Based on the article, which tissue do you think is affected by polio?

Answer:

A. The haploid set of n is ABC

B. ABC, ABc, AbC, aBC, abc, abC, aBc, Abc.

Explanation:

A diploid cell is a cell that comprises of two complete sets of chromosomes. This is double the haploid chromosome number. Each pair of chromosomes in a diploid cell is noted to be a homologous chromosome set. A homologous chromosome pair consists of one chromosome donated from one parent and one from the other parent. This number is represented as 2n. It varies in different organisms. A diploid cell replicates via mitosis. It preserves its diploid chromosome number by making an identical copy of its chromosomes and distributing its DNA in an equal manner between two daughter cells.

Haploid refers to a cell that contains a single set of chromosomes. Gametes are made up of half the chromosomes contained in normal diploid cells of the body, these are also known as somatic cells. Haploid gametes are produced during meiosis. Meiosis is a type of cell division that reduces the number of chromosomes in a parent diploid cell by half.

Therefore,

Karyotype of this organism is designated as AaBbCc.

AaBbCc is the diploid set (2n).

Hence ABC or abc are the haploid set (n).

B. The eight different combinations of these chromosomes that could possibly be produced by spermatogenesis in this male are:

ABC, ABc, AbC, aBC, abc, abC, aBc, Abc.

Answer:

the answer is useing energy

Explanation:

the answer is useing energy bc a lion is a natural enemy of zebras. lions are very fast running animals so by running at extremely high speed,they preform hunting of zebra. by doing so,they utilize energy in getting their prey which is there meal. hope this helps!!:)

A lion eating a zebra exemplifies the characteristics of life, including nutrition, growth, response to stimuli, reproduction, homeostasis, and the role of evolution in adaptation to its environment.

The act of a lion eating a zebra exemplifies several key characteristics of life, primarily pertaining to the lion as an organism:

1. Nutrition: The lion obtains nourishment by consuming the zebra, demonstrating its ability to ingest and digest food.

2. Growth and Development: The lion's body will process the zebra's nutrients, contributing to its own growth and maintenance, a fundamental aspect of life.

3. Response to Stimuli: The lion's predatory behavior, hunting, and eating are responses to stimuli such as hunger and the presence of prey.

4. Reproduction: While not directly observed in this scenario, the lion's capacity to reproduce is a vital aspect of its life cycle.

5. Homeostasis: After eating, the lion must maintain internal balance (homeostasis) in terms of temperature, blood sugar, and other physiological factors.

6. Evolution: Over generations, the lion's predatory traits and behaviors may evolve in response to environmental pressures and changes in prey availability, highlighting the role of evolution in life.

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The complete question is:

A lion eating a zebra is an example of which characteristics of life what?

Answer:

The answer is C hope this helps+

Answer:

The answer is C.

Explanation:

Don't forget that the human skeleton or backbone protects delicate organ such as the heart and lungs, which are protected by the ribs. It also protects structures such as the spinal cord, which is protected by the vertebral column.

Answer:

More than one gene is controlling the loss train of that eye. Thus, the Cross among the scarlet and grey species, F1 crossover has fractional eyes. Despite the fact that both sort of cavern fishes are visually impaired, yet cross between them has brought about offspring with fractional eyes. This shows multi-gene legacy for eye misfortune attribute. Blending of two genotypes has brought about incomplete characteristic in the descendants.

Answer:

Main protein in ending high fidelity in E. Coli is the Tus protein that binds to Ter sequences in order to prevent replication forks from passing through the end region. In the Ter sequences, the Tus protein blocks replication by establishing a close association with a particular G-C base pair.

The main protein in human cells is telomerase, which contains an RNA primer and is required to extend the synthesis of lagging strands in linear chromosomal telomeres.

The key E. coli protein involved in the termination of DNA replication is the Tus protein, while in humans, it is the telomerase enzyme.

