In order to caluclate the specific activity of the isolated
fractions, the protein concetration has to be determined. Why is
the protein concetration necessary for accurate comparison of the
enzyme activity of the fractions?

Answer:

the blood enters into the right atrium by inferior and superior vena cava--- left ventricle---pulmonary artery ---- lungs----left atrium----left ventricle---aorta----different body tissues.

Explanation:

The circulation in man is called the double circulation. Because in the heart the two types of blood circulate separately i.e. oxygenated blood/oxygen-poor blood and deoxygenated blood/oxygen-rich blood and they never mix.

The route of deoxygenated blood:

In the heart, the deoxygenated blood enters by the superior and inferior vena cava. Then this oxygen-poor blood stores in the right atrium and from there it passes into the left ventricle. The flow of blood from the right atrium to left ventricle occurs by the tricuspid valve.  

When the blood enters into the right ventricle the valves closed to stop the backflow of the blood.  

The blood from the right ventricle goes to the pulmonary artery by the pulmonary valve. From the pulmonary valve, the deoxygenated blood passes to the lungs. In the lungs, the deoxygenated blood/oxygen-poor blood gets purified.

The rout of oxygenated blood:

Now the blood becomes rich in oxygen and called oxygenated blood. This oxygenated blood enters into the left atrium by pulmonary vein. From the left atrium, the oxygen-rich blood moves into the left ventricle. Then the oxygenated blood reaches to the different parts of the body by the aorta.

In this way, the right half of the heart circulates the deoxygenated blood and the left half of the heart provides deoxygenated blood.

The one that is incorrectly paired is  DNA ligase—cutting DNA, creating sticky ends of restriction fragments. The correct option is B.

DNA ligase is a ligase enzyme that facilitates the joining of DNA strands by catalyzing the formation of a phosphodiester bond.

It aids living life forms repair single-strand cuts in duplex DNA, but some forms may specifically repair double-strand breaks.

By connecting cuts in the phosphodiester backbone of DNA that occur during replication and recombination, as well as as a result of DNA damage and repair, DNA ligases play an important role in maintaining genomic integrity.

Electrophoresis is used to separate DNA fragments, DNA polymerase is used to amplify sections of DNA, and reverse transcriptase is used to produce cDNA from mRNA.

Thus, the correct option is B.

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The correct answer is C. Interstitial cells of the testis.

Explanation:

The testosterone is one of the main sex hormones in males, due to this, this hormone is linked to sexual characteristics as well as behavior, mass muscle development, among others. In terms of testosterone production, the production of testosterone occurs mainly on the testis (reproductive glands or gonads) and more specifically in the interstitial cells of these called Leydig cells which are specialized cells that besides producing testosterone produce androstenedione and dehydroepiandrosterone. Additionally, the production of any hormone in this cell is stimulated by the luteinizing hormone. According to tho this, it can be concluded testosterone is produced in the interstitial cells of the testis.

Testosterone is produced in the interstitial cells of the testis.

Testosterone is produced in the interstitial cells of the testis. These cells, also known as Leydig cells, are located in the testes and are responsible for the production of testosterone. The hormone is then released into the bloodstream and has various effects on the body, including the development of male reproductive organs and secondary sexual characteristics.

The hormone Testosterone is primarily produced in the Interstitial cells of the testis, also known as Leydig cells. These cells are located in the connective tissue between the seminiferous tubules in the testes. The production of testosterone is stimulated by Luteinising Hormone (LH) which is produced in the anterior lobe of the pituitary gland in the brain. However, the pituitary gland does not produce testosterone itself. Hence, the correct answer to your question is C. Interstitial cells of the testis.

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Answer:

The correct answer will be option-NADH generated from the citric acid cycle.

Explanation:

During cellular respiration, few ATP molecules are synthesized by substrate-level phosphorylation in glycolysis and citric acid cycle.

The process also produces high energy equivalents which are NADH and FADH₂ during the citric acid cycle. Two pyruvate molecules generate 6 NADH and FADH₂ molecules. These molecules release electron and hydrogen during the electron transport chain which is utilized to generate the ATP.

Each NADH produces 3 ATP molecule whereas FADH₂ molecule produces 2 ATP molecules. Since 6 NADH (6x3=18 ATP) produced and 2  FADH₂ (2x2).

Thus, NADH provides the highest amount of ATP molecules.

Answer: (C) The production of tissue-specific proteins.

Explanation:

 The cell differentiation is the process in which the cell change from one cell place to another place. It basically occur due to the process of gene expression.

