) How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Answer: Explanation:

To emphasize culture's evolutionary basis

Answer:

By heat trap mechanisms...

Explanation:

The main green house gases in our atmosphere are Carbon dioxide, Nitrous oxide, water vapors and ozone. All of these gases have a property of heat trap that trap the heat waves coming from the sun and heat up our atmosphere as a result. This is the reason why Venus is farther from sun but is more hotter than mercury because of it's dense Carbon Dioxide atmosphere.

Answer:

Explanation:

(1) Remember that this occurred in a polylysine which consists only of L-Lys, the pKa of Lys residues might be different in polylysine as compared with the free amino acid in the solution.

The transition can also occur if more than 50% of the lysine residues need to  be charged in order to ‘break’ the helix. Note that there is a 50:50 ratio of the coordinated bond of protons and dissociated forms.

(2) Other residues that are positively charged at neutral pH. They are arginine and histidine.  

(3) Both arginine and histidine are more voluminous than Lys. Steric interference can prohibit the formation of an α-helices.

Dendrites

hope this helps

Answer: the correct answer is dendrite.

Explanation: Dendrite is actually what receive signals from axon of other neurons, synapse is an area which joins axon and Dendrite. Signals move across the axon in the form of action potential, coming from the cell body.

Answer:

12

12

24

Explanation:

Musca domestica is the scientific name of an animal popularly known as Housefly.  From the question, it is  known that this organism has a haploid chromosome number of 6.

In a diploid chromosomes, the number of chromatids that will be present will be twice of that in haploid.

Since haploid (n) = 6, diploid (2n) will be; 2 × 6 = 12

Somatic cells on the other hand also exhibit diploid number of chromosomes, again it means we will have (2n) = 2 × 6 = 12

In metaphase cell, cells do make sure they complete the S'phase of the cell cycle before cellular division.Therefore, the DNA present in the chromatids in the S'phase actively engage in chromatids doubling, as a result, 24 chromatids (i.e 12 × 2= 24) exists in the metaphase cell.

The chromatids should be present in a diploid, somatic, metaphase cell-

The scientific name of the fly Housefly is Musca domestica. It is known that this organism has a haploid chromosome number of 6. In the diploid cells, the number of chromatids that will be present will be twice that in haploid.

Diploid genome (2n)

= [tex]2\times6=12[/tex].

Somatic cells also exhibit diploidy (2n)

= 12

Metaphase stage cells contain 24 chromatids as cells before entering cell division complete S-phase of the cell cycle in which DNA present in chromatids become double due to replication.

Learn more about Metaphase:

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Answer: Fluoroscopy is used as positioning device prior to obtaining film radiographs. It is used for examination where the type of projection and habitus if the patient present difficulties

It is not an acceptable practice because it exposes the skin and underlying tissues to radiation induced injuries and there is a possibility of future occurances of radiation induced cancer in the future

Explanation:

Answer: (d) Membrane at the top

Explanation:

Pellicle: surface membrane produced by organisms floating on top of the medium. Sediment: organisms that sink to the bottom. Uniform fine turbidity: cloudy. Flocculent: Organisms that clump/suspended pieces or chunks.

(2) Other important factors apart from

physical growth too include would be the growth medium, incubation temperature and atmosphere (aerobic or anaerobic). Growth medium,incubation temperatureandatmosphere(aerobicoranaerobic).

A pellicle describing uniform fine turbidity means that the microorganisms are evenly dispersed throughout a medium, giving it a uniformly cloudy appearance. Apart from physical characteristics, environmental conditions, genetic predisposition, presence of competitors or predators, and availability of resources are crucial factors to consider when recording data about an organism.

The term 'pellicle' in biology refers to a thin layer supporting the cell membrane in various protozoa. When it describes a uniform fine turbidity, it refers to how the organisms in a liquid medium are dispersed. In this case, 'a. Evenly cloudy throughout' pertains to a uniform fine turbidity, meaning that the microorganisms are evenly dispersed throughout the medium, giving it a uniformly cloudy appearance.