 Explanation:

In Escherichia coli (E. coli), the termination of DNA replication is a highly regulated process that ensures the accurate duplication of the circular genome. The key protein involved in this process is the Tus (terminus utilization substance) protein. The Tus protein binds to specific sequences in the DNA called Ter sites (termination sites), which are located opposite the origin of replication on the circular chromosome. When the replication forks approach these Ter/Tus complexes, the Tus protein acts as a polar block to the progress of the replication machinery, causing the forks to stall and eventually terminate replication. This interaction is essential for the coordinated termination of replication and helps to prevent the formation of tangled or incompletely replicated DNA molecules.

 In contrast, human cells have linear chromosomes, and the termination of DNA replication is not as straightforward as in bacteria. The ends of linear chromosomes are capped by telomeres, which are repetitive nucleotide sequences that protect the chromosome ends from degradation and fusion. The key protein involved in maintaining telomeres is the telomerase enzyme. Telomerase is a reverse transcriptase that adds telomeric repeats to the ends of chromosomes, compensating for the shortening that occurs during DNA replication due to the end-replication problem. This ensures that the ends of the chromosomes are maintained and that the cell can continue to divide without losing critical genetic information. The high fidelity of DNA replication termination in humans is thus maintained by telomerase activity, which is crucial for chromosomal stability and cell viability.

  In summary, while the Tus protein in E. coli ensures the accurate termination of replication in a circular genome by blocking replication forks, the telomerase enzyme in humans maintains the integrity of linear chromosome ends, allowing for high-fidelity replication and preventing genomic instability. Both proteins are essential for the faithful duplication of their respective genomes."

Gram's iodine reagent must be added to the plate before examination of amylase production.

Explanation:

The starch hydrolysis test is conducted to test the presence of the enzyme amylase in the test medium.

These are confirmatory lab tests done to detect and identify the presence of bacteria which can hydrolyze starch like amylose with the help of enzymes like amylase.

The test medium for this test is agar medium in a petri dish, where soluble starch is first added to initiate the microbial growth. Once incubation period of the microbes is over, dilute iodine solution is added to the petri dish in increased quantity. Iodine is a dye which helps to clearly identify the areas which are hydrolyzed by amylase and those which are not with the help of its color.

Answer:

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example,

What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?

If 1 of 15 individuals are carriers for this disorder, this means that 1/15 are heterozygous, 2pq. So, 2pq = 1/15 = 0.066

Now we must calculate the allelic frequencies.

We know that 1 in 15 individuals are heterozygous, and we also know that there are no recessive homozygous, q², because they can not survive, so of the 15 individuals only one is heterozygous and the rest 14 individuals must be dominant homozygous, p².

The dominant homozygous genotypic frequency is

p²= 14/15 = 0.933

And by clearing the next equation we can get the allelic frequency for p

p²= 0.933

p = √0.933

p = 0.967

So now we know that the allelic frequency for p is 0.967  

This means that the allelic frequency for q or p is 0.033, which we deduce by clearing the equation p + q = 1

                          0.967 + q = 1

                         q = 1 - 0.967

                          q = 0.033

Answer:

In an antibiogram, the larger the inhibition halo, the more sensitive bacteria will be to the drug that is exposed, that is why if the halo is 25 mm, it is considered quite wide, so these drugs such as chloramphenicol and amoxicillin affect the bacterial development in marine culture both bactericidal or bacteriostatic.

Both have a 25mm halo, so both would be SAME as specified.

Explication:

suppose that one of the two drugs has a smaller inhibition halo in the culture, that drug will be less effective in treating this bacterium, therefore more would allow its development, on the contrary, the larger halo would be the most effective drug (This is an example for you to understand that the greater the length of the inhibition halo, the greater the efficacy of the drug, and the lower the halo, the less efficacy)

Answer:

Option B

Explanation:

In any food cycle or ecosystem, the top most organisms are the decomposers. They are kept above the tertiary species because when the tertiary species die, these decomposers or micro-organisms feed on them and convert organic material with in them into reusable form/nutrients/chemicals.  