The cell differentiation involve the production of the specific tissue protein known as muscle actin. In cell differentiation, the pluripotent stem cell basically go in the specific differentiation level and then reach in the state of fully differentiation.

The fully differentiation produced a specific function that the production of the protein. Therefore, option (C) is correct.

Cell differentiation involves the production of tissue-specific proteins. The correct option is C.

Thus, the process through which cells develop specialized properties and functions is referred to as cell differentiation. Cells go through a variety of alterations during this process that equip them to carry out certain tasks in a particular tissue or organ.

The creation of tissue-specific proteins is one of the crucial components of cell differentiation. These proteins, which are frequently unique to a given cell type or tissue, carry out the specialised duties of differentiated cells. They are essential in determining the differentiated cell's identity and purpose.

Thus, the ideal selection is option C.

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Answer:

This results are explained by the fact that these two genes are in the X chromosome and both mutations are recessive.

There is a estimated map distance of 5.76 cM.

Explanation:

First step is to know what are the initial phenotypes, they say "males from a true-breeding stock" that means that they are homozigous for both genes. Also, they say that in F1 generation, all females had the same phenotype but all males had a different phenotype than females, so this suggest a sex-linked inheritance. In the F1 generation, females had a wild type phenotype, but we know these females had to carry the mutant copies of the two genes, so mutant alleles must be recessive.

Therefore, if r is the allele for raspberry-colored eyes and s for sable-colored bodies, the genotypes of parents should be:

- Males with raspberry-colored eyes (wild type for body color):

[tex]X^{rS}Y^{}[/tex]

- Females with sable-colored bodies (wild type for eyes color):

[tex]X^{Rs}X^{Rs}[/tex]

If they are crossed, we obtain the results on the Punnet Square.

If there is not recombination, we expect F2 as in Punnter Square, but in the results, there are two phenotypes that were not expected: wild type males (23) and males with raspberry-colored eyes with sable-colored bodies (27). These two should be recombinants. To calculate the distantance between the two genes, we use:

[tex]Distance=\frac{recombinants}{Total }*100[/tex]

[tex]Distance=\frac{23+27}{868}*100=\frac{50}{868}*100=5.76 cM[/tex]

Therefore, distance between s and r is 5.76 centimorgans.

Answer:

The correct answer will be option-C.

Explanation:

Each DNA molecule exists in helical shape made by two strands of nucleotides. Each nucleotide is composed of a five-carbon sugar, a phosphate group and nitrogenous bases.

The sugar-phosphate unit bond via the phospho-diester bond and form the backbone of the DNA. The phosphate functional group is acidic in nature as it donates the proton while the formation of the phosphodiester bond which provides DNA with the acidic nature.

Thus, Option-C is the correct answer.

Answer:

a. add more of the enzyme.

Explanation:

Enzymes speed up the rate of reaction by lowering the activation energy for the reaction. They are not used themselves in the reaction. They are specific to substrate molecule. Substrate molecule binds to enzyme's active site and they undergo the reaction to form the product and release back the enzyme.

Rate of reaction depends on both substrate and enzyme concentration. Maximum rate of reaction is reached when all the active sites of enzyme molecules have been occupied by the substrate molecules which means that they are saturated. If more substrate is added at this point it wont have any effect on rate of reaction since there are no free active sites. Hence more enzyme is required to be added so that extra substrate can be utilized and rate of reaction can further be increased.  

Answer:

e. 2

Explanation:

The seed color in a pea plant is a discontinuous trait and is regulated by two alleles of a gene. The dominant allele imparts "yellow" color to the seeds while the recessive allele imparts them "green" color.  

If the dominant allele is represented by letter "Y" and the recessive allele is written as the lower case letter "y"; the homozygous dominant and heterozygous dominant genotypes exhibit "yellow" phenotype with respect to seed color while the homozygous recessive genotype express "green" seed color.

Answer:

d.condiophores is the correct answer.

Explanation:

condiophores is cells or structures are associated with asexual reproduction in fungi

Conidium, a kind of asexual spore-bearing bodies of fungi normally produced at the top of hyphae on specific spore-producing stalks called conidiophores.

The spores get released and detach when it is mature.Then spores germinate into a new hypha and hypha develops into a mycelium.

Thus the Axseual reproduction in Fungi(ascomycetes) occurs by the generation of conidia which are produced on a specialized structure called conidiophores.