Beyond physical growth characteristics, other factors are important to consider when recording data about an organism. These may include the organism's environment, its genetic predisposition, the presence of predators or competitors, availability of resources such as food and space and potential changes in these conditions over time. All these elements contribute to the overall understanding of the organism and its behavior.

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Answer:

2, 3, 4, 1

Explanation:

Inversion of DNA sequences refers to the rearrangement of the sequences where a segment is reversed end to end.

For instance in order for the gene of species to have the sequence STUVWX

The order of evolution will be

When XTSW segment is inverted WSTX

section UVWS is inverted to  SWVU

when WVUT undergoes inversion to TUVW

Answer:

Borneo Island was once home to one of the most majestic forests in the world. During the 80s and 90s, Borneo underwent a profound transformation. Their forests were demolished at an unprecedented speed. Its rainforests ended in countries like Japan and the United States in the form of garden furniture, paper pulp and chopsticks. Today, the remaining forests are threatened by the new biofuel market, especially palm oil. As a result, large tracts of land are being transformed to oil palm plantations.

Explanation:

The jungles of Borneo were considered one of the wildest and pristine jungles of the planet, home to nomadic tribes and important populations of orangutans, pygmy elephants and rhinos. Currently, the traditions of these tribes have disappeared, rhinos are almost extinct, and orangutans and elephants are in danger. Otherwise, the rainforests of Borneo have gone from being a net carbon sink, which absorbed greenhouse gases, to being a carbon source, thus contributing to climate change along with deforestation and fires.

Conservation is a priority in Borneo, especially in biologically diverse regions that have escaped intensive logging and fires. The initiative called "Heart of Borneo" is an example of what can be achieved. It is essential that forests be restored. The use of native tree species should be encouraged through financial incentives and education programs, especially with the help of external governments, NGOs and private foundations. In addition, there is a possibility that under future climate agreements, reforestation could pay direct dividends stimulating the local economy and entrepreneurship in the villages.

Answer:

E

Explanation:

some of the nematodes (non-segmented ) are parasites

Answer:

The statements which are correct about nematodes are options C, D and E:

C: Organisms from the phylum nematoda are animals, multicellular in nature, has tissues, have a complete digestive tract and exhibit bilateral symmetry.

D. Nematodes possess external structure called cuticle.

E. Some members of the phylum nematoda are parasitic, free living organisms that feed on other living materials.

Answer:22.180

Explanation:divide the two together then use the first three number after the decimal.

.

Answer:

Explanation:

Let's start by stating that the work done here would be determined by taking into account the following:

-External forces

-Change in kinetic energy

Work Energy theorem is fully going to guide the solution to this problem.

Let's go now:

Mass of car = 1475kg

Velocity of car= 66.5km/h

Velocity = 66.5km/h * 0.278

V = 18.472 m/s

Kinetic Energy (K.E) = 1/2(mv^2)

K.E = 0.5 * 1475 * 18.472^2

K.E = 737.5 * 341.214

K.E = 251645.90 kgm^2/s^2

1kgm^2/s^2 = 1 Joule

So therefore,

K.E = 251645.90 Joules

K.E = 60.144KCals

Answer:

All the options A,B,C,D and E statements are correct.

Chordates are animals that posses a notochord, dorsal, hollow nerve chord, pharyngeal slits and they are bilaterally symmetrical.

Answer:

[tex]1 - (\frac{61}{64})^n[/tex]

Explanation:

Given

There are three stop codon

TAA, TAG, TGA.

Out of 64 available codons, 3 are stop codon thus the remaining are non-stop codon.

So the probability of choosing a non stop codon is

[tex]\frac{61}{64}[/tex]

Let us suppose the number of trials are "x" then the probability of choosing a non stop codon is  

[tex](\frac{61}{64})^x\\[/tex]

Probability of choosing a stop codon is equal to

[tex]1 -[/tex] probability of choosing a non stop codon

[tex]1 - (\frac{61}{64})^n[/tex]

The chance of having a stop codon at any given position in a random sequence of codons can be calculated by dividing the number of stop codons by the total number of possible codons.

