigments.

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Answer:

by water and sun

Explanation:

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Answer:

The skydiver will stop accelerating, and will travel at a constant speed in one constant direction.

The consequence of treating the vector, before ligation, with calf intestinal phosphatase will be B. It would prevent the ends of the plasmids from being ligated.

Calf intestinal phosphatase is a phosphatase enzyme that is derived from the intestine of the calf. The function of this enzyme is to remove the phosphatases present in the 3' and the 5' of a DNA segment by cleaving them.

Vectors, such as plasmids, are treated with the calf intestinal enzyme in order to prevent the plasmid from being ligated again. In order to add our gene of interest to the vector, the calf intestinal phosphatase is added so that the vector binds to the gene of interest rather than itself.

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The correct difference between graded potentials and action potentials is that the magnitude of action potentials decreases with distance, while graded potentials do not exhibit this property.

The statement that is not a difference between graded potentials and action potentials is:

Greater stimulus intensity results in larger graded potentials, but not larger action potentials.

In reality, both graded potentials and action potentials can exhibit an increase in magnitude with a stronger stimulus. This property is known as "recruitment" or "summation," where stronger stimuli can lead to larger responses in both graded potentials and action potentials.

Here are the correct differences between graded potentials and action potentials:

Graded potentials can result from the opening of chemically gated channels, whereas action potentials require the opening of voltage-gated channels. Graded potentials are local changes in membrane potential that can be either depolarizing (excitatory) or hyperpolarizing (inhibitory) and occur in response to neurotransmitter binding or other chemical stimuli. In contrast, action potentials are rapid, all-or-nothing depolarizations that occur when the threshold is reached and voltage-gated ion channels are activated

The magnitude of action potentials decreases as the impulse travels further away from the start of the impulse, while graded potentials do not decrease in magnitude. Action potentials exhibit a characteristic "propagation" along the axon, where the strength of the signal diminishes with distance due to the passive spread of ions. Graded potentials, however, are localized changes in membrane potential and do not propagate along the membrane like action potentials.

Graded potentials occur along dendrites, whereas action potentials occur along axons. Graded potentials are typically generated in response to synaptic inputs at the dendrites or cell body of a neuron. They can summate spatially and temporally to influence the generation of an action potential. Action potentials, on the other hand, are the propagated electrical signals that travel along the axon of a neuron to transmit information over long distances.

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Answer:

Ribosome, this is because they are responsible for protein synthesis

Answer:

Explanation:

. And

It's important to note that the distribution and organization of nerve fibers within the skin are complex and can vary among individuals. Additionally, other factors such as receptor types, nerve fiber diameter, and central processing also influence the overall sensation experienced.

As an AI text-based model, I'm unable to directly examine or view images. However, I can provide some general information regarding the impact of nerve fiber location on sensation in the skin.

The cutaneous innervation refers to the distribution of nerves throughout the skin, which allows for the perception of various sensations such as touch, pressure, temperature, and pain. The location of nerve fibers within the skin can have a significant impact on sensation in several ways:

1. Sensory Threshold: Different areas of the skin have varying densities of nerve endings, leading to variations in sensitivity. For example, areas with a high concentration of nerve fibers, such as the fingertips or lips, tend to have a lower sensory threshold, meaning they can detect even subtle stimuli.

2. Sensory Discrimination: Nerve fibers are responsible for transmitting signals related to different sensory modalities. In specific regions, such as the fingertips, the nerve fibers are densely packed and finely tuned, allowing for precise discrimination of tactile stimuli or small differences in texture.

3. Receptive Field Size: Nerve fibers have distinct receptive fields, which are the specific areas of the skin where stimulation activates a particular nerve ending. Receptive field sizes can vary depending on the location in the skin. Smaller receptive fields, found in areas like the fingertips, allow for more precise localization of touch.

4. Sensory Adaptation: Nerve fibers exhibit different levels of adaptation to prolonged or repetitive stimulation. Rapidly adapting fibers are more responsive to changes in stimuli, while slowly adapting fibers maintain their response over a prolonged period. This adaptation can affect how sensations are perceived in different areas of the skin.

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Given data: Adult dosage of a substance is 280mg. Rule for calculating dosage for children between 1 to 12 years of age is given by the function, D(t) = a * t / t + 12. To find the rate of change of this function with respect to the child's age, we need to differentiate the function with respect to t.

Let's differentiate this function with respect to t. d/dt [ D(t) ]= d/dt [ a * t / t + 12 ]

Using quotient rule,= [ a * (t + 12) - a * t ] / ( t + 12 )²= a / ( t + 12 )²

Thus, the rate of change of child's dosage with respect to child's age is given by D'(t) = a / ( t + 12 )².

Hence, the required expression is D'(t) = 280 / ( t + 12 )².

Now, substituting t = 3,

we get, D'(3) = 280 / (3 + 12)²

= 280 / 225

= 1.244 mg/year

Substituting t = 12,

we get, D'(12) = 280 / (12 + 12)²

= 280 / 576

= 0.486 mg/year

Therefore, the rate of change of a child's dosage with respect to his or her age for a 3-yr-old child is 1.244 mg/year and for a 12-yr-old child is 0.486 mg/year. We are given the formula, D(t) = a * t / t + 12, which represents the dosage for a child as a function of their age, t. To find the rate of change of this function with respect to the child's age, we need to differentiate the function with respect to t.

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When the native proteins are separating and we shine the UV light on the gel, the bands are faint. The reason is that native proteins maintain their three-dimensional shape, which enables them to fold back on themselves and protects their hydrophobic core from interaction with the negatively charged SDS molecule.