Answer:

d. condiophores

Explanation:

As you may know, fungal spores are the main structure for the asexual reproduction of these organisms. These spores are formed in hyphae, but some fungi have special hyphae called conidiophores and their spores called conidia, which are structures for the asexual reproduction of these fungi. For this reason, we can say that, among the options given in the question above, "conidiophores" is the structure that is associated with asexual reproduction in fungi.

Answer:

Wax.

Explanation:

Hydrophobic material may be defined as the material that contains the non polar molecule and hydrophobic interaction in their structure. The hydrophobic molecule are soluble in non polar solvents.

Wax is made of lipids. The lipids are non polar molecule and are hydrophobic in nature. Wax are made of large hydrocarbon part that are insoluble in water and can be easily soluble in non polar solvents.

Thus, the correct answer is option (c).

The term hydrophobic refers to substances that repel water. Out of the mentioned options, wax best fits this description as it doesn't mix with water. Paper, table salt, and sugar dissolve in or absorb water and are thus hydrophilic, or attracted to water.

In the context of the question, which asks for the material that is hydrophobic, the correct answer is c. wax. Hydrophobic substances are those that don't mix with or repel water. For instance, wax is a hydrophobic substance you commonly encounter. When you pour water on wax, it simply rolls off or forms beads and doesn't mix, this is due to its hydrophobic characteristic. Materials like table salt, paper, and sugar, are all hydrophilic, they tend to dissolve in and interact with water.

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Answer:

a. malaria

Explanation:

Malaria is caused by a parasite that is transmitted by the bite of infected anopheles mosquitoes, entering the bloodstream, affecting the red blood cells. Therefore, Mosquito Dunks may be a good control of malaria in tropical countries.

On the other hand, mosquito dunk is an inappropriate control method for the other diseases listed, because bubonic plague was transmitted by the fleas of rats, legionellosis is an infectious disease caused by a bacterium (Legionella) that is found mainly in water bodies and toxoplasmosis is caused by a parasite present in undercooked contaminated meat and the feces of cats.

Answer:

b. The thylakoid lumen into the stroma

Explanation:

During light reaction of photosynthesis, when electrons which have high energy travel across electron transport chain some energy is released by them. This energy causes some H+ from stroma to move into thylakoid lumen. Due to this, H+ concentration becomes more in lumen as compared to stroma. Soon after that because of concentration gradient, these H+ then start moving from high concentration to low concentration i.e. from thylakoid lumen to stroma. There are some reverse pumps i.e. ATP synthase located in the thylakoid membrane which are responsible for generating energy in the form of ATP from the movement of H+. From the movement of 3 H+ , 1 ATP is generated.

Answer:

5' to 3'.

Explanation:

Transcription may be defined as the process of formation of RNA molecule from the template DNA with the help of enzymes and the various transcriptional factors.

The process of RNA synthesis occur in the 5' to 3' direction. The RNA polymerase reads in the 3' to 5' direction and leads to the synthesis in 5' to 3' direction. The transcription factors binds to the DNA promoter during RNA synthesis.

Thus, the correct answer is option (d).

Answer:

DNA:

Dexoyribonucleic acid is present as a genetic material in all organism except some virus. DNA is made up of polymers of nucleotides. They are made up of nitrogenous base (adenine, guanine, thymine and cytosine) , deoxyribose pentose sugar and phosphate group. The nitrogenous bases are linked with each other with hydrogen bonds and nucleotides are linked together with phosphodiester bond.

Protein:

Proteins are the main building block of the body. Proteins are made of the polymers of amino acids. The amino acid consist of amine group, carboxylic group, hydrogen and R group attached with the carbon. The amino acids are linked together through the peptide bond.

Answer:

Carbon dioxide is the prime waste component of aerobic respiration. Too little or too much carbon dioxide in the blood can result in extreme issues. In order to make this safe and sound, carbon dioxide is transported in the blood in three ways. That is, in the form of carbamino compounds, hydrogen carbonate, and in the form of a dissolved state.  

Approximately 30 percent of all the carbon dioxide is mediated in the form of carbamino compounds. At high quantity, carbon dioxide binds directly with amino acids and the amine groups of hemoglobin to produce carbaminohaemoglobin.  

Almost 60 percent, that is, the majority of the carbon dioxide is transported in the blood in the form of bicarbonate ions or hydrogen carbonate. The diffusion of carbon dioxide takes place in the RBCs and gets transformed into bicarbonate ions and hydrogen ions with the help of an enzyme known as carbonic anhydrase.  

