Briefly, in two to three sentences, contract the functions of innate, humoral, and cell-mediated immunity.

Answer:

e. both b. and c. are correct

Explanation:

According to Mendel's law of segregation, each genetic trait is regulated by a pair of factors, alleles. These genetic factors are present as distinct particles.

The two alleles of a gene occupy the corresponding position on the homologous chromosomes. The two alleles of a gene are separated from each other during gamete formation due to the separation of homologous chromosomes during anaphase-I of meiosis-I.

This leads to the formation of gametes carrying one allele for each gene. This is the reason that the law of segregation is also called the law of purity of gametes.

Answer: False

Explanation:

Food irradiation is a process in which the food is exposed to the treatment of radiations. This helps in increasing shelf life and decreasing the colonization of the pathogens on the food. This process does not make the food radioactive.

Answer:

Enzymes are named according to the reaction they catalyze. Polymers are made of subunits joined together by different types of bonds, forming a macromolecule.

Hydrolases are used by the organism to catalyze the hydrolysis of polymers so they can be easily manipulated as monomers. Hydrolysis means reacting with water, water can break the bonds of different polymers turning it into its constitutive monomers.

Answer:

All of the above are possible proximate causes of the behavior

Explanation:

Cats have reproductive organs, these organs go through a normal cycle, called heat cycle, that let reproduction to occur. During this cycle, they have hormonal fluctuations that mean that their body is ready to breed.

During the heat cycle, cats have several behaviors and signs such as spraying urine, rolling on the floor and scratching at doors begging to go outside. When a female cat is in heat, she is likely to urine more frequently and even spray urine on objects, to let any nearby male cats know that she is ready to be mated with. The urine contains pheromones and hormones, signals of her reproductive status to other male cats.

In the wild, females cats pee on trees or objects to find a male cat to mate with. This is why one might infer that female cats did this in the past and the domestic cats inherited this behavior.  

Answer:

Glycogen. (Ans. C)

Explanation:

Glycogen is known as the storage form of glucose in animal cells. It a polysaccharide of glucose which is multi-functional and serves as an energy form storage in animals. Liver cells primarily stored glycogen, some glycogen also stored in muscle cells for immediate use if needed.

Glycogen molecule composed of many glucose molecules that are linked together with the help of the alpha acetal group. To create energy currency (adenosine triphosphate) glucose is the primary source used by every cell.

The storage form of glucose in animal cells is glycogen, a complex carbohydrate stored mainly in the liver and muscle cells. This is utilised as a source of energy when required.

The storage form of glucose in animal cells is glycogen. This polysaccharide is a complex carbohydrate that animals and humans use to store glucose. It is found mainly in the liver and muscle cells. When your body requires energy, it transforms the stored glycogen into glucose which can be used as a source of energy. You can perceive it as animal's equivalent to plant's starch storage mechanism. So, the correct answer to your question is: c) Glycogen.

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Answer:

The sex steroids are the essential hormones for the proper function and development of the body, they monitor sexual differentiation, sexual behavior patterns, and the secondary sex characters. There are five prime categories of steroid hormones. These are estradiol (estrogen), testosterone (androgen), aldosterone, and cortisol.  

Of these testosterone, estradiol, and progesterone comes under the category of sex-steroids. The progesterone and estradiol are produced by the ovarian granulosa cells of the ovary. On the other hand, the testicular Leydig cells are the location of testosterone synthesis.  

Answer:

The correct answer is option b. "the citric acid cycle".

Explanation:

The citric acid cycle is one of the most important pathways of the cellular metabolism because of its role in energy production and biosynthesis. The citric acid cycle is where most of the CO2 from catabolism is released. This happens by the release of two molecules of CO2 per turn in the cycle, using the two atoms of carbon from acetyl CoA.

The majority of CO2 produced during catabolism is released during the citric acid cycle. Neither glycolysis, lactate fermentation, nor the electron transport chain, emit CO2.