These nutrients get mixed with the soil and are reused by the plants and trees (producers) for making food for all other organism in the food chain by entrapping solar energy.  

Hence, option B is correct

Its D. Sunlight. Since plants are producers they get their energy from the sun

Answer:

Scientists may study and discuss organisms more easily

Explanation:

Yes that's the point!

Answer:

Its C

Explanation:

I did the quiz!!

Answer:

Although all flowers are different, they have several things in common that make up their basic anatomy. The four main parts of a flower are the petals, sepals, stamen, and carpel (sometimes known as a pistil). If a flower has all four of these key parts, it is considered to be a complete flower.

Explanation:

(Hope you have a good day. Stay Safe!)

Answer:

e. meiosis → gametes → zygote → mitosis → embryo

Explanation:

In the vertebrates, the life cycle alternates between the diploid and haploid phase. The vertebrate body is diploid and the haploid phase is only observed during the gamete formation.

The sex organs of the vertebrates produce gametes from the gamete mother cell through meiosis, which reduces the chromosome number to half and produce four haploid gametes.

The male and female gametes undergo fertilization event and form a diploid zygote. The zygote develops into the embryo through mitosis as mitosis produces the cells with an equal number of chromosomes.

Thus, Option-E is correct.

Answer:

- many plants will die w/out sunlight

- some animals will adapt over time

- many species will eventually die off

Answer:

3,4 and 5

Explanation:)

Answer:

Yes, if the touch their snot and touch you or an object you touch. They can also sneeze their existing snot at you if they didn't wipe it.

Yes, someone with a runny nose can spread their snot to you just by being close. The common cold, which often causes a runny nose, is typically caused by viruses like rhinoviruses, coronaviruses, and adenoviruses. These viruses can be transmitted through direct contact and droplet transmission, such as when an infected person coughs or sneezes and produces infectious aerosols.

Yes, someone with a runny nose can spread their snot to you just by being close. The common cold, which often causes a runny nose, is typically caused by viruses like rhinoviruses, coronaviruses, and adenoviruses. These viruses can be transmitted through direct contact and droplet transmission, such as when an infected person coughs or sneezes and produces infectious aerosols.

When an infected person coughs or sneezes, mucus droplets are released into the air. These droplets can remain suspended in the air for some time and travel considerable distances. If you are close to the infected person, you can breathe in these droplets or they can land on surfaces that you touch, leading to potential transmission.

To minimize the risk of transmission, it is important to practice good respiratory hygiene, such as covering your mouth and nose with a tissue or your elbow when coughing or sneezing, and regularly washing your hands with soap and water or using hand sanitizer.

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Answer:

The chances of producing children that will be non-tasters is 1/4.

Explanation:

The ability to taste is determined by T which can either be TT or Tt. Non tasters can only be inherited in the recessive condition (tt).

A man who is a taster with a non taster mother is heterozygous for the trait (Tt). He marries a woman who is a taster (she is a TT or Tt since she has siblings that are non tasters). To have a higher percentage of non-tasters in her (the woman) family, it means, one of the parent is heterozygous for the trait and the other is homozygous recessive for the trait. Thus, the taster woman will be an heterozygote.

Now the taster man that is heterozygous for the trait marries this heterozygote woman, the chances of producing children that will be nontasters is 1/4.

Answer:

1/4 or 25%

Explanation:

A man who is a taster and has a non taster mother is heterozygous and will have the genotype Tt.

A woman who is a taster will have the genotype Tt or TT. However, the woman has 5 siblings out of which 3 are non-tasters. It means that her parent had 6 children out of which 3 are tasters and 3 are non-tasters (1:1).

A 1:1 result is usually from a cross involving one homozygous non-taster and one heterozygous taster. Hence;

Tt   x   tt = Tt, Tt, tt and tt

It thus means that the taster woman is heterozygous with genotype Tt.