In a given sequence of codons, the chance of having a stop codon at any given position can be calculated by dividing the number of stop codons by the total number of possible codons. The number of stop codons is given as 3, and the total number of possible codons is 64. Therefore, the chance of having a stop codon at any given position is 3/64, or about 0.046875, assuming that the sequence is random.

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Answer: This moth has evolved

Explanation:

Evolutionary patterns show different patterns. One of the patterns is that: one species gradually transform into another species.

The above mentioned pattern is responsible for the moth's ability to lay more eggs and have more surviving offspring than other moths in the area. The moth is gradually transforming, so also its offspring will inherit these new traits, and it would further demonstrate evolution of one species.

Thus, it can be concluded that the unique ability of this moth over other moths is because the moth has evolved.

The possible outcomes for the child's genotype are Rb, Rb, bR, and bb. The probability of the child having the Blond/Blond genotype is 25% while the probability of the child having red hair is 75%.

In this genetics-based question, we're dealing with a common system of inheritance where the red hair allele is dominant over the blond hair allele. Let's denote the red hair allele as 'R' and the blond hair allele as 'b'. Since both parents have the genotype Red / Blond, their genotypes can be represented as 'Rb'.

a. The different possible outcomes for their children's genotypes (which determine hair color) can be determined via a Punnett square. The possibilities are: Rb (Red/Blond), Rb (Red/Blond), bR (Red/Blond), and bb (Blond/Blond).

b. The probability that a child of these parents will have the genotype Blond/Blond (bb) is 1 out of 4, or 25%.

c. If a child has at least one red allele (R), they will have red hair, owing to the dominance of the red allele. In this scenario, there are 3 out of 4 possibilities where the child has at least one red allele, so the probability that the child will have red hair is 75%.

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Answer:

Biomass: can be defined as the total number of viable organisms, such as plants, animals in a given area.

While,

Standing crop: can be defined as the amount of biomass at a particular time or it is the amount of above-ground plant biomass.

Answer:

The DNA is a double helix or made up of two strands. The strands are separated during replication, each serving as a template to produce a complementary strand of each of the separated single strands. Therefore after replication, two double stranded DNA molecules will be present.

Final answer:

Two double-stranded DNA molecules are produced following DNA replication, with each new molecule containing one old and one new strand, reflecting the semiconservative nature of the replication process.

Explanation:

After DNA replication, two double-stranded DNA molecules exist. This is because DNA replication is semiconservative, meaning each new DNA molecule consists of one original and one newly synthesized strand. During DNA replication, the double helix of the parent DNA molecule is unwound by enzymes. Then, DNA polymerase synthesizes a new complementary strand for each of the two single strands from the parent molecule.

The entire replication process ensures that each daughter cell receives an exact copy of the DNA. Mechanisms such as DNA proofreading and the repair of mistakes, which involve enzymatic correction during replication, help maintain the accuracy of the genetic information passed on to each new cell.

Answer:

Yes it in not a muscle of the medial forelimb of the fetal pig

Explanation:

Rectus femoris is not the part of forelimb of the fetal pig.

However, rectus femoris in fetal pig originates from the hip and that to through the ilium and extends towards the knee and insert into its tendon. These are responsible for flexing the hip and extension of lower knee and legs. This muscle is a part of muscles of thigh and lies at its top. In quadriceps these muscles are the most anterior muscles.  

Answer: Rectus femoris is not a muscle of medial forelimb of the fetal pig.

Explanation:

Pigs are mammals like humans. Pigs have similar structures with humans but there are slight differences in these structures because pigs are quipedal and humans are bipedal.

Rectus femoris is not found in the medial forelimb of fetal pigs instead it is found in the medial hindlimbs of fetal pigs and is one of the quadriceps muscles. It starts from the hip through the ilium and extend to the knee.Rectus femoris function in knee extension, its extend the lower leg and it flexes the hip.