The majority of carbon dioxide (CO2) produced during catabolism, the process by which the body breaks down molecules to produce energy, is released during the citric acid cycle, also known as the Krebs cycle or the tricarboxylic acid (TCA) cycle. Glycolysis, the first step of cellular respiration, does not release CO2. It converts glucose into pyruvate, without releasing any CO2. In lactate fermentation, no CO2 is produced either. The electron transport chain, the last step of cellular respiration, also doesn’t produce CO2. It utilizes the electrons and hydrogens gained from the Krebs cycle to create water and ATP, but it doesn’t generate CO2.

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Answer:

False

Explanation:

There are a lot of studies corroborating that lightly pigmented people are at higher risk of develop skin cancer than heavily pigmented people.

Answer:

b. oxidative phosphorylation in cellular respiration.

In mechanism, photophosphorylation is most similar to oxidative phosphorylation in cellular respiration. Both processes involve establishing a proton gradient across a membrane to generate ATP.

In the biological process of energy production, photophosphorylation parallels most closely to oxidative phosphorylation in cellular respiration as option b suggests.

During oxidative phosphorylation, electron transport chain drives the creation of a proton gradient across a membrane, and the subsequent flow of protons down this gradient is used to power ATP production. Similarly, during photophosphorylation, the energy of sunlight is used to generate a proton gradient, which again drives the synthesis of ATP. So, in essence, both processes use a membrane-bound proton gradient to drive ATP synthesis.

This is different from substrate-level phosphorylation, carbon fixation and the reduction of NADP+. Substrate-level phosphorylation refers to direct ATP production during a specific reaction in the metabolic pathway, while carbon fixation is the process of converting CO2 into organic compounds; and the reduction of NADP+ is about adding electrons to NADP+ to convert it to its reduced form, NADPH.

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Answer:

a. If the b and c loci were unliked the expected phenotypic proportions would be:

- 1/4 Brown

- 1/4 Black

- 1/2 Albino

b. The estimated distance between the two loci is equal to 17 centimorgans (cM).

Explanation:

a. If the loci b and c are unlinked or they are not in the same chromosome, it means that b and c are segregated independently (following Mendel's laws).  So, the first step is to state the genotypes of the rabbits.

Albino rabbits are homozygous for c (cc)

Brown rabbits are homozygous for b (bb).

Black rabbits have a copy of B (Bb or BB).

But, rabbits have the two genes (b and c), so they could be:

Albino: BBcc/Bbcc/bbcc.

Brown: bbCC/bbCc

Black: BBCC/BBCc/BbCC/BbCc

A cross was made with true-breeding brown rabbits( that means "pure" or homocygotes) bbCC and albino BBcc.

So, we have :

bbCC X BBcc

This is a dihybrid cross with only one possible genotype in F1: BbCc. -> Black rabbits.

After this, these rabbits (BbCc) were crossed with double recessive (albino) rabbits, like this:

BbCc X bbcc

To know the possible outcomes of F2, it is necessary to make a Punnett square 4x4. Following the independent assortment principle, we will have the following possible gametes:

Black Rabbits: Cb, CB, cB,cb.

Double recessive rabbits: cb, cb, cb, cb.

After doing the Punnett square, we will have 4 possible genotypes and 3 possible phenotypes in F2:

1. bbCc = Brown  

2 BbCc = Black

3. Bbcc= Albino

4. bbcc= Albino

That means:

1. 1/4 Brown

2. 1/4 Black

3. 1/2 Albino

b. To estimate the distance between the two loci is necessary to use the data provided in the question.  

In this case, we are assuming that b and c are in the same chromosome, so they don't follow Mendel's laws. The first step is to know which are the initial genotypes.

The initial cross was between a brown rabbit (bbCC ) and an albino rabbit (BBcc). If the two genes are linked, we can say that in the brown rabbit bC are always together and in the albino Bc will be together, because they're linked.

F1 will be black rabbits bCBc, (remember bC and Bc go always together).

So, after this, a new cross was made with a double recessive:

bCBc X bcbc

The possible outcomes are:

bCbc Brown

bCbc Brown

Bcbc Albino

Bcbc Albino

As you see, there are not black rabbits, so to find black rabbits, we need to find a BCbc genotype. This genotype can be produced by recombination or crossingover in the same chromosome in parentals. So, the distance between two loci is equal to the proportion of individuals with a recombinant genotype, in this case, BCbc or black rabbits.