A marriage between Tt and Tt:

Tt   x   Tt = TT, 2Tt and tt

Recall that the trait is is a dominant one, hence TT and Tt are tasters while tt is a non-taster.

The chances of the children being a non-taster therefore is 1/4 or 25%.

Answer:

Explanation:

1. There are six lanes in total out of which three are being showing self-splicing activity and the remaining three are being provided with splicing enzymes. As can be seen in case of RNA A, in the absence of a nuclear extract or when no splicing enzymes are added, there would be no splicing which can be reflected in lane 1 showing formation of only a single band which indicates that splicing does not occur on its own, although when the nuclear extract containing the splicing enzyme, there will be separation of the exons and the introns leading to the formation of two bands. So, the first two lanes specific for RNA A will be having a Spliceosomal Intron.

2. Now the task remains to identify which out of the remaining RNA's will be either Group I or Group II introns. There is no data regarding the addition of Guanosine nucleotide which is usually required in case of Group 1 splicing. Now, as we know that the splicing mechanism in case of Group II and spliceosomal intron is similar being carried out in a way in which an Adenine which is located in the branch site shows binding to the 5'- splice site which shows a similar conformational change for both the groups of introns. So, the Ribosome 2 which will act as a ribozyme and specifies lane 3 and 4 should be Group II intron and thus the last two lanes namely 5 and 6 will specify Group I intron.

Answer:

Plants are mutiticelullar

Explanation:

All species of plants are mutiticelullar and just a few like mold,amoeba are unicellular.

Answer:

233 snakes are heterozygous for the banding allele

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example,

How many snakes are heterozygous for the banding allele?

The frequency of banded snakes refers to the genotypic frequency for the trait, which is bb= q2= 0.4.

If q2= 0.4, then q = √0.4 = 0.63

The allelic frequency for b is 0.63.  

This means that the allelic frequency for B or p is 0.37, which we deduce by clearing the equation p + q = 1

                          p + 0.63 = 1

                          p = 1 - 0.63

                          p = 0.37

The allelic frequency of B is 0,37, and the allelic frequency for b is 0,63. The population heterozygote frequency for this allele is 2 x p x q = 2 x 0,37 x 0,63 = 0.466. The percentage of the population that is heterozygous for this allele is 46%.

As the population size is 500 individuals, then we can calculate how many of these snakes are heterozygous. This is: 0.466 x 500 = 233

The number of heterozygous water snakes for the banding allele on an island in Lake Erie is determined using the Hardy-Weinberg principle, which calculates genotype frequencies based on allele frequencies.

The question asks for the number of water snakes that are heterozygous for the banding allele on an island in Lake Erie, given that the phenotype of banding is autosomal recessive and its frequency is 0.4 in a population of 500 snakes. To determine the number of heterozygous individuals, we can use the Hardy-Weinberg principle which assumes that allele frequencies in a population will remain constant from one generation to the next in the absence of other evolutionary influences. The Hardy-Weinberg formula is expressed as p² + 2pq + q² = 1, where p is the frequency of the dominant allele, q is the frequency of the recessive allele, p² represents the frequency of homozygous dominant individuals, 2pq represents the frequency of heterozygous individuals, and q² represents the frequency of homozygous recessive individuals. Since q² is given as 0.4 (the frequency of banded snakes), we can calculate q as the square root of 0.4, which is approximately 0.63. To calculate p, we use p = 1 - q, which would be approximately 0.37. Subsequently, the number of heterozygous individuals (2pq) can be calculated as 2 * 0.37 * 0.63, then multiplied by the total number of snakes (500) to give the final count of heterozygous snakes.

Answer:

The correct genotype of the  two pure lines and the F1 is:

A⁺A⁺B⁰B⁰C⁰C⁰D⁰D⁰   and A⁰A⁰B⁺B⁺C⁺C⁺D⁺D⁺

The number of additive alleles on each genotype are two and six respectively.

Explanation:

Locus( plural form . loci) are fixed point on a chromosome in which genes are located. These genes are specific genetic material or genotype.