Approximately 10 percent of all the carbon dioxide is transported in the form of a dissolved state in plasma. The concentration of gas dissolved in a liquid relies upon its partial pressure and its solubility. In spite of its solubility, only some of the carbon dioxide in the blood is in reality transported in the form of dissolved state in plasma.  

Answer:

CO2 is transported in three forms from tissues to the alveoli: a) In dissolved form through plasma; b) As bicarbonate ions; c) by RBC as HbCO2.

Explanation:

When oxygen reaches the body cells, oxidation of glucose takes place during which CO2 is released along with H2O and energy. CO2 diffuses out of the tissues into the blood capillaries where it is transported to the lungs in three forms:

1. In dissolves form in plasma: About 7% of CO2 gets transported in dissolved form.  It gets dissolved in the blood plasma and is carried to the lungs.

2. As bicarbonate ions: about 70% of CO2 is converted into bicarbonate ion and transported in plasma. Within the RBC there is an enzyme called carbonic anhydrase. When CO2 enters the RBC, it reacts with water to form carbonic acid which further dissociates into bicarbonate and hydrogen ion. Both these reactions occur in the presence of carbonic anhydrase enzyme. The HCO3- ion formed in the RBC quickly diffuses into the plasma, where they are carried to the lung.

3. By RBCs as carbaminohemoglobin: About 20-25% of CO2 is transported as HBCO2. CO2 binds with the amino group of globin protein which is a part of Hb.

Answer:

Simple diffusion and Facilitated diffusion

Explanation:

The simple diffusion process involves movement of molecules. This movement occurs from high concentration to low concentration. The facilitated diffusion process is a passive diffusion process. This process requires proteins for movement of molecules, the proteins are carrier and channel proteins.

The enzymes that promote membrane diffusion are carrier proteins. These proteins act as enzymes and allows transport by changing the conformation and opening of channels.

The two types of membrane diffusion are simple diffusion and facilitated diffusion. Simple diffusion does not involve any enzymes, as it's the movement of small and nonpolar molecules down their concentration gradient through the phospholipid bilayer directly. Facilitated diffusion may involve carrier proteins or channel proteins but does not involve classical enzymes that catalyze reactions; instead, these proteins facilitate the movement of substances across cell membranes that cannot easily pass through the lipid bilayer directly.

Diffusion is critical for the transport of materials across biological membranes, helping to maintain homeostasis within cells and organisms. Simple diffusion allows for the passive movement of substances like oxygen and carbon dioxide, relying on the concentration gradient to drive the process without energy input. Facilitated diffusion also moves substances along their concentration gradient but through specific carrier proteins or channels. These proteins provide a pathway for larger or polar molecules that cannot diffuse through the lipid bilayer unaided. Temperature is one of the factors that can affect the rate of diffusion; higher temperatures increase molecular kinetic energy, which generally speeds up diffusion rates.

Answer:

b. operator

Explanation:

An expression vector is usually a plasmid or a virus designed for gene expression in cells, it's designed to act as an enhancer and promoter to efficient transcription of the gene that it carries, meaning it is used for protein production.

As any other vector expression vector may have an origin of replication, a selectable marker, and multiple cloning sites, it also needs elements that are necessary for gene expression, a promoter, a ribosomal binding site, and a start, terminal codon and a transcription termination sequence.

An operator contains the code to begin the process of transcribing the DNA message of one or more structural genes into mRNA, it works as a repressor to regulate the gene expression, since the function of an expression vector is the protein production an operator is not specifically required.

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Answer:

A

Explanation:

The vast majority of bacteria that produce gastrointestinal symptoms are transmitted via fecal-oral transmission. This means that when sub-optimal hygiene is present, bacteria from the hands of the infected person that got there from contact with its own feces (after going to the bathroom, for example) is passed to the next person. This happens a lot in the food business, and it's the mode of transmission of the much famous Salmonella.

Answer:

d. You cannot eat too much fiber.

Explanation:

Fiber also called as roughage forms the part of plant based foods. Body is not able to digest it so it passes through the digestive system and makes the bulk of stool. It keeps the digestive system clean and eases bowel movements.