So, we have:

34 black

66 brown

100 albino

Total: 200

D[tex]Distance=\frac{Recombinants}{Total } *100= \frac{34}{200}*100= 17 cM[/tex]

Estimated distances are measured in centimorgans (cM).

Answer:

In in-vitro fertilization, the oocytes from women were taken and inseminated with the donor sperm, post the process of insemination, the oocytes were fertilized successfully and were further allowed to develop into a blastocyst stage.  

From the blastocyst, the inner cell masses were extracted, which further produces stem cell lines. From leftover in-vitro fertilization, the extraction of single blastomere becomes very easy.  

In SCNT, the nucleus of a somatic cell is administered into the cytoplasm of an enucleated egg that then develops into a zygote and is then permitted to further give rise to a blastocyst stage.  

Answer:

a. lycophytes is the correct answer.

Explanation:

Microphylls are found in lycophytes.

lycophyte is a vascular plant, they have a unique type of leaves called microphylls.

The lycophytes belong to the division of Lycophyta, they are seedless plants and have vascular tissue.

The leaves(microphylls) present in the lycophyte have a single vein that is unbranched and narrow, which provides the water to the leaf and supplies nutrients to other parts of the plant.

Microphylls, which are characterized by a single vascular vein, are found primarily in lycophytes. Lycophytes are an ancient group of vascular plants and include species like club mosses and quillworts.

Microphylls are a specific type of leaf that are typically characterized by only having a single vascular strand or vein. This contrasts with megaphylls, which usually have multiple veins. The answer to which plant group microphylls are found in is a. lycophytes.

Lycophytes are one of the oldest groups of vascular plants and are recognized by their microphyllous leaves. Notable examples of lycophytes include club mosses, quillworts, and spike mosses. The presence of microphylls is a defining characteristic of this plant group.

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Answer:

Explained

Explanation:

Type II response can be defined as an antibody-dependent process in which specific antibodies bind to antigens and results in tissue damage. Transfusion reaction is a type II response because here the mismatched RBC's are rapidly destroyed by specific preformed antibodies (anti-ABO or -Rh) and complement.

Transfusion reactions takes place when incompatible blood products are transfused into a patient's circulation. This triggers the patient's immune system and consequently donor RBC's are destroyed by antibodies in the recipient's circulation. This is usually seen when antigen-positive donor RBC's are transfused into a patient who has preformed antibodies to that antigen.

Answer:

Option (a).

Explanation:

Virus are the acellular organism and can only acts as living organism when present inside the host organism. Virus can have DNA or RNA as their genetic material.

The bacteria is infected with T2 phage protein coat and T4 phage DNA. As only DNA can acts as the genetic material and can be inherited to the next generation. The new phages has the T4 DNA as their genetic material and T2 phage protein coat.

Thus, the correct answer is option (a).

Answer:

Option (a).

Explanation:

Answer: Cellulose

Explanation:

In plant the cell wall is mainly composed of strong fiber made of carbohydrates polymer known as cellulose.

It is the major component of cotton fiber, wood and other plants. It also helps in paper production.

This substance helps in the protection of the plants from outer environment, also provides strength and rigidity to the plants.

Hence, the correct answer is cellulose.

The best way of solving this is to draw a Punnett square.

You know the F0 had one parent with singed bristles (s), and normal wings (L), and the other parent is normal bristles (S) with vestigial wings (l).

If you do the cross ssLL x SSll you'll find 100% of the offspring is F1: SsLl, this means, all of them show the dominant traits: normal wings and normal bristles.

If you cross two parents from F1 to have F2, you'll find:

SsLl x SsLl = SSLL + SslL + sSlL+ ssll = 25% SSLL,all dominant traits. 50% SsLl is a recessive trait carrier but shows dominant traits. 25% ssll this one has all recessive alleles, which means, it will show vestigial wings and signed bristles.