The question is incomplete. The complete question is as follows:

Which characteristics of DNA polymerase I raised doubts that its in vivo function is the synthesis of DNA leading to complete replication?

its composition of a single polypeptide chain .

deficiency of enzyme in some organisms that are still capable of DNA synthesis .

requirement of Mg2+ presence in order for the enzyme to work .

low stability under normal physiological conditions.

Answer:

Deficiency of enzyme in some organisms that are still capable of DNA synthesis

Explanation:

The DNA polymerase I may be defined as the important enzyme that play an important role in the DNA replication of prokaryotes. DNA pol I is the replicating enzyme, DNA repair enzyme and can also acts as the exonuclease.

DNA pol I has been studied invitro and Arthur Korenberg explain the discovery of the DNA pol I. This DNA pol I plays an important role in DNA repair rather than the replication process. This explained invivo by the fact that some in some organisms the deficiency of this enzyme do not halt the process of replication. If the DNA pol I acts as the main replaicating enzyme, the DNA synthesis must be stopped in the organisms that lack DNA synthesis.

Thus, the correct answer is option (2).

Answer:. Nondisjunction during meiosis I in either the male and female gamete

The failure of   homologous chromosome to separate at mitosis,  or failure of sister chromatids  to separate at Anaphase of  meiosis I or ii is called  non-disjunction.

 In this  cell division anomaly the  unsegregated chromosomes  are  separated  into  cells as single chromosomes, therefore extra copies are  therefore inherited.This  result is one of the cell having an extra copies of  chromosomes than others( chromosomes 21.

If this extra  chromosome copies  is segregated into a sperm or egg, the resulting zygote cells will   have extra  copy as  chromosome 21, and hence Down syndrome.

→It is more common in Meiosis 1 than ii,because of longer duration of the first stage of division than meiosis ii,given rooms to errors.

→Most common in women(with age increase)  than men because of exchange of telomeres in   them,which  worsen with age. Their Meiotic mechanism  which   weaken with age, and  more prone to errors compare to men,

Explanation:

The given question is not complete. The complete question is:

Watch the film “Liz Hadly Tracks the Impact of Climate Change in Yellowstone (Links to an external site.)Links to an external site.” from the Scientists at Work series. Prior to watching the film, read the questions below and think about how you might answer them. You do not need to turn in your answers from before you watch the film.

Questions

Explain what is happening to the whitebark pine in Yellowstone National Park as a result of climate change

Researchers often study ecosystems for a long period of time. Dr. Hadly has studied Yellowstone’s ecology for 30 years and the amphibians for 20 years. What is the value of long- term studies to advancing scientific understanding?

Answer:

1. As a consequence of climate change, the whitebark pine is been targeted by pine beetles, which can thrive in winter conditions. An attack led by various beetles for one to two days has generated holes in the pine leading to its destruction.  

2. The objective of long term studies is to find the connections between animals and plants and to witness how they are associating with each other in their micro-environment. Like in the documentary it was the association between the squirrel, pine trees and bear.  

The beer needs to consume high content nutritious seeds of the pine tree, which safeguarded them in the winters due to the high amount of fat present in them that offered warmth. However, the beers were not able to reach the pine trees as they were far from their reach and thus, they took help from squirrels, which serve the objective for the bears.  

Thus, it can be comprehended that all three creatures, that is, squirrel, tree and bear are associated with each other and developing a food web. All these studies need an ample amount of time and observation. Hence, Dr. Hardley took thirty years examining them and also took twenty years examining the influence of climate on amphibians living in the pond. Their species diversity declined substantially due to global warming that resulted in the drying up of the pond.  

The given question is incomplete. The complete question is:

Sort the molecules in the glycolysis pathway based on whether they are intermediates or products in the first half of the pathway that requires energy, or are intermediates in the second half of the pathway that produces energy. Drag the appropriate items to their respective bins.

glucose-6-phosphatefructose-6-phosphatefructose-1,6-bisphosphateglyceraldehyde-3-phosphatedihydroxyacetone phosphate1,3-bisphosphoglycerate3-phosphoglycerate2-phosphoglyceratephosphoenolpyruvate

Answer:

The procedure in which glucose gets dissociated into pyruvate and generates energy in the form of ATP is known as glycolysis. The process takes place in the cytosol and can get differentiated into two phases, that is, the preparatory phase that required energy in the form of ATP and the pay-off phase in which the production of ATP takes place.  