The daily required intake of fiber is 25-30 grams. Many people do not take this much fiber hence suffer from problems like constipation, upset stomach etc. But crossing the threshold of 70 grams of fiber can also lead to another set of problems like:

Answer:

Option C

Explanation:

Given ,

A1 allele is carried by [tex]70[/tex] % people

Let us assume A1 s dominant genotype

This means [tex]p= \frac{70}{100} = 0.7[/tex]

Thus, frequency of allele in the given population is [tex]0.7[/tex]

It is also given that the population is in Hardy-Weinberg equilibrium thus

[tex]p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3[/tex]

Frequency of fruit fly with genotype A2A2 will be

[tex]q^2\\= (0.3)^2\\= 0.09[/tex]

As per Hardy-Weinberg's second equation of equilibrium

[tex]p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42[/tex]

Hence, option C is correct

Answer: Uses glucose ;produces carbon dioxide

Explanation:

Cellular respiration can be defined as the process by which the chemical energy in the body is transferred into ATP and some of it is released in the form of heat.

During this process the glucose that is broken down and the energy released is converted into ATP.

Carbon dioxide is released as a by-product of this process.

Hence, the correct answer is option C and D

Answer:

The correct option is: e. All of the above

Explanation:

Cocaine is a habit-forming illegal recreational drug and a very strong nervous system stimulant. The effects of this drug include euphoria, hallucinations, paranoid delusions, increased heart rate, large pupils,  perspiration, itching and high body temperature.

Cocaine acts on the nervous systems of the human beings by blocking the reuptake of dopamine, serotonin, and norepinephrine neurotransmitters in the brain.

It also interferes with the action potential propagation by blocking the voltage-gated sodium channels.

Answer:

a. True, b. False, c.True, d. True

Explanation:

a. Base excision repair is started by a DNA glycosylase  that recognizes the changes and removes the altered base by cleavage of the glycosidic bond binding the base and the deoxyribose sugar together.

b. Nucleotide excision repair works by a cut-and patch mechanism that removes their heavy lesions, including pyrimidine dimers and nucleotides . Endonucleases are responsible for the lesion of the damaged strand.

c. Nucleotide excision repair  is initiated by the proteins namely UvrA, UvrC, and UvrB in Escherichia coli.

-UvrD (helicase II) later removes the damaged strand

-DNA polymerase I (PolI) fills in the resulting gap.

d. DNA glycolases removes the damaged nitrogenous base.

-It leaves the sugar-phosphate backbone intact and thus creating an  apurinic/apyrimidinic site, which is commonly referred to as an AP site.

e. Xeroderma pigmentosum complementation group A(XPA)

-This is an essential protein in the nucleotide excision repair pathway.

- It helps to make a pre-incision complex along with other proteins.                

Answer:

Innate immunity: first line of defense against infectious agents; Most pathogens can be controlled before a declared infection occurs.

Adaptive immunity: takes action when innate immunity fails. Make a specific response for each infectious agent and keep memory of it (it can prevent reinfection).

Explanation:

The immune system has evolved to be able to identify the strangers and develop a protective response to the latter (cognitive and destructive processes).

Innate or natural immunity: it is present at birth, being the first line against invasive microorganisms. Its characteristics are: it is present for life, it is not specific, it lacks memory and does not change intensity with exposure. It is useful against pyogenic microorganisms, fungi and multicellular parasites and includes three components: 1) physicochemical: skin, mucous membranes, secretions and cilia, which perform a washing and continuous cleaning, 2) humoral: complement, lectin binding to mannan and opsonins additional as C-reactive protein and proteolytic enzymes and 3) cellular: neutrophils, eosinophils, mast cells and natural killer lymphocytes.

Complement: it includes a high number of serum proteins that are produced mainly in the liver, form cascades so that each activated component catalyzes the activation of several molecules of the next component, amplifying the response. The consequences are cell lysis, the production of proinflammatory mediators and the solubilization of antigen-antibody complexes. Activation of the complement system occurs through three different pathways (alternating, classical and mannan-binding lectin) that converge in the final common pathway that provides most of the biological activity.

Cellular mechanisms: Neutrophils participate in the destruction of bacteria and fungi. Upon activation, adhesion molecules facilitate their entry into tissues, moving to chemical attractants and phagocytizing microorganisms. Destruction is mediated by oxygen dependent and independent pathways.

acquired immunity: acquired as part of development, increases with age and with repeated exposures, has specificity and memory for what is called adaptive. Its components are antibodies and cells (lymphocytes) and protects against bacteria (including those that produce intracellular infections), viruses and protozoa. In general, innate and acquired immune responses are not activated independently but complement each other.

Antigen-Antibodies: Antigens are structures that generate an anti-response from the immune system that has 3 elements of union and recognition of these;

1) antibodies that are soluble glycoproteins belonging to the group of immunoglobulins produced by B lymphocytes and plasma cells,

2) T-cell receptors that are large glycoproteins that interact with the peptide epitope preserved and presented by the third element

3) which are the major histocompatibility complex (CMH) molecules.