The recombinant F2 offspring in this genetic cross of Drosophila are those that do not exhibit the parental trait combinations. Therefore, singed bristles/normal wings and normal bristles/vestigial wings are considered recombinant offspring. Option C is correct.

Complete question as follows:

A researcher conducted crosses between two different strains of Drosophila. When true-breeding flies with singed bristles (s) and normal wings (L) were crossed to true-breeding flies with normal bristles (S) and vestigial wings (l), all F1 offspring had normal wings and normal bristles. The F1 offspring were crossed to flies with singed bristles and vestigial wings. Which F2 offspring is/are recombinant?

A.Singed bristles/vestigial wings only

B.Singed bristles/normal wings only

C.Singed bristles/normal wings and normal bristles/vestigial wings

D.Singed bristles/vestigial wings and normal bristles/normal wings

Answer:

There is various kind of proteins found in the human body, which exhibits an array of functions and applications. All the proteins are formed of similar monomers, that is, amino acids, though one can witness different kinds of proteins.  

This has been made possible because of the existence of different sequences of amino acids coded for in the DNA. Due to different sequences of amino acids, each protein will fold distinctly and therefore, will exhibit distinct functions inside the body.  

Thus, the distinct sequences of amino acids permit for the great diversity of proteins and their functions in the human body even though each of the protein comprises similar kinds of monomers.  

Answer:

Amino group (NH3)

Explanation:

The amino acids have a central carbon atom to which four functional groups are bonded. These are namely a carboxyl group (COOH), an amino group (NH3), a hydrogen atom and an R group (it varies for different amino acids).

Amino group (NH3) is common to all the standard 20 amino acids. NH3 is a polar group since higher electronegativity of nitrogen atom allows it to pull the shared electrons of the covalent bonds towards itself. It makes the nitrogen atom of amino group partially negative and the hydrogen atoms carry a partial positive charge (a dipole is present).

Amino group (NH3) accepts one proton and becomes positively charged (NH4+).

The polar functional group that can become positively charged on an amino acid is typically found in basic amino acids like lysine and arginine.

The functional group on an amino acid that is polar and can become positively charged is often found in basic amino acids. Basic amino acids such as lysine and arginine have side chains that contain a positive charge under physiological conditions. These amino acids are essential in the structure and biological activity of proteins. For example, in the classification of amino acids, those like lysine (Lys) and arginine (Arg) are recognized for the basic groups in their side chains which can participate in hydrogen bonding and other important biochemical interactions.

The functional group on an amino acid that is polar and can become positively charged is the amino group (-NH2). This group contains a nitrogen atom, which can accept a hydrogen ion (H+) to form a positively charged ammonium group (-NH3+). Amino acids that have an amino group in their side chain, such as lysine and arginine, are considered basic amino acids because they can donate a proton and become positively charged.

Answer: A (uppercase letter) is used to refer to dominant alleles.

a (lowecase letter) is used to refer to recessive alleles.

Explanation:

Alleles are different forms of a gene, and they can be dominant or recessive. Each gene codes for a specific trait.

Diploid organisms, such as human beings, have two alleles for each gene. If an organism is heterozygous for a certain trait, it means it has one dominant and one recessive allele. And since the dominant allele masks the effects of the other one, the dominant trait is expressed in the phenotype. So, dominant alleles show their effect even if the organisms only has one copy of it.

If an organisms is homozygous, it has two dominant alleles or two recessive alleles. Then it can be homozygous dominant, where both are dominant alleles; or homozygous recessive where both are recessive. Recessive alleles only show their effect if the organisms has two copies of it.

For example, if we want to study the gene that codes for hair color, we could say that dominat allele "A" codes for brown hair, and recessive allele "a" codes for blonde hair. If a person has brown hair, the genotype could be either AA or Aa, becuase AA is homozygous dominant and it has two copies of brown color. And Aa is heterozyogus, but the person only needs one copy of the allele to be expressed. On the other hand, if the person has blonde hair, the genotype can only be aa because recessive alleles only show their effect if there are two copies of the allele.

In genetics, alleles are symbolized by letters, with uppercase for dominant alleles and lowercase for recessive alleles. A homozygous dominant genotype is written with two uppercase letters (VV), a homozygous recessive with lowercase (vv), and a heterozygous genotype combines both (Vv).