Post glycolysis, the process of cellular respiration takes place in which the eventual outcomes, that is, two pyruvates are further used. Of the mentioned intermediates, the intermediates that are used in the first half of the pathway are glucose-6-phosphate, fructose-6-phosphate, fructose-1,6-bisphosphate, dihydroxyacetone phosphate and glyceraldehyde-3-phosphate.  

The intermediates used in the second half of the pathway are 1,3-bisphosphoglycerate, 3-phosphoglycerate, 2-phosphoglycerate, and phosphoenolpyruvate.  

Answer:All of the above

Explanation:

Membrane filters are very thin, highly porous media composed of foamed and/or stretched polymeric compounds. Because of their homogeneous structure they cannot contaminate the filtrate with fibres or particles from the membrane matrix.

Microfiltration membranes retain particles according to the size of the pore, and the affinity of the filter materials for the solute. While materials are held largely on the surface of the membrane, they can also retain within the matrix itself, as in depth filtration. Unlike depth filters, however, which bind solutes of nominal size ranges only, membrane filters retain particles with absolute accuracy due to their controlled, predetermined pore size. Some membranes also demonstrate retention of specific substances as for example in the case nitrocellulose membranes which bind proteins and nucleic acids in high concentration. This feature may be an advantage for some applications where exclusion based on chemical properties rather than porosity alone is required.

Membrane filters for microfiltration are used primarily for separating particulate materials or micro-organisms, larger than the rated pore size, from gases or liquids. On filtration of gases particles are also retained which are smaller than the pore size rated. Also they show a low particle capacity per unit area as compared to a depth filter. For some filtration applications it may be advisable to prefilter through a depth filter, eg, a glass fiber filter, prior to microporous filtration to prevent clogging. This may be necessary, because the solution is highly viscous, contain molecules which may precipitate in the membrane filter or are heavily contaminated by microorganisms and particulates.

This separation is carried out either for the purpose of cleaning to obtain highly pure or sterile products or recovering (concentrating) these materials in order to carry out further chemical, microscopic, microbiological or other analyses on the separated sample.

Typical applications include Sterile filtration

–Dialysis

–Fluid clarification/purification

–Gas filtration/particle control

–Microbiological investigations

–HPLC solvent filtration and sample preparation

What phenotypic ratio would you expect as a result of a test cross between a dihybrid organism and one that is homozygous recessive for alleles at two independent loci?

a. 3:1

b. 9:3:3:1

c. 1:1:1:1

d. 1:2:1

e. 9:4:2:1

Answer:

c. 1:1:1:1

Explanation:

When a heterozygous individual for two genes is test crossed with a double homozygous recessive individual, the progeny is obtained in 1:1:1:1 phenotypic ratio. This occurs as the heterozygous dominant individual forms four types of gametes in 1:1:1:1 ratio while the homozygous recessive individual would form only one type of gamete having one recessive allele for each gene.  

For example, a test cross between TtRr (tall and red) and ttrr (short and white) would produce a progeny in following ratio=

1 tall, red: 1 tall, white: 1 short, red: 1 short, white

Here, T= tall, t= short, R= red, r= white

Final answer:

The genotypic ratio expected from the mating of two individuals heterozygous for a recessive lethal allele expressed in utero is 2:1 (homozygous dominant: heterozygous) among surviving offspring because the homozygous recessive genotype results in death before birth.