Antibodies perform many functions and have numerous uses as biological and clinical instruments

Answer:

The correct answer will be option-D.

Explanation:

Before Citric acid cycle or Krebs cycle, an intermediate reaction takes place which converts the pyruvate into Acetyl CoA. This reaction is known as pyruvate decarboxylation as it produces Carbon dioxide.  

Coenzyme A reacts with pyruvate which causes the release of two oxygen atoms and one carbon to form CO₂ along with the reduction of  NAD+ to NADH and produce "Acetyl CoA."

Thus,  option-D is the correct answer.

The intermediary metabolite formed by the removal of CO2 from pyruvate and that enters the Krebs cycle is acetyl CoA. This occurs in the phase of cellular respiration known as decarboxylation.

The intermediary metabolite that enters the Krebs cycle and is formed, in part, by the removal of a CO2 molecule from a molecule of pyruvate is acetyl CoA (option D).

The process occurs during cellular respiration. Specifically, one carbon atom from pyruvate is released as one molecule of CO2 in a step called decarboxylation, facilitated by an enzyme complex in the mitochondrial matrix. This remaining two-carbon molecule then combines with Coenzyme A to form Acetyl CoA, which enters the Krebs cycle.

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Answer:

Eukaryotic transcription refers to an elaborate procedure, which is used by the eukaryotic cells to duplicate the genetic information present in DNA into the units of RNA replica. The transcription of the gene takes place in both prokaryotic and eukaryotic cells. The expression of eukaryotic genes is mainly monitored at the level of the beginning of transcription, though in certain cases, transcription may be monitored and attenuated at subsequent steps.  

Like in bacteria, the transcription in eukaryotic cells is monitored by the proteins, which combines with the unique regulatory sequences and modulate the activity of RNA polymerase. The intricate task of monitoring the expression of gene in various kinds of cells of multicellular species is achieved mainly by the combined activities of various different transcriptional regulatory proteins.  

In supplementation, the packaging of DNA into chromatin and its variation by methylation produce further levels of complexity to the control of eukaryotic gene expression. The transcription in bacteria is monitored by the combination of proteins to the cis-acting sequences, which regulate the transcription of adjacent genes.  

The same kind of cis-acting sequence monitors the expression of eukaryotic genes. These sequences have been determined in mammalian cells mainly by the application of gene transfer assays to examine the activity of suspected regulatory regions of cloned genes.  

Answer:

a. normally leads to formation of head structures.

Explanation:

The product of the bicoid gene in Drosophila predominantly influences the formation of anterior body parts, suggesting that it primarily leads to the formation of head structures.

Based on the provided information, bicoid mRNA's absence from a Drosophila egg causes the absence of anterior (head/front) body parts and duplicates posterior (tail/back) parts. This indicates that the bicoid gene's product primarily influences the formation of anterior parts. It can be inferred that the bicoid gene's product is crucial in the formation of the head structures. So, the correct answer is, (a) the bicoid gene's product normally leads to the formation of head structures.

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Answer:

a) 1 b) 2 c) 1 d) 2 e) 1 f) 2

Explanation:

a) A coactivator CBP  contains a subunit that has histone acetyltransferase (HAT) activity.  These enzymes transfer acetyl groups from an acetyl CoA donor to the  amino groups of specific lysine residues on histone proteins.  

-Gradually this leads to activation of several other coactivators and as a result initiation of transcription takes place.

b) HDACs are associated with transcriptional repression. HDACs are present as subunits of larger  complexes described as corepressors.

-Corepressors are similar to  coactivators, except that they are recruited to specific genetic  loci by transcriptional factors (repressors) that cause the targeted gene to be silenced rather than activated.

c) Acetylation of histone  residues is thought to prevent chromatin fibers from folding into compact structures, which helps to maintain active, euchromatic regions.

-On a finer scale, histone acetylation increases access of specific regions of the DNA template to interacting proteins, which promotes transcriptional activation.

d) Deacetylation of the the histone tails makes the DNA more tightly wrapped around the histone cores, making it difficult for transcription factors to bind to the DNA. This leads to decreased levels of gene expression.

e) Decreased attraction is directly  proportional to loosening of the DNA from histone core hence accessible to the transcriptional factors which  further leads to transcription and gene expression  

f) Increased attraction means DNA gets tightly wound around the histone core and inaccessible to the transcription factor hence decreased gene expression.