Understanding Allele Symbols in Genetics

In genetics, the symbols for alleles are represented by letters, with the dominant allele typically denoted by an uppercase letter and the recessive allele by a lowercase letter. For instance, in the example of pea plant flower color, if violet is the dominant trait, then 'V' would symbolize the dominant allele, while 'v' would represent the recessive allele for a white flower color. Therefore, a homozygous dominant pea plant with violet flowers would have the genotype 'VV', a homozygous recessive plant with white flowers would be 'vv', and a heterozygous pea plant with violet flowers would have the genotype 'Vv'.

Dominance was discovered by Mendel, who observed that certain traits, like the height in pea plants, showed a dominant and recessive pattern. A plant with genotype 'TT' (tall) would be homozygous dominant, while 'tt' (dwarf) would indicate a homozygous recessive plant. A heterozygous plant with the genotype 'Tt' would also be tall, displaying the dominant trait. Furthermore, when two heterozygous pea plants are crossed (Tt x Tt), Mendel deduced a characteristic 3:1 ratio of dominant to recessive phenotypes in the offspring.

Answer:

b. several transcription factors have bound to the promoter.

Explanation:

The initiation of RNA synthesis in eukaryotes includes assembly of RNA polymerase and several transcription factors at the promoter. The binding of TATA-binding proteins (TBP) to the TATA box serves to stabilize the TFIIB-TBP complex at the promoter.

The TFIIB is a transcription factor that is bound to both transcription factor binding proteins and DNA. This is followed by the binding of transcription factor TFIIF and the RNA Pol II enzyme to the TFIIB-TBP complex. Then the other transcription factors such as TFIIE, TFIIB and TFIIH also join the complex. The result is the formation of a closed complex.

Here, the function of TFIIA is to stabilize the TFIIB and TBP at promoter while TFIIB serves in the recruitment of RNA Pol II enzyme to the promoter.

TFIIE facilitates binding of TFIIH and has helicase activity to unwind the DNA duplex while TFIIF serves to prevent the binding of Pol II enzyme to the DNA sequences other than promoters.

Finally, the transcription factor TFIIH phosphorylates the enzyme RNA polymerase II and brings about a conformational change in the whole complex to facilitate the start of transcription.

In eukaryotic cells, the process of transcription, the creation of an RNA copy of a gene sequence, cannot begin until transcription factors have bound to the promoter region of the DNA.

In eukaryotic cells, transcription, which is the process of making an RNA copy of a gene sequence, cannot start until several transcription factors have bound to the promoter. This process takes place in the cell nucleus. Preparatory steps include the DNA strands unwinding and the promoter region being exposed, but transcription only actually commences when transcription factors bind to the promoter, effectively marking the start site for RNA synthesis.

The options 'the 5 ′ caps are removed from the mRNA' and 'the DNA introns are removed from the template' pertain to post-transcriptional modifications and mRNA processing, not the start of transcription itself.

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Answer:

Linkage may be defined as the transmission of chromosome to the next generation that are located on same chromosome. Linkage results in the formation of recombination progeny.

PDS (parental ditypes) and non parental diypes can be used to find linkage in the Neurospora and other organism Parental ditypes has the genotype same as parental genotype whereas non parental ditypes contains recombinant progeny. If the ratio and arrangement of PDS and NPDs is 2:4:2 or 2:2:2:2. This means linkage is present between the gene and centromere of an organism.

Answer:

True

Explanation:

Unlike eukaryotic cells, which have a nucleus that contains the genome and is separated from the cytoplasm by a membrane, the prokaryotic nucleoid is not membrane-bound and is not considered an organelle. The nucleoid is simply the area within a prokaryiotic cell where its DNA is located.

To calculate magnification on a microscope, multiply the magnification of the objective lens by the magnification of the eyepiece lens. To focus a microscope using the high-power lens, start by locating the specimen with the low-power lens and then switch to the high-power lens for clearer focus. Safety precautions include proper handling, avoiding touching the lenses, and caution with the fine focus knob.