Explanation:

When two individuals that are heterozygous for a recessive lethal allele expressed in utero are mated, the expected genotypic ratio (homozygous dominant:heterozygous: homozygous recessive) among the offspring would be 2:1. This is because the homozygous recessive genotype, which would normally make up one-quarter of the offspring, results in death before birth due to the lethal allele. Therefore, the live offspring genotypic ratio becomes 1:2:0, as the homozygous recessive individuals do not survive. This contrasts with a typical Mendelian 3:1 phenotypic ratio observed in monohybrid crosses of non-lethal traits. The Punnett square for such a cross shows that out of the four possible genotypes, one (homozygous recessive) does not survive, leaving a genotypic ratio of homozygous dominant:heterozygous at 1:2 among surviving offspring.

Answer: Option C.

Energy flows in one direction in ecosystem.

Explanation:

In the ecosystem, energy flow in one direction. Energy is transfered from one trophic level to another.

Energy is transfered from producers, majorly plants,convert sunlight to chemical energy during photosynthesis and transfer it to herbivores( goat, sheep ) when they eat plants,the herbivores after using the energy transfer energy to the carnivores when the carnivores feed on the herbivores and to decomposer who feed on dead animals.

Energy flows in one direction through an ecosystem, from producers to consumers, and eventually dissipates as heat, while matter is recycled, such as through biogeochemical cycles.

Energy in ecosystems behaves differently than matter. It flows in only one direction: from the sun or inorganic chemicals to producers, and then on to consumers. Producers, such as plants, are capable of capturing energy directly through photosynthesis and chemosynthesis. Consumers then acquire energy by feeding on these producers or on other consumers. This flow of energy is depicted using models like food chains and food webs. In contrast, matter is recycled within ecosystems, ensuring that elements such as carbon and nitrogen are continuously available in one form or another.

It is also pertinent to highlight the process of biogeochemical cycles, such as the water cycle, carbon cycle, and nitrogen cycle, whereby nutrients and water are constantly recycled, interacting with both biotic components and abiotic factors of an ecosystem. While matter recycles and maintains a certain balance, energy does not recycle but rather dissipates as heat with each transfer through the trophic levels, eventually exiting the ecosystem.

Answer:

(a) (Glu)zo or(Phe-Met)3 at pH 7.0

O (Glu)zo ✔

O (Phe-Met)s ❌

(b) (Gly) zo or (Lys-Ala)3 at pH 7.0:

O (Gly12) ❌

O (Lys-Ala)✔

(c) (Ala-Asp-Gly)s or (Asn-Ser-His)s at pH 3.0:

O (Asn-Ser-His)s ✔

O (Ala-Asp-Gly)s ❌

(d) (Ala-Ser-Gly)s or (Asn-Ser-His)s at pH 6.0:

O(Ala-Ser-Gly)s ❌

O (Asn-Ser-Hish)s ✔

Explanation:

Polypeptides that has polar or charged side chains are more soluble than polypeptides with nonpolar side chains.

(a) At ph 7.0

(Glu)20 is negatively charged at pH 7 and more soluble

(Phe-Met)3 is observed to be less polar and less soluble

(b)At ph 7.0

(Lys-Ala)3 is positively charged (polar) and more soluble

(Gly)20 is uncharged as only the amino- and carboxyl-terminal groups are charged as its less polar and less soluble too.

(c) At pH 6.0

(Asn-Ser-His)5 has polar Asn side chains and partially protonated His side chains and it's more soluble unlike the (Ala-Asp-Gly)s at that pH.

(d) At pH 3.0

(Asn-Ser-His)s as partially protonated carboxylate groups of Asp residues and it is also neutral but the imidazole groups of His residues are fully protonated and positively charged. Hence it's more soluble than the (Ala-Ser-Gly)s at that particular pH.

Answer:

15.21 %

Explanation:

If we recall the basic formula of Hardy-Weinberg's equilibrium ; we have the following below:

p + q = 1

p² + 2pq + q² = 1

where;

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p² = percentage of homozygous dominant individuals

q² = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Given that p= 0.68 and q = 0.39

the percentage of the homozygous recessive genotype (q² ) will be

(0.39)² = 0.1521

= 0.1521 × 100

= 15.21 %

∴ the percentage of the population that has a homozygous recessive genotype = 15.21 %

Approximately 15.21% of the population has a homozygous recessive genotype.