The SDS-PAGE buffer has a denaturing agent, which causes the protein to denature, lose its three-dimensional shape, and have a uniform negative charge distribution.The migration of proteins that are exposed to a denaturing agent, such as sodium dodecyl sulfate, is based on the length of their polypeptide chains. The bands are darker when proteins are denatured because the denaturation process eliminates their tertiary structure and results in a uniform negative charge distribution, making the SDS-protein complex size-dependent. As a result, the negatively charged SDS binds to the protein, giving it a negative charge, which causes it to migrate through the gel at a rate proportional to its length and charge-density.

Hence, we can conclude that native proteins maintain their three-dimensional shape, and the SDS-PAGE buffer has a denaturing agent that causes the protein to denature, lose its three-dimensional shape, and have a uniform negative charge distribution, which is the main reason for the difference in what we see in the denatured protein lane compared to the native protein lane when we shine the UV light on the gel.

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An elastic lamina on both sides of the tunica media is a characteristic of C. muscular arteries.

In the blood vessel hierarchy, muscular arteries (also known as distributing arteries) are characterized by having an elastic lamina on both sides of the tunica media.

This structural feature allows them to better regulate blood flow and pressure within the circulatory system.

Summary: The presence of an elastic lamina on both sides of the tunica media is a distinct characteristic of muscular arteries, making option C the correct answer.

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Answer:

C Able to live through changes in temperature

Explanation:

A primer must bind to the template DNA strand to serve as a starting point for polymerization of a new strand.

Answer:

Cell walls are structures that surround and protect the cells of some organisms. According to the web page context, cell walls can be found in:

Cell walls are not found in animals or most protists.

a. The overall colonization rate of the entire metapopulation. The colonization rate of the entire metapopulation is the answer.

The metapopulation equation: dP/dt=[cP(1−P)]−ePHere, cP(1-P) is the birth term, which is the product of the colonization rate and the number of individuals in patches not occupied yet. eP is the death term, which is the product of the extinction rate and the number of individuals in patches occupied but about to go extinct. Thus, the equation gives us the number of patches that are empty and the patches with checkerspot butterflies. Therefore, the overall colonization rate is 2.3. b. The overall extinction rate of the entire metapopulation. The extinction rate of the entire metapopulation is 1.3. This is because 1.3 patches of the occupied 9 patches are going to become extinct.

c. Are the colonization and extinction rates likely to be at equilibrium with each other? How do you know? Both colonization and extinction rates are unlikely to be in equilibrium with each other because the colonization rate is higher than the extinction rate. Therefore, the occupied patches' population will continue to increase until the patches are fully occupied, while the number of empty patches will continue to decrease until the patches are fully occupied.

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Answer:

The answer is A.

Explanation:

Corn has gone through many genetic mutations and adaptations from both humans and from their environment.

Answer:

A is the answer

Explanation:

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Restriction enzymes, also known as restriction endonucleases, cut DNA at specific sites called recognition sites.

These recognition sites have distinct nucleotide sequences, typically 4-8 base pairs long, that the enzyme can identify.

When the restriction enzyme encounters its target sequence, it binds to the DNA and cleaves the phosphodiester bonds between the nucleotides, resulting in the separation of the DNA strands.

This process is highly specific, ensuring that the enzyme only cuts at the intended recognition site rather than cutting DNA randomly or at sequences with a high concentration of adenine bases.

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In the given scenario of a unique species of mouse where two genes c and t control the overall fur phenotype and both the genes are X-linked and located in the heterochromatin region, the proposed patterns of heterochromatin are represented as wavy lines.

The c+ allele produces the wild type brown coat color and c allele produces white coat color. The t+ allele produces the wild type narrow stripes, and the t allele produces thick stripes. Now, let's match each coat pattern to the expected coat phenotype of the mice as follows:Answer:A mouse with genotype C+T+ produces brown coat color with narrow stripes.A mouse with genotype C+T produces brown coat color with thick stripes.

A mouse with genotype CT+ produces white coat color with narrow stripes.A mouse with genotype CT produces white coat color with thick stripes.Therefore, the pattern of heterochromatin shown in the figure (d) would theoretically never be observed because the c gene is located in the region of heterochromatin and, hence, it will be inaccessible to regulatory proteins. Hence, the genes will not get expressed in the expected way.

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Answer:

Urbanization has environmental and economic advantages for humans, but also substantial disadvantages, including increased problems with physical and mental health/stress, a modified climate, water and energy management, and air pollution.

Answer:

The degree of genetic variety within a population and the conditions that drive genetic change are referred to as the population's generic makeup.

Explanation:

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Answer: mitosis

Explanation: Binary fission is similar to mitosis in the way that the process ultimately leads to the production of two identical daughter cells. However, they differ in many aspects. While binary fission is for reproductive purposes mitosis is primarily for growth in multicellular organisms.

Answer:

A decrease in all the prey populations.

Answer:

a decrease in all the prey populations

Explanation:

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When stimulus control is not developing, an intervention procedure called differential reinforcement of alternative behavior (DRA) can be used.

DRA is a type of behavior modification that is used to reinforce an appropriate behavior while not reinforcing an inappropriate behavior. This procedure is typically used when a behavior is not responding to other forms of reinforcement or is not being shaped by antecedent manipulation. The goal of DRA is to increase the likelihood of the appropriate behavior occurring in the presence of the antecedent while decreasing the likelihood of the inappropriate behavior. DRA can also be done along with other procedures like shaping, classical or operant conditioning, etc. A therapist or a professional in behavior analysis should be consulted to design and implement an effective DRA program.

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