To calculate magnification on a microscope, you need to multiply the magnification of the objective lens by the magnification of the eyepiece lens. For example, if the objective lens has a magnification of 10x and the eyepiece lens has a magnification of 20x, the total magnification would be 10x multiplied by 20x, which equals 200x.

To focus a microscope using the high-power lens, start by placing the slide on the stage and using the low-power lens to locate the specimen. Then, adjust the fine focus knob to bring the specimen into sharp focus. Once the specimen is in focus, switch to the high-power lens and use the fine focus knob again to make any necessary adjustments for a clearer image. Safety issues to be aware of include ensuring proper handling of the microscope to prevent damage, avoiding touching the lenses with fingers, and using caution when manipulating the fine focus knob to avoid breaking the slide.

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To calculate the magnification on a microscope, you multiply the power of the objective lens by that of the eyepiece. When focusing a microscope using a high power lens, adjust the coarse and fine focus knobs and avoid physical contact with the lenses to prevent damage.

To calculate magnification on a microscope, you multiply the magnification power of the objective lens by the magnification power of the eyepiece lens. For instance, if the objective lens is 10x and the eyepiece is 40x, the total magnification is 400x.

Focusing a microscope using the high power lens, requires adjusting the coarse and fine knobs while observing through the eyepiece. Start with the lowest power lens and gradually switch to the high power lens. Ensure that you don't touch the lens with your fingers and avoid knocking the microscope to prevent damage. Additionally, use oil immersion with high power objectives for enhancing the resolving power.

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Answer:

DNA is also called deoxyribonucleic acid which is made up of two chains which wind around each other to form a double helix model. The 2 DNA strands are also called polynucleotides and they are made up of monomeric units known as nucleotides. These nucleotides are made up of one of four nitrogen-containing nucleobases: cytosine, guanine, adenine and thymine, a phosphate group, and sugar known as deoxyribose.  

Nitrogen bases present on the two separate polynucleotides strands are bound together with the help of base pairing (such as adenine with Thymine) and with hydrogen bonds to form double-stranded DNA.

So, adenine in DNA is complementary to thymine.

Answer:

DNA is the genetic material of all the living organism except some viruses. The structure of DNA consists of nitrogenous base, pentose sugar and the phosphate group.

Tube 1 - The nucleotide are linked together by phosphodiester bond. As these bonds are broken, the structure of DNA contains the individual complementary nucleotide are linked together by hydrogen bonds.

Tube 2 - The bond between the sugar and the bases are broken. The DNA structure consists of the phosphate group chain only with any pairing with the base.

Tube 3 - The DNA nitrogenous bases are adenine, guanine, thymine and cytosine that are linked together by hydrogen bonds. The agents break the hydrogen bonds. Now, the DNA structure consists of the single strand only.

Answer:

Megaspore—2n.

Explanation:

Angiosperms are the fruit bearing plants and reproduce by the process of sexual reproduction. The chromosome number are specific at each stage of the cell cycle of the angiosperms.

Microspores , egg are haploid. Zygote is diploid in nature. Megaspores get germinate into the female gametophytes and these are haploid in nature. Megaspores are also haploid in angiosperms.

Thus, the correct answer is option (b).

The description 'megaspore—2n' is INCORRECTLY paired with its chromosome count.

In plants (including angiosperms) the egg cell is the haploid (n) female gamete.

After fertilization, the egg cell forms a diploid (2n) zygote that subsequently develops the embryo inside the ovule.

A microspore is a haploid (n) cell that gives a male gametophyte.

In conclusion, the description 'megaspore—2n' is INCORRECTLY paired with its chromosome count.

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Explanation:

Three enzymes for lactose metabolism are grouped in the lac operon: lacZ (β-galactosidase), lacY (permease) and lacA (trans-acetylase). The transcription of this operon occurs only when lactose is available to digest, presumably to avoid wasting energy.  Apart from these protein-coding genes we have P(promoter), O(operator and CBS(Cap-binding site), sequences that work as binding sites for transcriptional regulation.