Now;

If we decide to designate the allele of the four loci into either additive (⁺) or non-additive(⁰); we have the following :

Let's the allele of the four loci to be

A⁺/A⁰, B⁺/B⁰, C⁺/C⁰ and D⁺/D⁰

However, from the diagram below; we deduce that the correct genotype for the two pure lines and the F1 is as follows:

A⁺A⁺B⁰B⁰C⁰C⁰D⁰D⁰   and A⁰A⁰B⁺B⁺C⁺C⁺D⁺D⁺ and the number of additive alleles on each genotype are two and six respectively.

The cross between both F1 traits will yield an heterozygous individual for the offspring. i.e A⁺A⁰B⁺B⁰C⁺C⁰D⁺D⁰ with only four additive allele

The two pure line genotypes are ++++ and 0000 respectively, with four and zero additive alleles. In the F1 generation, the genotype will be +0+0, indicating individuals possess two additive alleles.

Let's consider two pure lines: one with additive alleles on all four loci (++++), and another with non-additive alleles on all four loci (0000). Thus, the first genotype (++++) has 4 additive alleles, and the second genotype (0000) has zero. If these two lines are crossed, in the F1 generation, every individual will inherit two additive alleles from the first parent and two non-additive alleles from the second, leading to a +0+0 genotype. Therefore, the F1 genotype will have 2 additive alleles.

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Potassium has the greatest influence on the resting membrane potential of most neurons due to the membrane's higher permeability to potassium ions through open non-gated potassium channels during the resting state.

The ion with the greatest influence on the resting membrane potential of most neurons is Potassium (K+). During the resting state, non-gated (leak) potassium channels are open, allowing potassium ions to move across the neuron's membrane. This movement of potassium is critical in establishing the resting membrane potential because the membrane is much more permeable to potassium than to sodium due to the higher number of open potassium channels. Essentially, these channels enable potassium to diffuse out of the neuron, influencing the membrane potential significantly. Although sodium (Na+) and chloride (Cl-) channels are also present, they are fewer in number, meaning that while these ions do play a role, their influence on the resting membrane potential is not as great as that of potassium. The resting membrane potential of a neuron is closer to the equilibrium potential of potassium, typically between -60 to -70 mV. Furthermore, the sodium-potassium pump helps to maintain the gradients of these ions, but the passive permeability largely dictates the resting membrane potential.

* The complete question is:

Which ion has the greatest influence on the resting membrane potential of most neurons?

Multiple Choice

Potassium

Sodium

Answer:

A) Oxidation of fats releases metabolic water

B) [tex]1.54[/tex] L of water per kg of tripalmitoylglycerol

Explanation:

A) During the oxidation of organic substances (fats) releases metabolic water along with release of energy

For example –  

 Oxidation of one mole of Palmitic acid (C16H32O7) acid releases 146 mole of H2O. In Tripalmitoylglycerol (fat C51H98O6) , there are 3 palmitic acid residues connected to glycerol molecule via ester bonds.

Oxidation of Palmitoyl CoA:

almitoyl-CoA + 23O2 + 108Pi + 108 ADP => 88nCoA + 108 ATP + 16CO2 + 23 H2O

B)

Molecular weight of tripalmitate [tex]= 807[/tex] g/mol

Number of moles in one Kg of fat  

[tex]= \frac{1000}{807}\\[/tex]

[tex]= 1.24[/tex] mole

Mass of water

[tex]1.24 * 69 * 18.02[/tex] ([tex]69[/tex] moles of water and weight of water [tex]= 18[/tex] grams)

[tex]= 1541.8[/tex] grams

Density of water [tex]= 1[/tex] g/mL

Volume of water [tex]= 1.541[/tex] L

Camels do not store water in their humps, but the humps store fat which is metabolized to water and carbon dioxide through oxidation. This provides hydration to the camel. For instance, 1.0kg of fat can produce about 1.1 liters of water.