To calculate the percent of the population with a homozygous recessive genotype, we need to square the frequency of the recessive allele. In this case, the frequency of the recessive allele is 0.39, so squaring it gives us 0.39 * 0.39 = 0.1521. To convert this to a percentage, we multiply by 100, giving us 15.21%. Therefore, approximately 15.21% of the population has a homozygous recessive genotype.

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Answer:

a) [tex]0.58[/tex]

b) [tex]0.598[/tex]

c) [tex]0[/tex]

Explanation:

Given -

Total sample i.e n [tex]= 18[/tex]

Probability (p) [tex]= 3[/tex] % [tex]= 0.03[/tex]

We will use binomial distribution theory for determining the probability of mutated sample

Let X be the number of mutated sample

a)  No samples are mutated i.e [tex]X = 0[/tex]

[tex]P(X=0) = 0.03^0 * 0.97^{18}\\= 0.5779 = 0.58\\[/tex]

[tex]0.58[/tex]

b) At most one sample is mutated

[tex]P(X=0) = 0.58 + 0.03^1 * 0.97^{17}\\= 0.598[/tex]

c) More than half the samples are mutated.

[tex]P(X = 10) + ........+ P(X = 18) = 0[/tex]

Answer:

Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group.

Explanation:

Phospholipid is a unique form of lipid. The  bonding of the glycerol backbone to the polar phosphate group makes phospholipid to have dual solubility unlike general triglycerides.

The polar head is said to be hydrophillic that is water loving, while the two carbon chains  that  retained lipid features are hydrophobic water hating.

Therefore if a phopholipid is placed in water, in relation  to its functions as component of cell membrane, it forms a bi-layer in which the water loving portion hydrophilic head points into the surrounding watery medium, while the hydrophobic layer points inwards far away from the watery medium into the internal cellular  layer to form an impermeable barrier to hydrophilic (polar) substances.

This forms the basis of   the  phospholipd  bilayer  of the  cell membrane. And it  controls the permeability of the cell membrane to  influx  substances  into the cells.

Phospholipids consist of a glycerol backbone bonded to two fatty acid chains and a small polar group, making it amphiphilic. They form the plasma membrane of cells, with the polar heads facing the water environment and the non-polar tails facing away from the water.

The correct answer is Option A: Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group. Phospholipids consist of a three-carbon glycerol backbone with two fatty acid molecules attached to carbons 1 and 2, and a phosphate-containing group attached to the third carbon. This structure gives the phospholipid molecule an amphiphilic, or "dual-loving," characteristic with a polar (charged) head and non-polar (uncharged) tail. Phospholipids play a crucial role in forming the plasma membrane of cells, where the hydrophilic (water-loving) head faces towards the water environment and the hydrophobic (water-avoiding) tail faces away from the water. Examples of phospholipids include Phosphatidylcholine and Phosphatidylserine.

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Answer:

C. causes cell lysis

Explanation:

The slow release of the bacteriophage progeny from a bacterial host cell causes bacterial cell lysis and further kills the host cell which is the bacteria, this results to the liberation of progeny viruses which then infect new bacterial cells.

Only a type of bacteriophage called the filamentous bacteriophage reproduces without killing the bacterial host cell.  The filamentous phages infect majorly gram negative bacteria.

The correct statements regarding the slow release of bacteriophage progeny are that it is characteristic of filamentous phages and does not kill the host cell. Therefore, the correct answer is E. A and B are correct.

The slow release of bacteriophage progeny from a bacterial host cell is a feature particular to filamentous phages and does not result in cell lysis. Therefore, the correct answer to the question is E. A and B are correct. Filamentous phages can exit the host cell without lysing it through a process called extrusion, where new virions are released from the cell membrane without causing cell death. Unlike lytic phages, which replicate and cause lysis of the host cell, filamentous phages exhibit a different life cycle that allows the host cell to survive the release of new phage particles. Therefore, the statement that slow release of bacteriophage progeny causes cell lysis is incorrect, because this slow release is characteristic of filamentous phages and does not kill the host cell.