Negative components: An important regulator is lacI when four of these molecules assemble they form a repressor, this will bind to the promoter, this won't allow the operon to be transcribed as long as the operator is occupied by a repressor. LacI can also prevent lactose to bind to the operator by binding with it and changing its shape.

Positive components: cAMP binding protein is another molecule that when is bound to CBS improves the chance of RNA polymerase to bind to the promoter to initiate transcription.

I hope you find this information useful! Good luck!

I adden an image that ilustrates what I explained.

Answer:c. DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand.

Explanation:

The double stranded molecule of DNA having the pairing strands in the anti-parallel condition. One strand is remain in configuration as the 5’ to 3’ end, the complementary strand is positioned 3’ to 5’ end.

New DNA strand synthesis occurs from the 3’ to 5’ end as the enzyme DNA polymerase attach itself to the 3’ end of a DNA strand. It will add new nucleotides to the 3' end of the DNA.

Hence, the option (c) is the correct answer.

The primary reason for the difference in the synthesis of leading and lagging strands of DNA is that the DNA polymerase can only add new nucleotides to the 3′ ends of a pre-existing strand. DNA replication always occurs in the 5' to 3' direction, which influences the synthesis of both strands due to the anti-parallel structure of DNA.

The basis for the difference in how the leading and lagging strands of DNA molecules are synthesized lies in the manner in which the DNA polymerase enzyme works. The key factor is that DNA polymerase can add new nucleotides only to the 3′ end of a pre-existing DNA strand.

This signifies that DNA replication always takes place in the 5′ -> 3′ direction. Considering the DNA double helix's anti-parallel structure, one strand (leading strand) is synthesized in a continuous fashion, as its orientation allows adding nucleotides at the 3′ end.

Conversely, the other strand (lagging strand) being oriented in 3′ -> 5′ direction, is synthesized discontinuously in small fragments called Okazaki fragments, as nucleotides can't be added to its 5' end. The enzyme DNA ligase then joins these pieces. Options a, b, and d may influence the process, but they're not the main reason for the leading and lagging strands' different synthesis.

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Answer:

Explanation:

A unicellular organism has only a single cell thus is unable to perform the diversity of functions. The complexity in multi-cellular organisms is more in terms of structure as well as in function thus the evolution of the multi-cellular organism is significant as as the time passes the changes in the morphology, physiology and genetic make up of the organisms will become appreciable.    

Answer:

The correct answer is option b. "The phosphodiester linkages of the polynucleotide backbone would be broken".

Explanation:

The phosphodiester linkages of the polynucleotide backbone is what binds each nucleotide to each other in the DNA molecules. These linkages are covalent bonds that take place between 3' carbon atom of one sugar molecule and the 5' carbon atom of another. The enzymes that break down DNA catalyze the hydrolysis of the phosphodiester linkage, which results in DNA cleavage within the backbone at specific or unspecific nucleotides.

Final answer:

Enzymes that break down DNA cleave the phosphodiester bonds in the DNA polynucleotide backbone, leading to the degradation of the DNA molecule into its constituent nucleotides. The correct choice is b - the phosphodiester linkages would be broken.

Explanation:

When enzymes that break down DNA are applied to the DNA molecule, they catalyze the hydrolysis of the covalent bonds that join nucleotides together. Specifically, these enzymes target the phosphodiester bonds in the polynucleotide backbone. Hydrolysis of these bonds would lead to a cleavage of the polynucleotide chain. This would result in the degradation of the DNA molecule into its individual nucleotides, where each phosphate group would be associated with a deoxyribose sugar and a nitrogenous base.

The correct answer to the student's question is: b. The phosphodiester linkages of the polynucleotide backbone would be broken. This process is different from denaturation, where hydrogen bonds between complementary bases break; breaking phosphodiester bonds involves a chemical reaction that cleaves the DNA backbone itself.

Enzymes such as nucleases perform this function, specifically endonucleases which cleave phosphodiester linkages within a DNA strand. Conversely, enzymes like DNA ligases can repair these breaks, forming a new phosphodiester linkage, though in this context, we are discussing the breaking of the DNA molecule by hydrolysis, not the joining of fragmented DNA.