When a camel's body uses these fat deposits for nutrition during periods without food, a reaction called oxidation happens. Oxidation of fat is a metabolic process in which the fat molecules are broken down into carbon dioxide and water using the oxygen that animals breathe in. If we consider that the complete oxidation of 1 gram of fat releases about 1.1 grams of water, then 1.0 kg (or 1000 grams) of fat can produce about 1100 grams of water. Because the density of water is 1 gram per milliliter, these 1100 grams are equivalent to approximately 1.1 liters of water.

Therefore, camels can stay hydrated for longer periods in the desert because of the water produced by the oxidation of their fat stores.

To summarize, though camels don't store water in their humps, the fat stored there can be metabolically converted to water in their bodies, helping them survive in harsh desert conditions.

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Answer:

A) One of the genotype i.e true breed white is lethal

B) No

Explanation:

A) Two hamsters with white spots are crossed, 2/3 of the offspring possess white spots and 1/3 has no spots.

This means that the two white hamsters would be carrier for the trait of no spot.  

The ratio for offspring signifies that one of the offspring dies because of lethal genotype.

B) No, because the true breed for white spotting is lethal.

However, heterozygous white spotting can be detected.

The genetic basis of white spotting in Chinese hamsters is a simple Mendelian pattern of inheritance involving a single gene with two alleles. It is not possible to produce Chinese hamsters that breed true for white spotting unless both parents are homozygous for the allele.

The genetic basis of white spotting in Chinese hamsters is a simple Mendelian pattern of inheritance involving a single gene with two alleles. The allele for white spotting (W) is dominant over the allele for no spots (w). When a Chinese hamster with white spots (Ww) is crossed with a hamster without spots (ww), approximately half of the offspring will have white spots (Ww) and half will have no spots (ww). When two hamsters with white spots (Ww) are crossed, two-thirds of the offspring will have white spots (Ww) and one-third will have no spots (ww).

The possibility of producing Chinese hamsters that breed true for white spotting depends on the genotype of the parents. If both parents are homozygous for the allele for white spotting (WW), then all of their offspring will also have white spots and they will breed true. However, if one or both of the parents are heterozygous for white spotting (Ww), there is a chance that some of their offspring will not have white spots. Therefore, it is not possible to produce Chinese hamsters that breed true for white spotting unless both parents are homozygous for the allele.

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Answer:

c. 1/2

Explanation:

Cross:

Parents    [tex]M^{D} M^{d}[/tex]  X  [tex]W^{d}W^{d}[/tex]

Gametes  [tex]M^{D}, M^{d}[/tex] X  [tex]W^{d}, W^{d}[/tex]

Offspring 2 [tex]M^{D} W^{d}[/tex], 2 [tex]M^{d}W^{d}[/tex]

Probability of polydactyl child = 2/4 or 1/2 or 50%

Answer:

6

Explanation:

Answer:

A key example is that DNA is double stranded where as RNA is single stranded.

Explanation:

Answer:

Explanation:

Like we all know, the purine ring of IMP is made up of a nine membered compound . they are heterocyclic aromatic organic compound that consist of a pyrimidine ring fused to an imidazole ring. there are four nitrogen atoms and five carbon atom.

The ring is imidazole ring is made up of N1, C2, N3, C4, C5, C6 with the pyrimidine sharing C4 and C5 with the imidazole ring and also made up of N7, C8, N9.

the direct source of N1 is from aspartate

the direct source of C2 and C8 is from N 10 ‑formyltetrahydrofolate (THF)

N3 and N9 is derived from the amide group of Glutamine

C4, C5 and N7 is derivd from Glycine

C6 is derived from bicarbonate

Answer:chase the engine

Because of the engine

Explanation:

Answer:The two systems work together in several ways. First, the respiratory system brings oxygen in the air to the alveoli in the lungs. The circulatory system then delivers the oxygen to the cells. Second, the circulatory system picks up carbon dioxide from the cells and carries it to the lungs, where it is released when we exhale.

Explanation:its the sample writing.